3.1324 \(\int \frac {\cos ^4(c+d x) \cot ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=183 \[ -\frac {2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 b^4 d}+\frac {a x \left (a^2-3 b^2\right )}{b^4}+\frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{b^3 d}+\frac {b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac {a x}{2 b^2}-\frac {\cot (c+d x)}{a d}-\frac {\cos ^3(c+d x)}{3 b d}+\frac {\cos (c+d x)}{b d} \]

[Out]

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

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Rubi [A]  time = 0.26, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {2897, 3770, 3767, 8, 2638, 2635, 2633, 2660, 618, 204} \[ \frac {\left (a^2-3 b^2\right ) \cos (c+d x)}{b^3 d}-\frac {2 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 b^4 d}+\frac {a x \left (a^2-3 b^2\right )}{b^4}+\frac {b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {a \sin (c+d x) \cos (c+d x)}{2 b^2 d}+\frac {a x}{2 b^2}-\frac {\cot (c+d x)}{a d}-\frac {\cos ^3(c+d x)}{3 b d}+\frac {\cos (c+d x)}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/(2*b^2) + (a*(a^2 - 3*b^2)*x)/b^4 - (2*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]
])/(a^2*b^4*d) + (b*ArcTanh[Cos[c + d*x]])/(a^2*d) + Cos[c + d*x]/(b*d) + ((a^2 - 3*b^2)*Cos[c + d*x])/(b^3*d)
- Cos[c + d*x]^3/(3*b*d) - Cot[c + d*x]/(a*d) - (a*Cos[c + d*x]*Sin[c + d*x])/(2*b^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 2633

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

Rule 2635

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

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

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Mathematica [A]  time = 1.41, size = 208, normalized size = 1.14 \[ -\frac {-12 a^5 c-12 a^5 d x+3 a^3 b^2 \sin (2 (c+d x))+30 a^3 b^2 c+30 a^3 b^2 d x+a^2 b^3 \cos (3 (c+d x))-3 a^2 b \left (4 a^2-9 b^2\right ) \cos (c+d x)+24 \left (a^2-b^2\right )^{5/2} \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )-6 a b^4 \tan \left (\frac {1}{2} (c+d x)\right )+6 a b^4 \cot \left (\frac {1}{2} (c+d x)\right )+12 b^5 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )-12 b^5 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{12 a^2 b^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(-12*a^5*c + 30*a^3*b^2*c - 12*a^5*d*x + 30*a^3*b^2*d*x + 24*(a^2 - b^2)^(5/2)*ArcTan[(b + a*Tan[(c + d*
x)/2])/Sqrt[a^2 - b^2]] - 3*a^2*b*(4*a^2 - 9*b^2)*Cos[c + d*x] + a^2*b^3*Cos[3*(c + d*x)] + 6*a*b^4*Cot[(c + d
*x)/2] - 12*b^5*Log[Cos[(c + d*x)/2]] + 12*b^5*Log[Sin[(c + d*x)/2]] + 3*a^3*b^2*Sin[2*(c + d*x)] - 6*a*b^4*Ta
n[(c + d*x)/2])/(a^2*b^4*d)

