\(\int \csc ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\) [60]

Optimal result

Integrand size = 23, antiderivative size = 182 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {163 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{64 d}-\frac {163 a^3 \cot (c+d x)}{64 d \sqrt {a+a \sin (c+d x)}}-\frac {163 a^3 \cot (c+d x) \csc (c+d x)}{96 d \sqrt {a+a \sin (c+d x)}}-\frac {17 a^3 \cot (c+d x) \csc ^2(c+d x)}{24 d \sqrt {a+a \sin (c+d x)}}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d} \]

[Out]

-163/64*a^(5/2)*arctanh(cos(d*x+c)*a^(1/2)/(a+a*sin(d*x+c))^(1/2))/d-163/64*a^3*cot(d*x+c)/d/(a+a*sin(d*x+c))^
(1/2)-163/96*a^3*cot(d*x+c)*csc(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)-17/24*a^3*cot(d*x+c)*csc(d*x+c)^2/d/(a+a*sin(d
*x+c))^(1/2)-1/4*a^2*cot(d*x+c)*csc(d*x+c)^3*(a+a*sin(d*x+c))^(1/2)/d

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {2841, 3059, 2851, 2852, 212} \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {163 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a \sin (c+d x)+a}}\right )}{64 d}-\frac {163 a^3 \cot (c+d x)}{64 d \sqrt {a \sin (c+d x)+a}}-\frac {17 a^3 \cot (c+d x) \csc ^2(c+d x)}{24 d \sqrt {a \sin (c+d x)+a}}-\frac {163 a^3 \cot (c+d x) \csc (c+d x)}{96 d \sqrt {a \sin (c+d x)+a}}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x) \sqrt {a \sin (c+d x)+a}}{4 d} \]

[In]

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

[Out]

(-163*a^(5/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/Sqrt[a + a*Sin[c + d*x]]])/(64*d) - (163*a^3*Cot[c + d*x])/(64*d*
Sqrt[a + a*Sin[c + d*x]]) - (163*a^3*Cot[c + d*x]*Csc[c + d*x])/(96*d*Sqrt[a + a*Sin[c + d*x]]) - (17*a^3*Cot[
c + d*x]*Csc[c + d*x]^2)/(24*d*Sqrt[a + a*Sin[c + d*x]]) - (a^2*Cot[c + d*x]*Csc[c + d*x]^3*Sqrt[a + a*Sin[c +
d*x]])/(4*d)

Rule 212

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

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c
+ a*d))), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1
)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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] && GtQ[m, 1] && LtQ[n, -1
] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2851

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp
[(b*c - a*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 - d^2)*Sqrt[a + b*Sin[e + f*x]])), x]
+ Dist[(2*n + 3)*((b*c - a*d)/(2*b*(n + 1)*(c^2 - d^2))), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n
+ 1), 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] &
& LtQ[n, -1] && NeQ[2*n + 3, 0] && IntegerQ[2*n]

Rule 2852

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]

