3.1326 \(\int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=197 \[ \frac {b \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 b^2 d}-\frac {b \left (3 a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\left (3 a^2-b^2\right ) \cot (c+d x)}{a^3 d}+\frac {a x}{b^2}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d}+\frac {\cos (c+d x)}{b d} \]

[Out]

a*x/b^2-2*(a^2-b^2)^(5/2)*arctan((b+a*tan(1/2*d*x+1/2*c))/(a^2-b^2)^(1/2))/a^4/b^2/d+1/2*b*arctanh(cos(d*x+c))
/a^2/d-b*(3*a^2-b^2)*arctanh(cos(d*x+c))/a^4/d+cos(d*x+c)/b/d-cot(d*x+c)/a/d+(3*a^2-b^2)*cot(d*x+c)/a^3/d-1/3*
cot(d*x+c)^3/a/d+1/2*b*cot(d*x+c)*csc(d*x+c)/a^2/d

________________________________________________________________________________________

Rubi [A]  time = 0.28, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 9, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {2897, 3770, 3767, 8, 3768, 2638, 2660, 618, 204} \[ -\frac {2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^4 b^2 d}+\frac {\left (3 a^2-b^2\right ) \cot (c+d x)}{a^3 d}-\frac {b \left (3 a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {b \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}+\frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {a x}{b^2}-\frac {\cot ^3(c+d x)}{3 a d}-\frac {\cot (c+d x)}{a d}+\frac {\cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + b*Sin[c + d*x]),x]

[Out]

(a*x)/b^2 - (2*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/(a^4*b^2*d) + (b*ArcTanh[Co
s[c + d*x]])/(2*a^2*d) - (b*(3*a^2 - b^2)*ArcTanh[Cos[c + d*x]])/(a^4*d) + Cos[c + d*x]/(b*d) - Cot[c + d*x]/(
a*d) + ((3*a^2 - b^2)*Cot[c + d*x])/(a^3*d) - Cot[c + d*x]^3/(3*a*d) + (b*Cot[c + d*x]*Csc[c + d*x])/(2*a^2*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 2897

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Int[ExpandTrig[(d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m*(1 - sin[e + f*x]^2)^(p/2), x], x]
/; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && IntegersQ[m, 2*n, p/2] && (LtQ[m, -1] || (EqQ[m, -1] && G
tQ[p, 0]))