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fricas [A]  time = 1.28, size = 549, normalized size = 3.00 \[ \left [\frac {3 \, a^{3} b^{2} \cos \left (d x + c\right )^{3} + 3 \, b^{5} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, b^{5} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 3 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\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 (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right ) - {\left (2 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x - 6 \, {\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, a^{2} b^{4} d \sin \left (d x + c\right )}, \frac {3 \, a^{3} b^{2} \cos \left (d x + c\right )^{3} + 3 \, b^{5} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 3 \, b^{5} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) + 6 \, {\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\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 (a^{3} b^{2} + 2 \, a b^{4}\right )} \cos \left (d x + c\right ) - {\left (2 \, a^{2} b^{3} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, a^{5} - 5 \, a^{3} b^{2}\right )} d x - 6 \, {\left (a^{4} b - 2 \, a^{2} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, a^{2} b^{4} d \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)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.19, size = 302, normalized size = 1.65 \[ -\frac {\frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {3 \, {\left (2 \, a^{3} - 5 \, a b^{2}\right )} {\left (d x + c\right )}}{b^{4}} - \frac {3 \, {\left (2 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - a\right )}}{a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )} + \frac {12 \, {\left (a^{6} - 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} - b^{6}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )}}{\sqrt {a^{2} - b^{2}} a^{2} b^{4}} - \frac {2 \, {\left (3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 18 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 12 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, a^{2} - 14 \, b^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} b^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 3*tan(1/2*d*x + 1/2*c)/a - 3*(2*a^3 - 5*a*b^2)*(d*x + c)/b^4 -
3*(2*b*tan(1/2*d*x + 1/2*c) - a)/(a^2*tan(1/2*d*x + 1/2*c)) + 12*(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^2*b
^4) - 2*(3*a*b*tan(1/2*d*x + 1/2*c)^5 + 6*a^2*tan(1/2*d*x + 1/2*c)^4 - 18*b^2*tan(1/2*d*x + 1/2*c)^4 + 12*a^2*
tan(1/2*d*x + 1/2*c)^2 - 24*b^2*tan(1/2*d*x + 1/2*c)^2 - 3*a*b*tan(1/2*d*x + 1/2*c) + 6*a^2 - 14*b^2)/((tan(1/
2*d*x + 1/2*c)^2 + 1)^3*b^3))/d

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maple [B]  time = 0.44, size = 557, normalized size = 3.04 \[ \frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {a \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {6 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 a^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {8 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,b^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 a^{2}}{d \,b^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {14}{3 d b \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{3}}{d \,b^{4}}-\frac {5 a \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,b^{2}}-\frac {1}{2 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {2 a^{4} \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 \,b^{4} \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 ) 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 {2 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}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*csc(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

1/2/a/d*tan(1/2*d*x+1/2*c)+1/d/b^2/(1+tan(1/2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)^5+2/d/b^3/(1+tan(1/2*d*x+1/
2*c)^2)^3*tan(1/2*d*x+1/2*c)^4*a^2-6/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^4+4/d/b^3/(1+tan(1/2*d*
x+1/2*c)^2)^3*a^2*tan(1/2*d*x+1/2*c)^2-8/d/b/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^2-1/d/b^2/(1+tan(1/
2*d*x+1/2*c)^2)^3*a*tan(1/2*d*x+1/2*c)+2/d/b^3/(1+tan(1/2*d*x+1/2*c)^2)^3*a^2-14/3/d/b/(1+tan(1/2*d*x+1/2*c)^2
)^3+2/d/b^4*arctan(tan(1/2*d*x+1/2*c))*a^3-5/d/b^2*a*arctan(tan(1/2*d*x+1/2*c))-1/2/a/d/tan(1/2*d*x+1/2*c)-1/d
/a^2*b*ln(tan(1/2*d*x+1/2*c))-2/d*a^4/b^4/(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-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))+2/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))

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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)^2/(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?

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mupad [B]  time = 13.10, size = 4892, normalized size = 26.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6/(sin(c + d*x)^2*(a + b*sin(c + d*x))),x)

[Out]

