3.83 \(\int \frac{\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=174 \[ -\frac{35 \cot (c+d x)}{16 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac{115 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{15 \cot (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}+\frac{\cot (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \]

[Out]

(5*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(5/2)*d) - (115*ArcTanh[(Sqrt[a]*Cos[c + d*x])
/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Cot[c + d*x]/(4*d*(a + a*Sin[c + d*x])^(5/2)) +
(15*Cot[c + d*x])/(16*a*d*(a + a*Sin[c + d*x])^(3/2)) - (35*Cot[c + d*x])/(16*a^2*d*Sqrt[a + a*Sin[c + d*x]])

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Rubi [A]  time = 0.506849, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {2766, 2978, 2984, 2985, 2649, 206, 2773} \[ -\frac{35 \cot (c+d x)}{16 a^2 d \sqrt{a \sin (c+d x)+a}}+\frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a \sin (c+d x)+a}}\right )}{a^{5/2} d}-\frac{115 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{15 \cot (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}}+\frac{\cot (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[c + d*x]^2/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(5*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(a^(5/2)*d) - (115*ArcTanh[(Sqrt[a]*Cos[c + d*x])
/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + Cot[c + d*x]/(4*d*(a + a*Sin[c + d*x])^(5/2)) +
(15*Cot[c + d*x])/(16*a*d*(a + a*Sin[c + d*x])^(3/2)) - (35*Cot[c + d*x])/(16*a^2*d*Sqrt[a + a*Sin[c + d*x]])

Rule 2766

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dis
t[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*
(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d,
0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] &&  !GtQ[n, 0] && (IntegersQ[2*m, 2*n] || (IntegerQ
[m] && EqQ[c, 0]))

Rule 2978

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*
x])^(n + 1))/(a*f*(2*m + 1)*(b*c - a*d)), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
0])

Rule 2984

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 2985

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])), x_Symbol] :> Dist[(A*b - a*B)/(b*c - a*d), Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[(
B*c - A*d)/(b*c - a*d), Int[Sqrt[a + b*Sin[e + f*x]]/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f,
A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 206

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

Rule 2773

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(-2*
b)/f, Subst[Int[1/(b*c + a*d - d*x^2), x], x, (b*Cos[e + f*x])/Sqrt[a + b*Sin[e + f*x]]], x] /; FreeQ[{a, b, c
, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rubi steps

\begin{align*} \int \frac{\csc ^2(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\frac{\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{\int \frac{\csc ^2(c+d x) \left (5 a-\frac{5}{2} a \sin (c+d x)\right )}{(a+a \sin (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac{\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac{\int \frac{\csc ^2(c+d x) \left (\frac{35 a^2}{2}-\frac{45}{4} a^2 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{8 a^4}\\ &=\frac{\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac{35 \cot (c+d x)}{16 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{\int \frac{\csc (c+d x) \left (-20 a^3+\frac{35}{4} a^3 \sin (c+d x)\right )}{\sqrt{a+a \sin (c+d x)}} \, dx}{8 a^5}\\ &=\frac{\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac{35 \cot (c+d x)}{16 a^2 d \sqrt{a+a \sin (c+d x)}}-\frac{5 \int \csc (c+d x) \sqrt{a+a \sin (c+d x)} \, dx}{2 a^3}+\frac{115 \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx}{32 a^2}\\ &=\frac{\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac{35 \cot (c+d x)}{16 a^2 d \sqrt{a+a \sin (c+d x)}}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{a^2 d}-\frac{115 \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{16 a^2 d}\\ &=\frac{5 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{a^{5/2} d}-\frac{115 \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}+\frac{\cot (c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac{15 \cot (c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac{35 \cot (c+d x)}{16 a^2 d \sqrt{a+a \sin (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 0.588044, size = 509, normalized size = 2.93 \[ \frac{\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right ) \left (8 \sin \left (\frac{1}{2} (c+d x)\right )+\frac{8 \sin \left (\frac{1}{4} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}{\cos \left (\frac{1}{4} (c+d x)\right )-\sin \left (\frac{1}{4} (c+d x)\right )}-\frac{8 \sin \left (\frac{1}{4} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4}{\sin \left (\frac{1}{4} (c+d x)\right )+\cos \left (\frac{1}{4} (c+d x)\right )}+8 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4-19 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3+38 \sin \left (\frac{1}{2} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2-4 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+40 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \log \left (-\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-40 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \log \left (\sin \left (\frac{1}{2} (c+d x)\right )-\cos \left (\frac{1}{2} (c+d x)\right )+1\right )-4 \tan \left (\frac{1}{4} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4-4 \cot \left (\frac{1}{4} (c+d x)\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4+(115+115 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{16 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[c + d*x]^2/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(8*Sin[(c + d*x)/2] - 4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 38*Sin[
(c + d*x)/2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - 19*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 + 8*(Cos[(c
+ d*x)/2] + Sin[(c + d*x)/2])^4 + (115 + 115*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/
4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 4*Cot[(c + d*x)/4]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + 40
*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 - 40*Log[1 - Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (8*Sin[(c + d*x)/4]*(Cos[(c + d*x)/2] + Si
n[(c + d*x)/2])^4)/(Cos[(c + d*x)/4] - Sin[(c + d*x)/4]) - (8*Sin[(c + d*x)/4]*(Cos[(c + d*x)/2] + Sin[(c + d*
x)/2])^4)/(Cos[(c + d*x)/4] + Sin[(c + d*x)/4]) - 4*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4*Tan[(c + d*x)/4]))
/(16*d*(a*(1 + Sin[c + d*x]))^(5/2))