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \cot ^4(c+d x)}{a+b \sin (c+d x)} \, dx &=\int \left (\frac {a}{b^2}+\frac {\left (3 a^2 b-b^3\right ) \csc (c+d x)}{a^4}+\frac {\left (-3 a^2+b^2\right ) \csc ^2(c+d x)}{a^3}-\frac {b \csc ^3(c+d x)}{a^2}+\frac {\csc ^4(c+d x)}{a}-\frac {\sin (c+d x)}{b}-\frac {\left (a^2-b^2\right )^3}{a^4 b^2 (a+b \sin (c+d x))}\right ) \, dx\\ &=\frac {a x}{b^2}+\frac {\int \csc ^4(c+d x) \, dx}{a}-\frac {\int \sin (c+d x) \, dx}{b}-\frac {b \int \csc ^3(c+d x) \, dx}{a^2}-\frac {\left (a^2-b^2\right )^3 \int \frac {1}{a+b \sin (c+d x)} \, dx}{a^4 b^2}-\frac {\left (3 a^2-b^2\right ) \int \csc ^2(c+d x) \, dx}{a^3}+\frac {\left (b \left (3 a^2-b^2\right )\right ) \int \csc (c+d x) \, dx}{a^4}\\ &=\frac {a x}{b^2}-\frac {b \left (3 a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\cos (c+d x)}{b d}+\frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d}-\frac {b \int \csc (c+d x) \, dx}{2 a^2}-\frac {\operatorname {Subst}\left (\int \left (1+x^2\right ) \, dx,x,\cot (c+d x)\right )}{a d}-\frac {\left (2 \left (a^2-b^2\right )^3\right ) \operatorname {Subst}\left (\int \frac {1}{a+2 b x+a x^2} \, dx,x,\tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 b^2 d}+\frac {\left (3 a^2-b^2\right ) \operatorname {Subst}(\int 1 \, dx,x,\cot (c+d x))}{a^3 d}\\ &=\frac {a x}{b^2}+\frac {b \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {b \left (3 a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d}+\frac {\left (3 a^2-b^2\right ) \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d}+\frac {\left (4 \left (a^2-b^2\right )^3\right ) \operatorname {Subst}\left (\int \frac {1}{-4 \left (a^2-b^2\right )-x^2} \, dx,x,2 b+2 a \tan \left (\frac {1}{2} (c+d x)\right )\right )}{a^4 b^2 d}\\ &=\frac {a x}{b^2}-\frac {2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {b+a \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{a^4 b^2 d}+\frac {b \tanh ^{-1}(\cos (c+d x))}{2 a^2 d}-\frac {b \left (3 a^2-b^2\right ) \tanh ^{-1}(\cos (c+d x))}{a^4 d}+\frac {\cos (c+d x)}{b d}-\frac {\cot (c+d x)}{a d}+\frac {\left (3 a^2-b^2\right ) \cot (c+d x)}{a^3 d}-\frac {\cot ^3(c+d x)}{3 a d}+\frac {b \cot (c+d x) \csc (c+d x)}{2 a^2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 6.20, size = 379, normalized size = 1.92 \[ \frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}-\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d}+\frac {\left (5 a^2 b-2 b^3\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}+\frac {\left (2 b^3-5 a^2 b\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 a^4 d}-\frac {2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (a \sin \left (\frac {1}{2} (c+d x)\right )+b \cos \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {a^2-b^2}}\right )}{a^4 b^2 d}+\frac {\csc \left (\frac {1}{2} (c+d x)\right ) \left (7 a^2 \cos \left (\frac {1}{2} (c+d x)\right )-3 b^2 \cos \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^3 d}+\frac {\sec \left (\frac {1}{2} (c+d x)\right ) \left (3 b^2 \sin \left (\frac {1}{2} (c+d x)\right )-7 a^2 \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 a^3 d}+\frac {a (c+d x)}{b^2 d}-\frac {\cot \left (\frac {1}{2} (c+d x)\right ) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{24 a d}+\frac {\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{24 a d}+\frac {\cos (c+d x)}{b d} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]^2*Cot[c + d*x]^4)/(a + b*Sin[c + d*x]),x]

[Out]

(a*(c + d*x))/(b^2*d) - (2*(a^2 - b^2)^(5/2)*ArcTan[(Sec[(c + d*x)/2]*(b*Cos[(c + d*x)/2] + a*Sin[(c + d*x)/2]
))/Sqrt[a^2 - b^2]])/(a^4*b^2*d) + Cos[c + d*x]/(b*d) + ((7*a^2*Cos[(c + d*x)/2] - 3*b^2*Cos[(c + d*x)/2])*Csc
[(c + d*x)/2])/(6*a^3*d) + (b*Csc[(c + d*x)/2]^2)/(8*a^2*d) - (Cot[(c + d*x)/2]*Csc[(c + d*x)/2]^2)/(24*a*d) +
((-5*a^2*b + 2*b^3)*Log[Cos[(c + d*x)/2]])/(2*a^4*d) + ((5*a^2*b - 2*b^3)*Log[Sin[(c + d*x)/2]])/(2*a^4*d) -
(b*Sec[(c + d*x)/2]^2)/(8*a^2*d) + (Sec[(c + d*x)/2]*(-7*a^2*Sin[(c + d*x)/2] + 3*b^2*Sin[(c + d*x)/2]))/(6*a^
3*d) + (Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/(24*a*d)

