3.1344 \(\int \frac {\csc ^2(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=128 \[ \frac {a \tan (c+d x)}{d \left (a^2-b^2\right )}-\frac {b \sec (c+d x)}{d \left (a^2-b^2\right )}-\frac {2 b^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}+\frac {b \tanh ^{-1}(\cos (c+d x))}{a^2 d}-\frac {\cot (c+d x)}{a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.29, antiderivative size = 150, normalized size of antiderivative = 1.17, number of steps used = 13, number of rules used = 11, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.379, Rules used = {2898, 2622, 321, 207, 2620, 14, 2696, 12, 2660, 618, 204} \[ -\frac {2 b^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 d \left (a^2-b^2\right )^{3/2}}-\frac {b^2 \sec (c+d x) (b-a \sin (c+d x))}{a^2 d \left (a^2-b^2\right )}-\frac {b \sec (c+d x)}{a^2 d}+\frac {b \tanh ^{-1}(\cos (c+d x))}{a^2 d}+\frac {\tan (c+d x)}{a d}-\frac {\cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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 207

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

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

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 2620

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(1 + x^2)^((
m + n)/2 - 1)/x^m, x], x, Tan[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n)/2]

Rule 2622

Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[1/(f*a^n), Subst[Int
[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n
+ 1)/2] &&  !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])

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 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

Rule 2898

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 0])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 1.03, size = 205, normalized size = 1.60 \[ \frac {-\frac {4 b^4 \tan ^{-1}\left (\frac {a \tan \left (\frac {1}{2} (c+d x)\right )+b}{\sqrt {a^2-b^2}}\right )}{a^2 \left (a^2-b^2\right )^{3/2}}-\frac {2 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}+\frac {2 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a^2}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right )}{(a-b) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}+\frac {\tan \left (\frac {1}{2} (c+d x)\right )}{a}-\frac {\cot \left (\frac {1}{2} (c+d x)\right )}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^2*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 1.02, size = 582, normalized size = 4.55 \[ \left [\frac {\sqrt {-a^{2} + b^{2}} b^{4} \cos \left (d x + c\right ) \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 ) + 2 \, a^{5} - 2 \, a^{3} b^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right )}, \frac {2 \, \sqrt {a^{2} - b^{2}} b^{4} \arctan \left (-\frac {a \sin \left (d x + c\right ) + b}{\sqrt {a^{2} - b^{2}} \cos \left (d x + c\right )}\right ) \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 2 \, a^{5} - 2 \, a^{3} b^{2} + {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - {\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} \cos \left (d x + c\right ) \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) \sin \left (d x + c\right ) - 2 \, {\left (2 \, a^{5} - 3 \, a^{3} b^{2} + a b^{4}\right )} \cos \left (d x + c\right )^{2} - 2 \, {\left (a^{4} b - a^{2} b^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (a^{6} - 2 \, a^{4} b^{2} + a^{2} b^{4}\right )} d \cos \left (d x + c\right ) \sin \left (d x + c\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.21, size = 259, normalized size = 2.02 \[ -\frac {\frac {12 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (a) + \arctan \left (\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + b}{\sqrt {a^{2} - b^{2}}}\right )\right )} b^{4}}{{\left (a^{4} - a^{2} b^{2}\right )} \sqrt {a^{2} - b^{2}}} + \frac {6 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}\right )}{a^{2}} - \frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a} - \frac {2 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 3 \, a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 10 \, a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, a^{3} - 3 \, a b^{2}}{{\left (a^{4} - a^{2} b^{2}\right )} {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(12*(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)))*b^4/
((a^4 - a^2*b^2)*sqrt(a^2 - b^2)) + 6*b*log(abs(tan(1/2*d*x + 1/2*c)))/a^2 - 3*tan(1/2*d*x + 1/2*c)/a - (2*a^2
*b*tan(1/2*d*x + 1/2*c)^3 - 2*b^3*tan(1/2*d*x + 1/2*c)^3 - 15*a^3*tan(1/2*d*x + 1/2*c)^2 + 3*a*b^2*tan(1/2*d*x
+ 1/2*c)^2 + 10*a^2*b*tan(1/2*d*x + 1/2*c) + 2*b^3*tan(1/2*d*x + 1/2*c) + 3*a^3 - 3*a*b^2)/((a^4 - a^2*b^2)*(
tan(1/2*d*x + 1/2*c)^3 - tan(1/2*d*x + 1/2*c))))/d

________________________________________________________________________________________

maple [A]  time = 0.44, size = 169, normalized size = 1.32 \[ -\frac {1}{d \left (a +b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}-\frac {1}{2 a d \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2}}-\frac {1}{d \left (a -b \right ) \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}-\frac {2 b^{4} \arctan \left (\frac {2 a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+2 b}{2 \sqrt {a^{2}-b^{2}}}\right )}{d \,a^{2} \left (a -b \right ) \left (a +b \right ) \sqrt {a^{2}-b^{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/d/(a+b)/(tan(1/2*d*x+1/2*c)-1)+1/2/a/d*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))-1/d/(a-b)/(tan(1/2*d*x+1/2*c)+1)-2/d/a^2*b^4/(a-b)/(a+b)/(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))

________________________________________________________________________________________

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(csc(d*x+c)^2*sec(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?