tan(c/2 + (d*x)/2)/(2*a*d) - (3*tan(c/2 + (d*x)/2)^4 + (8*tan(c/2 + (d*x)/2)^3*(2*a*b^2 - a^3))/b^3 + (4*tan(c
/2 + (d*x)/2)^5*(3*a*b^2 - a^3))/b^3 + (tan(c/2 + (d*x)/2)^2*(2*a^2 + 3*b^2))/b^2 - (tan(c/2 + (d*x)/2)^6*(2*a
^2 - b^2))/b^2 + (4*tan(c/2 + (d*x)/2)*(7*a*b^2 - 3*a^3))/(3*b^3) + 1)/(d*(2*a*tan(c/2 + (d*x)/2) + 6*a*tan(c/
2 + (d*x)/2)^3 + 6*a*tan(c/2 + (d*x)/2)^5 + 2*a*tan(c/2 + (d*x)/2)^7)) - (b*log(tan(c/2 + (d*x)/2)))/(a^2*d) -
(atan((((-(a + b)^5*(a - b)^5)^(1/2)*((8*(10*a^2*b^12 - 14*a^14 + 136*a^4*b^10 - 386*a^6*b^8 + 467*a^8*b^6 -
309*a^10*b^4 + 105*a^12*b^2))/(a^2*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(16*b^15 - 52*a^2*b^13 + 50*a^4*b^
11 + 86*a^6*b^9 - 155*a^8*b^7 + 76*a^10*b^5 - 12*a^12*b^3))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(84*a^2*b^18 - 3
60*a^4*b^16 + 1020*a^6*b^14 - 1264*a^8*b^12 + 661*a^10*b^10 - 140*a^12*b^8 + 8*a^14*b^6))/(a^3*b^12) + ((-(a +
b)^5*(a - b)^5)^(1/2)*((8*(32*a^2*b^14 - 64*a^4*b^12 + 50*a^6*b^10 - 14*a^8*b^8))/(a^2*b^8) + (((8*(16*a^4*b^
13 - 12*a^6*b^11))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(64*a^4*b^18 - 68*a^6*b^16 + 8*a^8*b^14))/(a^3*b^12))*(-(
a + b)^5*(a - b)^5)^(1/2))/(a^2*b^4) + (8*tan(c/2 + (d*x)/2)*(64*a^2*b^19 - 148*a^4*b^17 + 152*a^6*b^15 - 80*a
^8*b^13 + 16*a^10*b^11))/(a^3*b^12)))/(a^2*b^4)))/(a^2*b^4) + (8*tan(c/2 + (d*x)/2)*(4*b^19 - 24*a^2*b^17 + 46
0*a^4*b^15 - 1300*a^6*b^13 + 1769*a^8*b^11 - 1408*a^10*b^9 + 652*a^12*b^7 - 160*a^14*b^5 + 16*a^16*b^3))/(a^3*
b^12))*1i)/(a^2*b^4) + ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(10*a^2*b^12 - 14*a^14 + 136*a^4*b^10 - 386*a^6*b^8 +
467*a^8*b^6 - 309*a^10*b^4 + 105*a^12*b^2))/(a^2*b^8) - ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(16*b^15 - 52*a^2*b
^13 + 50*a^4*b^11 + 86*a^6*b^9 - 155*a^8*b^7 + 76*a^10*b^5 - 12*a^12*b^3))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(
84*a^2*b^18 - 360*a^4*b^16 + 1020*a^6*b^14 - 1264*a^8*b^12 + 661*a^10*b^10 - 140*a^12*b^8 + 8*a^14*b^6))/(a^3*
b^12) - ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(32*a^2*b^14 - 64*a^4*b^12 + 50*a^6*b^10 - 14*a^8*b^8))/(a^2*b^8) -
(((8*(16*a^4*b^13 - 12*a^6*b^11))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(64*a^4*b^18 - 68*a^6*b^16 + 8*a^8*b^14))/
(a^3*b^12))*(-(a + b)^5*(a - b)^5)^(1/2))/(a^2*b^4) + (8*tan(c/2 + (d*x)/2)*(64*a^2*b^19 - 148*a^4*b^17 + 152*