________________________________________________________________________________________

fricas [A]  time = 1.95, size = 801, normalized size = 4.07 \[ \left [\frac {4 \, {\left (7 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a^{2} + b^{2}} \log \left (\frac {{\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2} + 2 \, {\left (a \cos \left (d x + c\right ) \sin \left (d x + c\right ) + b \cos \left (d x + c\right )\right )} \sqrt {-a^{2} + b^{2}}}{b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}}\right ) \sin \left (d x + c\right ) + 3 \, {\left (5 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (5 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 12 \, {\left (2 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right ) + 6 \, {\left (2 \, a^{5} d x \cos \left (d x + c\right )^{2} + 2 \, a^{4} b \cos \left (d x + c\right )^{3} - 2 \, a^{5} d x - {\left (2 \, a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{4} b^{2} d \cos \left (d x + c\right )^{2} - a^{4} b^{2} d\right )} \sin \left (d x + c\right )}, \frac {4 \, {\left (7 \, a^{3} b^{2} - 3 \, a b^{4}\right )} \cos \left (d x + c\right )^{3} - 12 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4} - {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a^{2} - b^{2}} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 3 \, {\left (5 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, {\left (5 \, a^{2} b^{3} - 2 \, b^{5} - {\left (5 \, a^{2} b^{3} - 2 \, b^{5}\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 12 \, {\left (2 \, a^{3} b^{2} - a b^{4}\right )} \cos \left (d x + c\right ) + 6 \, {\left (2 \, a^{5} d x \cos \left (d x + c\right )^{2} + 2 \, a^{4} b \cos \left (d x + c\right )^{3} - 2 \, a^{5} d x - {\left (2 \, a^{4} b + a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{12 \, {\left (a^{4} b^{2} d \cos \left (d x + c\right )^{2} - a^{4} b^{2} d\right )} \sin \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

[1/12*(4*(7*a^3*b^2 - 3*a*b^4)*cos(d*x + c)^3 - 6*(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c
)^2)*sqrt(-a^2 + b^2)*log(((2*a^2 - b^2)*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2 + 2*(a*cos(d*x + c)*s
in(d*x + c) + b*cos(d*x + c))*sqrt(-a^2 + b^2))/(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2))*sin(d*x
+ c) + 3*(5*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) -
3*(5*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^5)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 12*(2*a
^3*b^2 - a*b^4)*cos(d*x + c) + 6*(2*a^5*d*x*cos(d*x + c)^2 + 2*a^4*b*cos(d*x + c)^3 - 2*a^5*d*x - (2*a^4*b + a
^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^4*b^2*d*cos(d*x + c)^2 - a^4*b^2*d)*sin(d*x + c)), 1/12*(4*(7*a^3*b^2
- 3*a*b^4)*cos(d*x + c)^3 - 12*(a^4 - 2*a^2*b^2 + b^4 - (a^4 - 2*a^2*b^2 + b^4)*cos(d*x + c)^2)*sqrt(a^2 - b^2
)*arctan(-(a*sin(d*x + c) + b)/(sqrt(a^2 - b^2)*cos(d*x + c)))*sin(d*x + c) + 3*(5*a^2*b^3 - 2*b^5 - (5*a^2*b^
3 - 2*b^5)*cos(d*x + c)^2)*log(1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 3*(5*a^2*b^3 - 2*b^5 - (5*a^2*b^3 - 2*b^
5)*cos(d*x + c)^2)*log(-1/2*cos(d*x + c) + 1/2)*sin(d*x + c) - 12*(2*a^3*b^2 - a*b^4)*cos(d*x + c) + 6*(2*a^5*
d*x*cos(d*x + c)^2 + 2*a^4*b*cos(d*x + c)^3 - 2*a^5*d*x - (2*a^4*b + a^2*b^3)*cos(d*x + c))*sin(d*x + c))/((a^
4*b^2*d*cos(d*x + c)^2 - a^4*b^2*d)*sin(d*x + c))]