Rule 3059

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.
) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*(B*c - A*d)*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n
+ 1)*(b*c + a*d)*Sqrt[a + b*Sin[e + f*x]])), x] + Dist[(A*b*d*(2*n + 3) - B*(b*c - 2*a*d*(n + 1)))/(2*d*(n +
1)*(b*c + a*d)), Int[Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(n + 1), 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] && LtQ[n, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {a^2 \cot (c+d x) \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {1}{4} a \int \csc ^4(c+d x) \left (-\frac {17 a}{2}-\frac {13}{2} a \sin (c+d x)\right ) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {17 a^3 \cot (c+d x) \csc ^2(c+d x)}{24 d \sqrt {a+a \sin (c+d x)}}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {1}{48} \left (163 a^2\right ) \int \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {163 a^3 \cot (c+d x) \csc (c+d x)}{96 d \sqrt {a+a \sin (c+d x)}}-\frac {17 a^3 \cot (c+d x) \csc ^2(c+d x)}{24 d \sqrt {a+a \sin (c+d x)}}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {1}{64} \left (163 a^2\right ) \int \csc ^2(c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {163 a^3 \cot (c+d x)}{64 d \sqrt {a+a \sin (c+d x)}}-\frac {163 a^3 \cot (c+d x) \csc (c+d x)}{96 d \sqrt {a+a \sin (c+d x)}}-\frac {17 a^3 \cot (c+d x) \csc ^2(c+d x)}{24 d \sqrt {a+a \sin (c+d x)}}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}+\frac {1}{128} \left (163 a^2\right ) \int \csc (c+d x) \sqrt {a+a \sin (c+d x)} \, dx \\ & = -\frac {163 a^3 \cot (c+d x)}{64 d \sqrt {a+a \sin (c+d x)}}-\frac {163 a^3 \cot (c+d x) \csc (c+d x)}{96 d \sqrt {a+a \sin (c+d x)}}-\frac {17 a^3 \cot (c+d x) \csc ^2(c+d x)}{24 d \sqrt {a+a \sin (c+d x)}}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d}-\frac {\left (163 a^3\right ) \text {Subst}\left (\int \frac {1}{a-x^2} \, dx,x,\frac {a \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{64 d} \\ & = -\frac {163 a^{5/2} \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {a+a \sin (c+d x)}}\right )}{64 d}-\frac {163 a^3 \cot (c+d x)}{64 d \sqrt {a+a \sin (c+d x)}}-\frac {163 a^3 \cot (c+d x) \csc (c+d x)}{96 d \sqrt {a+a \sin (c+d x)}}-\frac {17 a^3 \cot (c+d x) \csc ^2(c+d x)}{24 d \sqrt {a+a \sin (c+d x)}}-\frac {a^2 \cot (c+d x) \csc ^3(c+d x) \sqrt {a+a \sin (c+d x)}}{4 d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(370\) vs. \(2(182)=364\).

Time = 6.09 (sec) , antiderivative size = 370, normalized size of antiderivative = 2.03 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {a^2 \csc ^{13}\left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sin (c+d x))} \left (-1030 \cos \left (\frac {1}{2} (c+d x)\right )+3102 \cos \left (\frac {3}{2} (c+d x)\right )-326 \cos \left (\frac {5}{2} (c+d x)\right )-978 \cos \left (\frac {7}{2} (c+d x)\right )+1467 \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-1956 \cos (2 (c+d x)) \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+489 \cos (4 (c+d x)) \log \left (1+\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-1467 \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+1956 \cos (2 (c+d x)) \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-489 \cos (4 (c+d x)) \log \left (1-\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+1030 \sin \left (\frac {1}{2} (c+d x)\right )+3102 \sin \left (\frac {3}{2} (c+d x)\right )+326 \sin \left (\frac {5}{2} (c+d x)\right )-978 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{192 d \left (1+\cot \left (\frac {1}{2} (c+d x)\right )\right ) \left (\csc ^2\left (\frac {1}{4} (c+d x)\right )-\sec ^2\left (\frac {1}{4} (c+d x)\right )\right )^4} \]

[In]

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

[Out]

-1/192*(a^2*Csc[(c + d*x)/2]^13*Sqrt[a*(1 + Sin[c + d*x])]*(-1030*Cos[(c + d*x)/2] + 3102*Cos[(3*(c + d*x))/2]
- 326*Cos[(5*(c + d*x))/2] - 978*Cos[(7*(c + d*x))/2] + 1467*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 1
956*Cos[2*(c + d*x)]*Log[1 + Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 489*Cos[4*(c + d*x)]*Log[1 + Cos[(c + d*x)
/2] - Sin[(c + d*x)/2]] - 1467*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 1956*Cos[2*(c + d*x)]*Log[1 - Co
s[(c + d*x)/2] + Sin[(c + d*x)/2]] - 489*Cos[4*(c + d*x)]*Log[1 - Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 1030*
Sin[(c + d*x)/2] + 3102*Sin[(3*(c + d*x))/2] + 326*Sin[(5*(c + d*x))/2] - 978*Sin[(7*(c + d*x))/2]))/(d*(1 + C
ot[(c + d*x)/2])*(Csc[(c + d*x)/4]^2 - Sec[(c + d*x)/4]^2)^4)