a^6*b^15 - 80*a^8*b^13 + 16*a^10*b^11))/(a^3*b^12)))/(a^2*b^4)))/(a^2*b^4) + (8*tan(c/2 + (d*x)/2)*(4*b^19 - 2
4*a^2*b^17 + 460*a^4*b^15 - 1300*a^6*b^13 + 1769*a^8*b^11 - 1408*a^10*b^9 + 652*a^12*b^7 - 160*a^14*b^5 + 16*a
^16*b^3))/(a^3*b^12))*1i)/(a^2*b^4))/((16*(10*b^13 - 14*a^12*b + 36*a^2*b^11 - 231*a^4*b^9 + 391*a^6*b^7 - 297
*a^8*b^5 + 105*a^10*b^3))/(a^2*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(10*a^2*b^12 - 14*a^14 + 136*a^4*b^10
- 386*a^6*b^8 + 467*a^8*b^6 - 309*a^10*b^4 + 105*a^12*b^2))/(a^2*b^8) + ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(16*
b^15 - 52*a^2*b^13 + 50*a^4*b^11 + 86*a^6*b^9 - 155*a^8*b^7 + 76*a^10*b^5 - 12*a^12*b^3))/(a^2*b^8) + (8*tan(c
/2 + (d*x)/2)*(84*a^2*b^18 - 360*a^4*b^16 + 1020*a^6*b^14 - 1264*a^8*b^12 + 661*a^10*b^10 - 140*a^12*b^8 + 8*a
^14*b^6))/(a^3*b^12) + ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(32*a^2*b^14 - 64*a^4*b^12 + 50*a^6*b^10 - 14*a^8*b^8
))/(a^2*b^8) + (((8*(16*a^4*b^13 - 12*a^6*b^11))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(64*a^4*b^18 - 68*a^6*b^16
+ 8*a^8*b^14))/(a^3*b^12))*(-(a + b)^5*(a - b)^5)^(1/2))/(a^2*b^4) + (8*tan(c/2 + (d*x)/2)*(64*a^2*b^19 - 148*
a^4*b^17 + 152*a^6*b^15 - 80*a^8*b^13 + 16*a^10*b^11))/(a^3*b^12)))/(a^2*b^4)))/(a^2*b^4) + (8*tan(c/2 + (d*x)
/2)*(4*b^19 - 24*a^2*b^17 + 460*a^4*b^15 - 1300*a^6*b^13 + 1769*a^8*b^11 - 1408*a^10*b^9 + 652*a^12*b^7 - 160*
a^14*b^5 + 16*a^16*b^3))/(a^3*b^12)))/(a^2*b^4) - ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(10*a^2*b^12 - 14*a^14 + 1
36*a^4*b^10 - 386*a^6*b^8 + 467*a^8*b^6 - 309*a^10*b^4 + 105*a^12*b^2))/(a^2*b^8) - ((-(a + b)^5*(a - b)^5)^(1
/2)*((8*(16*b^15 - 52*a^2*b^13 + 50*a^4*b^11 + 86*a^6*b^9 - 155*a^8*b^7 + 76*a^10*b^5 - 12*a^12*b^3))/(a^2*b^8
) + (8*tan(c/2 + (d*x)/2)*(84*a^2*b^18 - 360*a^4*b^16 + 1020*a^6*b^14 - 1264*a^8*b^12 + 661*a^10*b^10 - 140*a^
12*b^8 + 8*a^14*b^6))/(a^3*b^12) - ((-(a + b)^5*(a - b)^5)^(1/2)*((8*(32*a^2*b^14 - 64*a^4*b^12 + 50*a^6*b^10
- 14*a^8*b^8))/(a^2*b^8) - (((8*(16*a^4*b^13 - 12*a^6*b^11))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(64*a^4*b^18 -
68*a^6*b^16 + 8*a^8*b^14))/(a^3*b^12))*(-(a + b)^5*(a - b)^5)^(1/2))/(a^2*b^4) + (8*tan(c/2 + (d*x)/2)*(64*a^2
*b^19 - 148*a^4*b^17 + 152*a^6*b^15 - 80*a^8*b^13 + 16*a^10*b^11))/(a^3*b^12)))/(a^2*b^4)))/(a^2*b^4) + (8*tan
(c/2 + (d*x)/2)*(4*b^19 - 24*a^2*b^17 + 460*a^4*b^15 - 1300*a^6*b^13 + 1769*a^8*b^11 - 1408*a^10*b^9 + 652*a^1