________________________________________________________________________________________

giac [A]  time = 0.22, size = 317, normalized size = 1.61 \[ \frac {\frac {24 \, {\left (d x + c\right )} a}{b^{2}} + \frac {a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 27 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{3}} + \frac {48}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )} b} + \frac {12 \, {\left (5 \, a^{2} b - 2 \, b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{4}} - \frac {48 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{4} b^{2}} - \frac {110 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 44 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 12 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a^{3}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/24*(24*(d*x + c)*a/b^2 + (a^2*tan(1/2*d*x + 1/2*c)^3 - 3*a*b*tan(1/2*d*x + 1/2*c)^2 - 27*a^2*tan(1/2*d*x + 1
/2*c) + 12*b^2*tan(1/2*d*x + 1/2*c))/a^3 + 48/((tan(1/2*d*x + 1/2*c)^2 + 1)*b) + 12*(5*a^2*b - 2*b^3)*log(abs(
tan(1/2*d*x + 1/2*c)))/a^4 - 48*(a^6 - 3*a^4*b^2 + 3*a^2*b^4 - b^6)*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(a) +
arctan((a*tan(1/2*d*x + 1/2*c) + b)/sqrt(a^2 - b^2)))/(sqrt(a^2 - b^2)*a^4*b^2) - (110*a^2*b*tan(1/2*d*x + 1/
2*c)^3 - 44*b^3*tan(1/2*d*x + 1/2*c)^3 - 27*a^3*tan(1/2*d*x + 1/2*c)^2 + 12*a*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a
^2*b*tan(1/2*d*x + 1/2*c) + a^3)/(a^4*tan(1/2*d*x + 1/2*c)^3))/d

________________________________________________________________________________________

maple [B]  time = 0.46, size = 442, normalized size = 2.24 \[ \frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d a}-\frac {\left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b}{8 d \,a^{2}}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}+\frac {b^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{3}}+\frac {2}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}+\frac {2 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}-\frac {1}{24 d a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {9}{8 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b^{2}}{2 d \,a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}+\frac {b}{8 d \,a^{2} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}+\frac {5 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 d \,a^{2}}-\frac {b^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}-\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) a^{2}}{d \,b^{2} \sqrt {a^{2}-b^{2}}}+\frac {6 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \sqrt {a^{2}-b^{2}}}-\frac {6 b^{2} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{2} \sqrt {a^{2}-b^{2}}}+\frac {2 \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right ) b^{4}}{d \,a^{4} \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^4/(a+b*sin(d*x+c)),x)

[Out]

1/24/d/a*tan(1/2*d*x+1/2*c)^3-1/8/d/a^2*tan(1/2*d*x+1/2*c)^2*b-9/8/a/d*tan(1/2*d*x+1/2*c)+1/2/d/a^3*b^2*tan(1/
2*d*x+1/2*c)+2/d/b/(1+tan(1/2*d*x+1/2*c)^2)+2/d/b^2*a*arctan(tan(1/2*d*x+1/2*c))-1/24/d/a/tan(1/2*d*x+1/2*c)^3
+9/8/a/d/tan(1/2*d*x+1/2*c)-1/2/d/a^3/tan(1/2*d*x+1/2*c)*b^2+1/8/d/a^2*b/tan(1/2*d*x+1/2*c)^2+5/2/d/a^2*b*ln(t
an(1/2*d*x+1/2*c))-1/d/a^4*b^3*ln(tan(1/2*d*x+1/2*c))-2/d/b^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*
c)+2*b)/(a^2-b^2)^(1/2))*a^2+6/d/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))-6/d*
b^2/a^2/(a^2-b^2)^(1/2)*arctan(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))+2/d/a^4/(a^2-b^2)^(1/2)*arcta
n(1/2*(2*a*tan(1/2*d*x+1/2*c)+2*b)/(a^2-b^2)^(1/2))*b^4

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^4/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more details)Is 4*b^2-4*a^2 positive or negative?

________________________________________________________________________________________

mupad [B]  time = 15.57, size = 4712, normalized size = 23.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(sin(c + d*x)^4*(a + b*sin(c + d*x))),x)

[Out]