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Maple [B]  time = 0.839, size = 356, normalized size = 2.1 \begin{align*} -{\frac{1}{ \left ( 32+32\,\sin \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) \cos \left ( dx+c \right ) d} \left ( 115\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}{a}^{2}+32\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{3/2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}+230\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}-160\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{3}{a}^{2}+148\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{3/2}\sin \left ( dx+c \right ) -38\, \left ( -a \left ( \sin \left ( dx+c \right ) -1 \right ) \right ) ^{3/2}\sqrt{a}\sin \left ( dx+c \right ) +115\,\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \sin \left ( dx+c \right ){a}^{2}-320\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}{a}^{2}+32\,\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{3/2}-160\,{\it Artanh} \left ({\frac{\sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }}{\sqrt{a}}} \right ) \sin \left ( dx+c \right ){a}^{2} \right ) \sqrt{-a \left ( \sin \left ( dx+c \right ) -1 \right ) }{a}^{-{\frac{9}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x)

[Out]

-1/32*(115*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c)^3*a^2+32*(-a*(sin(d*x+c)-
1))^(1/2)*a^(3/2)*sin(d*x+c)^2+230*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1/2)/a^(1/2))*sin(d*x+c)^2
*a^2-160*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^3*a^2+148*(-a*(sin(d*x+c)-1))^(1/2)*a^(3/2)*sin
(d*x+c)-38*(-a*(sin(d*x+c)-1))^(3/2)*a^(1/2)*sin(d*x+c)+115*2^(1/2)*arctanh(1/2*(-a*(sin(d*x+c)-1))^(1/2)*2^(1
/2)/a^(1/2))*sin(d*x+c)*a^2-320*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)^2*a^2+32*(-a*(sin(d*x+c)
-1))^(1/2)*a^(3/2)-160*arctanh((-a*(sin(d*x+c)-1))^(1/2)/a^(1/2))*sin(d*x+c)*a^2)*(-a*(sin(d*x+c)-1))^(1/2)/a^
(9/2)/(1+sin(d*x+c))/sin(d*x+c)/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (d x + c\right )^{2}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(csc(d*x + c)^2/(a*sin(d*x + c) + a)^(5/2), x)

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Fricas [B]  time = 2.02892, size = 1683, normalized size = 9.67 \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)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/64*(115*sqrt(2)*(cos(d*x + c)^4 - 2*cos(d*x + c)^3 - 5*cos(d*x + c)^2 - (cos(d*x + c)^3 + 3*cos(d*x + c)^2 -
2*cos(d*x + c) - 4)*sin(d*x + c) + 2*cos(d*x + c) + 4)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a*sin(
d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x +
c) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 80*(cos(d*x + c)^4 - 2*cos(
d*x + c)^3 - 5*cos(d*x + c)^2 - (cos(d*x + c)^3 + 3*cos(d*x + c)^2 - 2*cos(d*x + c) - 4)*sin(d*x + c) + 2*cos(
d*x + c) + 4)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 + (cos(d*x + c) + 3)*sin(
d*x + c) - 2*cos(d*x + c) - 3)*sqrt(a*sin(d*x + c) + a)*sqrt(a) - 9*a*cos(d*x + c) + (a*cos(d*x + c)^2 + 8*a*c
os(d*x + c) - a)*sin(d*x + c) - a)/(cos(d*x + c)^3 + cos(d*x + c)^2 + (cos(d*x + c)^2 - 1)*sin(d*x + c) - cos(
d*x + c) - 1)) + 4*(35*cos(d*x + c)^3 - 20*cos(d*x + c)^2 - (35*cos(d*x + c)^2 + 55*cos(d*x + c) + 4)*sin(d*x
+ c) - 51*cos(d*x + c) + 4)*sqrt(a*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^4 - 2*a^3*d*cos(d*x + c)^3 - 5*a^3*d
*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c) + 4*a^3*d - (a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*co
s(d*x + c) - 4*a^3*d)*sin(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)**2/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \mathit{sage}_{2} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(d*x+c)^2/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

sage2