________________________________________________________________________________________

mupad [B]  time = 13.68, size = 778, normalized size = 6.08 \[ -\frac {\left (a\,b^6\,1{}\mathrm {i}-a^7\,\cos \left (2\,c+2\,d\,x\right )\,2{}\mathrm {i}-a^3\,b^4\,2{}\mathrm {i}+a^5\,b^2\,1{}\mathrm {i}+a\,b^6\,\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}-a^6\,b\,\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}-a^2\,b^5\,\sin \left (c+d\,x\right )\,2{}\mathrm {i}+a^4\,b^3\,\sin \left (c+d\,x\right )\,4{}\mathrm {i}-a^3\,b^4\,\cos \left (2\,c+2\,d\,x\right )\,4{}\mathrm {i}+a^5\,b^2\,\cos \left (2\,c+2\,d\,x\right )\,5{}\mathrm {i}+b^7\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}-a^2\,b^5\,\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+a^4\,b^3\,\sin \left (2\,c+2\,d\,x\right )\,2{}\mathrm {i}-a^6\,b\,\sin \left (c+d\,x\right )\,2{}\mathrm {i}-a^6\,b\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+2\,b^4\,\sin \left (2\,c+2\,d\,x\right )\,\mathrm {atan}\left (\frac {a^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}\,1{}\mathrm {i}+b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}\,4{}\mathrm {i}-a^2\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}\,3{}\mathrm {i}+a\,b^3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}\,2{}\mathrm {i}-a^3\,b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^7+2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^6\,b-3\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^5\,b^2-7\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^4\,b^3+4\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^3\,b^4+9\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a^2\,b^5-2\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,a\,b^6-4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,b^7}\right )\,\sqrt {-a^6+3\,a^4\,b^2-3\,a^2\,b^4+b^6}-a^2\,b^5\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sin \left (2\,c+2\,d\,x\right )\,3{}\mathrm {i}+a^4\,b^3\,\ln \left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,\sin \left (2\,c+2\,d\,x\right )\,3{}\mathrm {i}\right )\,1{}\mathrm {i}}{a^2\,d\,\sin \left (2\,c+2\,d\,x\right )\,\left (a^2-b^2\right )\,\left (a^4-2\,a^2\,b^2+b^4\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^2*sin(c + d*x)^2*(a + b*sin(c + d*x))),x)

[Out]

-((a*b^6*1i - a^7*cos(2*c + 2*d*x)*2i - a^3*b^4*2i + a^5*b^2*1i + a*b^6*cos(2*c + 2*d*x)*1i - a^6*b*sin(2*c +
2*d*x)*1i - a^2*b^5*sin(c + d*x)*2i + a^4*b^3*sin(c + d*x)*4i - a^3*b^4*cos(2*c + 2*d*x)*4i + a^5*b^2*cos(2*c
+ 2*d*x)*5i + b^7*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x)*1i - a^2*b^5*sin(2*c + 2*d*x)*1i
+ a^4*b^3*sin(2*c + 2*d*x)*2i - a^6*b*sin(c + d*x)*2i - a^6*b*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(
2*c + 2*d*x)*1i + 2*b^4*sin(2*c + 2*d*x)*atan((a^4*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2
)*1i + b^4*sin(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*4i - a^2*b^2*sin(c/2 + (d*x)/2)*(b^6 -
a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*3i + a*b^3*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*2i
- a^3*b*cos(c/2 + (d*x)/2)*(b^6 - a^6 - 3*a^2*b^4 + 3*a^4*b^2)^(1/2)*1i)/(a^7*cos(c/2 + (d*x)/2) - 4*b^7*sin(
c/2 + (d*x)/2) - 2*a*b^6*cos(c/2 + (d*x)/2) + 2*a^6*b*sin(c/2 + (d*x)/2) + 4*a^3*b^4*cos(c/2 + (d*x)/2) - 3*a^
5*b^2*cos(c/2 + (d*x)/2) + 9*a^2*b^5*sin(c/2 + (d*x)/2) - 7*a^4*b^3*sin(c/2 + (d*x)/2)))*(b^6 - a^6 - 3*a^2*b^
4 + 3*a^4*b^2)^(1/2) - a^2*b^5*log(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x)*3i + a^4*b^3*log(si
n(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2))*sin(2*c + 2*d*x)*3i)*1i)/(a^2*d*sin(2*c + 2*d*x)*(a^2 - b^2)*(a^4 + b^4 -
2*a^2*b^2))

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________