2*b^7 - 160*a^14*b^5 + 16*a^16*b^3))/(a^3*b^12)))/(a^2*b^4) - (16*tan(c/2 + (d*x)/2)*(32*a^18 - 500*a^4*b^14 +
2500*a^6*b^12 - 5260*a^8*b^10 + 6036*a^10*b^8 - 4080*a^12*b^6 + 1624*a^14*b^4 - 352*a^16*b^2))/(a^3*b^12)))*(
-(a + b)^5*(a - b)^5)^(1/2)*2i)/(a^2*b^4*d) - (a*atan(((a*(2*a^2 - 5*b^2)*((8*(10*a^2*b^12 - 14*a^14 + 136*a^4
*b^10 - 386*a^6*b^8 + 467*a^8*b^6 - 309*a^10*b^4 + 105*a^12*b^2))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(4*b^19 -
24*a^2*b^17 + 460*a^4*b^15 - 1300*a^6*b^13 + 1769*a^8*b^11 - 1408*a^10*b^9 + 652*a^12*b^7 - 160*a^14*b^5 + 16*
a^16*b^3))/(a^3*b^12) - (a*(2*a^2 - 5*b^2)*((8*(16*b^15 - 52*a^2*b^13 + 50*a^4*b^11 + 86*a^6*b^9 - 155*a^8*b^7
+ 76*a^10*b^5 - 12*a^12*b^3))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(84*a^2*b^18 - 360*a^4*b^16 + 1020*a^6*b^14 -
1264*a^8*b^12 + 661*a^10*b^10 - 140*a^12*b^8 + 8*a^14*b^6))/(a^3*b^12) - (a*(2*a^2 - 5*b^2)*((8*(32*a^2*b^14
- 64*a^4*b^12 + 50*a^6*b^10 - 14*a^8*b^8))/(a^2*b^8) - (a*((8*(16*a^4*b^13 - 12*a^6*b^11))/(a^2*b^8) + (8*tan(
c/2 + (d*x)/2)*(64*a^4*b^18 - 68*a^6*b^16 + 8*a^8*b^14))/(a^3*b^12))*(2*a^2 - 5*b^2)*1i)/(2*b^4) + (8*tan(c/2
+ (d*x)/2)*(64*a^2*b^19 - 148*a^4*b^17 + 152*a^6*b^15 - 80*a^8*b^13 + 16*a^10*b^11))/(a^3*b^12))*1i)/(2*b^4))*
1i)/(2*b^4)))/(2*b^4) + (a*(2*a^2 - 5*b^2)*((8*(10*a^2*b^12 - 14*a^14 + 136*a^4*b^10 - 386*a^6*b^8 + 467*a^8*b
^6 - 309*a^10*b^4 + 105*a^12*b^2))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(4*b^19 - 24*a^2*b^17 + 460*a^4*b^15 - 13
00*a^6*b^13 + 1769*a^8*b^11 - 1408*a^10*b^9 + 652*a^12*b^7 - 160*a^14*b^5 + 16*a^16*b^3))/(a^3*b^12) + (a*(2*a
^2 - 5*b^2)*((8*(16*b^15 - 52*a^2*b^13 + 50*a^4*b^11 + 86*a^6*b^9 - 155*a^8*b^7 + 76*a^10*b^5 - 12*a^12*b^3))/
(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(84*a^2*b^18 - 360*a^4*b^16 + 1020*a^6*b^14 - 1264*a^8*b^12 + 661*a^10*b^10
- 140*a^12*b^8 + 8*a^14*b^6))/(a^3*b^12) + (a*(2*a^2 - 5*b^2)*((8*(32*a^2*b^14 - 64*a^4*b^12 + 50*a^6*b^10 - 1
4*a^8*b^8))/(a^2*b^8) + (a*((8*(16*a^4*b^13 - 12*a^6*b^11))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(64*a^4*b^18 - 6
8*a^6*b^16 + 8*a^8*b^14))/(a^3*b^12))*(2*a^2 - 5*b^2)*1i)/(2*b^4) + (8*tan(c/2 + (d*x)/2)*(64*a^2*b^19 - 148*a
^4*b^17 + 152*a^6*b^15 - 80*a^8*b^13 + 16*a^10*b^11))/(a^3*b^12))*1i)/(2*b^4))*1i)/(2*b^4)))/(2*b^4))/((16*(10
*b^13 - 14*a^12*b + 36*a^2*b^11 - 231*a^4*b^9 + 391*a^6*b^7 - 297*a^8*b^5 + 105*a^10*b^3))/(a^2*b^8) - (16*tan