cos(c + d*x)/(32*a*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + (3*sin(c + d*x))/(32*b*d*((3*sin(c + d*x))
/32 - sin(3*c + 3*d*x)/32)) - (7*cos(3*c + 3*d*x))/(96*a*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + sin(
2*c + 2*d*x)/(32*b*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - sin(3*c + 3*d*x)/(32*b*d*((3*sin(c + d*x))
/32 - sin(3*c + 3*d*x)/32)) - sin(4*c + 4*d*x)/(64*b*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + (b^2*cos
(3*c + 3*d*x))/(32*a^3*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - (b^2*cos(c + d*x))/(32*a^3*d*((3*sin(c
+ d*x))/32 - sin(3*c + 3*d*x)/32)) + (b*sin(2*c + 2*d*x))/(32*a^2*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/3
2)) + (15*b*sin(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(64*a^2*d*((3*sin(c + d*x))/32 - sin(3*c
+ 3*d*x)/32)) - (3*a*sin(c + d*x)*atan((2*a^5*cos(c/2 + (d*x)/2) - 2*b^5*sin(c/2 + (d*x)/2) + 5*a^2*b^3*sin(c/
2 + (d*x)/2))/(2*b^5*cos(c/2 + (d*x)/2) + 2*a^5*sin(c/2 + (d*x)/2) - 5*a^2*b^3*cos(c/2 + (d*x)/2))))/(16*b^2*d
*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - (5*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*
x))/(64*a^2*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) - (3*b^3*sin(c + d*x)*log(sin(c/2 + (d*x)/2)/cos(c/
2 + (d*x)/2)))/(32*a^4*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + (a*sin(3*c + 3*d*x)*atan((2*a^5*cos(c/
2 + (d*x)/2) - 2*b^5*sin(c/2 + (d*x)/2) + 5*a^2*b^3*sin(c/2 + (d*x)/2))/(2*b^5*cos(c/2 + (d*x)/2) + 2*a^5*sin(
c/2 + (d*x)/2) - 5*a^2*b^3*cos(c/2 + (d*x)/2))))/(16*b^2*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x)/32)) + (b^3
*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(3*c + 3*d*x))/(32*a^4*d*((3*sin(c + d*x))/32 - sin(3*c + 3*d*x
)/32)) + (sin(3*c + 3*d*x)*atan((a^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5
*a^8*b^2)^(3/2)*8i + a^18*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(
1/2)*8i + b^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*32i + a
*b^7*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*16i - a^7*b*cos(
c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*12i - a^17*b*cos(c/2 + (d
*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*2i - a^3*b^5*cos(c/2 + (d*x)/2)*(
b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*52i + a^5*b^3*cos(c/2 + (d*x)/2)*(b^10 -
a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*45i - a^7*b^11*cos(c/2 + (d*x)/2)*(b^10 - a^10 -
5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*2i + a^9*b^9*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b
^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*12i - a^11*b^7*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 1
0*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*14i - a^13*b^5*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*
b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*12i + a^15*b^3*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 -
10*a^6*b^4 + 5*a^8*b^2)^(1/2)*15i - a^2*b^6*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*
b^4 + 5*a^8*b^2)^(3/2)*114i + a^4*b^4*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 +
5*a^8*b^2)^(3/2)*121i - a^6*b^2*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*
b^2)^(3/2)*50i + a^2*b^16*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(
1/2)*2i - a^4*b^14*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*17
i + a^6*b^12*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*60i - a^
8*b^10*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*115i + a^10*b^
8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*162i - a^12*b^6*sin
(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*199i + a^14*b^4*sin(c/2
+ (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*146i - a^16*b^2*sin(c/2 + (d*
x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*50i)/(32*b^23*sin(c/2 + (d*x)/2) +
16*a*b^22*cos(c/2 + (d*x)/2) - 10*a^22*b*sin(c/2 + (d*x)/2) - 172*a^3*b^20*cos(c/2 + (d*x)/2) + 825*a^5*b^18*
