3.1345 \(\int \frac{\csc ^3(c+d x) \sec ^2(c+d x)}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=181 \[ \frac{2 b^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{3/2}}-\frac{b \tan (c+d x)}{d \left (a^2-b^2\right )}+\frac{\left (3 a^2-b^2\right ) \sec (c+d x)}{2 a d \left (a^2-b^2\right )}-\frac{\left (3 a^2+2 b^2\right ) \tanh ^{-1}(\cos (c+d x))}{2 a^3 d}+\frac{b \cot (c+d x)}{a^2 d}-\frac{\csc ^2(c+d x) \sec (c+d x)}{2 a d} \]

[Out]

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

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Rubi [A]  time = 0.359732, antiderivative size = 212, normalized size of antiderivative = 1.17, number of steps used = 17, number of rules used = 12, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.414, Rules used = {2898, 2622, 321, 207, 2620, 14, 288, 2696, 12, 2660, 618, 204} \[ \frac{2 b^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 d \left (a^2-b^2\right )^{3/2}}+\frac{b^2 \sec (c+d x)}{a^3 d}-\frac{b^2 \tanh ^{-1}(\cos (c+d x))}{a^3 d}+\frac{b^3 \sec (c+d x) (b-a \sin (c+d x))}{a^3 d \left (a^2-b^2\right )}-\frac{b \tan (c+d x)}{a^2 d}+\frac{b \cot (c+d x)}{a^2 d}+\frac{3 \sec (c+d x)}{2 a d}-\frac{3 \tanh ^{-1}(\cos (c+d x))}{2 a d}-\frac{\csc ^2(c+d x) \sec (c+d x)}{2 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Csc[c + d*x]^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

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

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])

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 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 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 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 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 288

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*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

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 12

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

Rubi steps

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

Mathematica [A]  time = 3.00455, size = 261, normalized size = 1.44 \[ \frac{\frac{16 b^5 \tan ^{-1}\left (\frac{a \tan \left (\frac{1}{2} (c+d x)\right )+b}{\sqrt{a^2-b^2}}\right )}{a^3 \left (a^2-b^2\right )^{3/2}}+\frac{4 \left (3 a^2+2 b^2\right ) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}-\frac{4 \left (3 a^2+2 b^2\right ) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a^3}-\frac{4 b \tan \left (\frac{1}{2} (c+d x)\right )}{a^2}+\frac{4 b \cot \left (\frac{1}{2} (c+d x)\right )}{a^2}+\frac{8 \sin \left (\frac{1}{2} (c+d x)\right )}{(a+b) \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{8 \sin \left (\frac{1}{2} (c+d x)\right )}{(a-b) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}-\frac{\csc ^2\left (\frac{1}{2} (c+d x)\right )}{a}+\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right )}{a}}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Csc[c + d*x]^3*Sec[c + d*x]^2)/(a + b*Sin[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.136, size = 227, normalized size = 1.3 \begin{align*}{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}-{\frac{b}{2\,d{a}^{2}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{1}{d \left ( a+b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+2\,{\frac{{b}^{5}}{d{a}^{3} \left ( a-b \right ) \left ( a+b \right ) \sqrt{{a}^{2}-{b}^{2}}}\arctan \left ( 1/2\,{\frac{2\,a\tan \left ( 1/2\,dx+c/2 \right ) +2\,b}{\sqrt{{a}^{2}-{b}^{2}}}} \right ) }+{\frac{1}{d \left ( a-b \right ) } \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{1}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-2}}+{\frac{3}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{{b}^{2}}{d{a}^{3}}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) }+{\frac{b}{2\,d{a}^{2}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 5.02943, size = 1931, normalized size = 10.67 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)**3*sec(d*x+c)**2/(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.19246, size = 331, normalized size = 1.83 \begin{align*} \frac{\frac{16 \,{\left (\pi \left \lfloor \frac{d x + c}{2 \, \pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (a\right ) + \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^{5}}{{\left (a^{5} - a^{3} b^{2}\right )} \sqrt{a^{2} - b^{2}}} + \frac{16 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a\right )}}{{\left (a^{2} - b^{2}\right )}{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}} + \frac{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{2}} + \frac{4 \,{\left (3 \, a^{2} + 2 \, b^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) \right |}\right )}{a^{3}} - \frac{18 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2}}{a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^3*sec(d*x+c)^2/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

1/8*(16*(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^5/(
(a^5 - a^3*b^2)*sqrt(a^2 - b^2)) + 16*(b*tan(1/2*d*x + 1/2*c) - a)/((a^2 - b^2)*(tan(1/2*d*x + 1/2*c)^2 - 1))
+ (a*tan(1/2*d*x + 1/2*c)^2 - 4*b*tan(1/2*d*x + 1/2*c))/a^2 + 4*(3*a^2 + 2*b^2)*log(abs(tan(1/2*d*x + 1/2*c)))
/a^3 - (18*a^2*tan(1/2*d*x + 1/2*c)^2 + 12*b^2*tan(1/2*d*x + 1/2*c)^2 - 4*a*b*tan(1/2*d*x + 1/2*c) + a^2)/(a^3
*tan(1/2*d*x + 1/2*c)^2))/d