(c/2 + (d*x)/2)*(32*a^18 - 500*a^4*b^14 + 2500*a^6*b^12 - 5260*a^8*b^10 + 6036*a^10*b^8 - 4080*a^12*b^6 + 1624
*a^14*b^4 - 352*a^16*b^2))/(a^3*b^12) - (a*(2*a^2 - 5*b^2)*((8*(10*a^2*b^12 - 14*a^14 + 136*a^4*b^10 - 386*a^6
*b^8 + 467*a^8*b^6 - 309*a^10*b^4 + 105*a^12*b^2))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(4*b^19 - 24*a^2*b^17 + 4
60*a^4*b^15 - 1300*a^6*b^13 + 1769*a^8*b^11 - 1408*a^10*b^9 + 652*a^12*b^7 - 160*a^14*b^5 + 16*a^16*b^3))/(a^3
*b^12) - (a*(2*a^2 - 5*b^2)*((8*(16*b^15 - 52*a^2*b^13 + 50*a^4*b^11 + 86*a^6*b^9 - 155*a^8*b^7 + 76*a^10*b^5
- 12*a^12*b^3))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(84*a^2*b^18 - 360*a^4*b^16 + 1020*a^6*b^14 - 1264*a^8*b^12
+ 661*a^10*b^10 - 140*a^12*b^8 + 8*a^14*b^6))/(a^3*b^12) - (a*(2*a^2 - 5*b^2)*((8*(32*a^2*b^14 - 64*a^4*b^12 +
50*a^6*b^10 - 14*a^8*b^8))/(a^2*b^8) - (a*((8*(16*a^4*b^13 - 12*a^6*b^11))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*
(64*a^4*b^18 - 68*a^6*b^16 + 8*a^8*b^14))/(a^3*b^12))*(2*a^2 - 5*b^2)*1i)/(2*b^4) + (8*tan(c/2 + (d*x)/2)*(64*
a^2*b^19 - 148*a^4*b^17 + 152*a^6*b^15 - 80*a^8*b^13 + 16*a^10*b^11))/(a^3*b^12))*1i)/(2*b^4))*1i)/(2*b^4))*1i
)/(2*b^4) + (a*(2*a^2 - 5*b^2)*((8*(10*a^2*b^12 - 14*a^14 + 136*a^4*b^10 - 386*a^6*b^8 + 467*a^8*b^6 - 309*a^1
0*b^4 + 105*a^12*b^2))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(4*b^19 - 24*a^2*b^17 + 460*a^4*b^15 - 1300*a^6*b^13
+ 1769*a^8*b^11 - 1408*a^10*b^9 + 652*a^12*b^7 - 160*a^14*b^5 + 16*a^16*b^3))/(a^3*b^12) + (a*(2*a^2 - 5*b^2)*
((8*(16*b^15 - 52*a^2*b^13 + 50*a^4*b^11 + 86*a^6*b^9 - 155*a^8*b^7 + 76*a^10*b^5 - 12*a^12*b^3))/(a^2*b^8) +
(8*tan(c/2 + (d*x)/2)*(84*a^2*b^18 - 360*a^4*b^16 + 1020*a^6*b^14 - 1264*a^8*b^12 + 661*a^10*b^10 - 140*a^12*b
^8 + 8*a^14*b^6))/(a^3*b^12) + (a*(2*a^2 - 5*b^2)*((8*(32*a^2*b^14 - 64*a^4*b^12 + 50*a^6*b^10 - 14*a^8*b^8))/
(a^2*b^8) + (a*((8*(16*a^4*b^13 - 12*a^6*b^11))/(a^2*b^8) + (8*tan(c/2 + (d*x)/2)*(64*a^4*b^18 - 68*a^6*b^16 +
8*a^8*b^14))/(a^3*b^12))*(2*a^2 - 5*b^2)*1i)/(2*b^4) + (8*tan(c/2 + (d*x)/2)*(64*a^2*b^19 - 148*a^4*b^17 + 15
2*a^6*b^15 - 80*a^8*b^13 + 16*a^10*b^11))/(a^3*b^12))*1i)/(2*b^4))*1i)/(2*b^4))*1i)/(2*b^4)))*(2*a^2 - 5*b^2))
/(b^4*d)

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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)**2/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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