cos(c/2 + (d*x)/2) - 2334*a^7*b^16*cos(c/2 + (d*x)/2) + 4332*a^9*b^14*cos(c/2 + (d*x)/2) - 5528*a^11*b^12*cos(
c/2 + (d*x)/2) + 4906*a^13*b^10*cos(c/2 + (d*x)/2) - 2975*a^15*b^8*cos(c/2 + (d*x)/2) + 1175*a^17*b^6*cos(c/2
+ (d*x)/2) - 275*a^19*b^4*cos(c/2 + (d*x)/2) + 30*a^21*b^2*cos(c/2 + (d*x)/2) - 352*a^2*b^21*sin(c/2 + (d*x)/2
) + 1734*a^4*b^19*sin(c/2 + (d*x)/2) - 5060*a^6*b^17*sin(c/2 + (d*x)/2) + 9738*a^8*b^15*sin(c/2 + (d*x)/2) - 1
2976*a^10*b^13*sin(c/2 + (d*x)/2) + 12156*a^12*b^11*sin(c/2 + (d*x)/2) - 7922*a^14*b^9*sin(c/2 + (d*x)/2) + 34
70*a^16*b^7*sin(c/2 + (d*x)/2) - 960*a^18*b^5*sin(c/2 + (d*x)/2) + 150*a^20*b^3*sin(c/2 + (d*x)/2)))*(b^10 - a
^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*1i)/(16*a^4*b^2*d*((3*sin(c + d*x))/32 - sin(3*c
+ 3*d*x)/32)) - (sin(c + d*x)*atan((a^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4
+ 5*a^8*b^2)^(3/2)*8i + a^18*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2
)^(1/2)*8i + b^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*32i
+ a*b^7*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*16i - a^7*b*c
os(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*12i - a^17*b*cos(c/2 +
(d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*2i - a^3*b^5*cos(c/2 + (d*x)/2
)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*52i + a^5*b^3*cos(c/2 + (d*x)/2)*(b^10
- a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(3/2)*45i - a^7*b^11*cos(c/2 + (d*x)/2)*(b^10 - a^1
0 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*2i + a^9*b^9*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^
2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*12i - a^11*b^7*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8
+ 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*14i - a^13*b^5*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a
^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*12i + a^15*b^3*cos(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6
- 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*15i - a^2*b^6*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a
^6*b^4 + 5*a^8*b^2)^(3/2)*114i + a^4*b^4*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4
+ 5*a^8*b^2)^(3/2)*121i - a^6*b^2*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a
^8*b^2)^(3/2)*50i + a^2*b^16*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2
)^(1/2)*2i - a^4*b^14*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)
*17i + a^6*b^12*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*60i -
a^8*b^10*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*115i + a^10
*b^8*sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*162i - a^12*b^6*
sin(c/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*199i + a^14*b^4*sin(c
/2 + (d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*146i - a^16*b^2*sin(c/2 +
(d*x)/2)*(b^10 - a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*50i)/(32*b^23*sin(c/2 + (d*x)/2
) + 16*a*b^22*cos(c/2 + (d*x)/2) - 10*a^22*b*sin(c/2 + (d*x)/2) - 172*a^3*b^20*cos(c/2 + (d*x)/2) + 825*a^5*b^
18*cos(c/2 + (d*x)/2) - 2334*a^7*b^16*cos(c/2 + (d*x)/2) + 4332*a^9*b^14*cos(c/2 + (d*x)/2) - 5528*a^11*b^12*c
os(c/2 + (d*x)/2) + 4906*a^13*b^10*cos(c/2 + (d*x)/2) - 2975*a^15*b^8*cos(c/2 + (d*x)/2) + 1175*a^17*b^6*cos(c
/2 + (d*x)/2) - 275*a^19*b^4*cos(c/2 + (d*x)/2) + 30*a^21*b^2*cos(c/2 + (d*x)/2) - 352*a^2*b^21*sin(c/2 + (d*x
)/2) + 1734*a^4*b^19*sin(c/2 + (d*x)/2) - 5060*a^6*b^17*sin(c/2 + (d*x)/2) + 9738*a^8*b^15*sin(c/2 + (d*x)/2)
- 12976*a^10*b^13*sin(c/2 + (d*x)/2) + 12156*a^12*b^11*sin(c/2 + (d*x)/2) - 7922*a^14*b^9*sin(c/2 + (d*x)/2) +
3470*a^16*b^7*sin(c/2 + (d*x)/2) - 960*a^18*b^5*sin(c/2 + (d*x)/2) + 150*a^20*b^3*sin(c/2 + (d*x)/2)))*(b^10
- a^10 - 5*a^2*b^8 + 10*a^4*b^6 - 10*a^6*b^4 + 5*a^8*b^2)^(1/2)*3i)/(16*a^4*b^2*d*((3*sin(c + d*x))/32 - sin(3
*c + 3*d*x)/32))

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*csc(d*x+c)**4/(a+b*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________