Maple [A] (verified)

Time = 75.63 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.89

method	result	size
default	\(-\frac {\left (1+\sin \left (d x +c \right )\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, \left (1047 \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}\, a^{\frac {11}{2}}-2303 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {3}{2}} a^{\frac {9}{2}}+1793 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {5}{2}} a^{\frac {7}{2}}-489 \left (-a \left (\sin \left (d x +c \right )-1\right )\right )^{\frac {7}{2}} a^{\frac {5}{2}}+489 \,\operatorname {arctanh}\left (\frac {\sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{\sqrt {a}}\right ) a^{6} \left (\sin ^{4}\left (d x +c \right )\right )\right )}{192 a^{\frac {7}{2}} \sin \left (d x +c \right )^{4} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(162\)

[In]

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

[Out]

-1/192*(1+sin(d*x+c))*(-a*(sin(d*x+c)-1))^(1/2)*(1047*(-a*(sin(d*x+c)-1))^(1/2)*a^(11/2)-2303*(-a*(sin(d*x+c)-
1))^(3/2)*a^(9/2)+1793*(-a*(sin(d*x+c)-1))^(5/2)*a^(7/2)-489*(-a*(sin(d*x+c)-1))^(7/2)*a^(5/2)+489*arctanh((-a
*(sin(d*x+c)-1))^(1/2)/a^(1/2))*a^6*sin(d*x+c)^4)/a^(7/2)/sin(d*x+c)^4/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 473 vs. \(2 (158) = 316\).

Time = 0.31 (sec) , antiderivative size = 473, normalized size of antiderivative = 2.60 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\frac {489 \, {\left (a^{2} \cos \left (d x + c\right )^{5} + a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{3} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right ) + a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - 4 \, {\left (\cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right ) + 3\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 3\right )} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} - 9 \, a \cos \left (d x + c\right ) + {\left (a \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) - a\right )} \sin \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 1}\right ) + 4 \, {\left (489 \, a^{2} \cos \left (d x + c\right )^{4} + 326 \, a^{2} \cos \left (d x + c\right )^{3} - 836 \, a^{2} \cos \left (d x + c\right )^{2} - 374 \, a^{2} \cos \left (d x + c\right ) + 299 \, a^{2} + {\left (489 \, a^{2} \cos \left (d x + c\right )^{3} + 163 \, a^{2} \cos \left (d x + c\right )^{2} - 673 \, a^{2} \cos \left (d x + c\right ) - 299 \, a^{2}\right )} \sin \left (d x + c\right )\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{768 \, {\left (d \cos \left (d x + c\right )^{5} + d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{3} - 2 \, d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )} \sin \left (d x + c\right ) + d\right )}} \]

[In]

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

[Out]

1/768*(489*(a^2*cos(d*x + c)^5 + a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^3 - 2*a^2*cos(d*x + c)^2 + a^2*cos(d*
x + c) + a^2 + (a^2*cos(d*x + c)^4 - 2*a^2*cos(d*x + c)^2 + a^2)*sin(d*x + c))*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*cos(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*(489*a^2*cos(d*x + c)^4 +
326*a^2*cos(d*x + c)^3 - 836*a^2*cos(d*x + c)^2 - 374*a^2*cos(d*x + c) + 299*a^2 + (489*a^2*cos(d*x + c)^3 + 1
63*a^2*cos(d*x + c)^2 - 673*a^2*cos(d*x + c) - 299*a^2)*sin(d*x + c))*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c
)^5 + d*cos(d*x + c)^4 - 2*d*cos(d*x + c)^3 - 2*d*cos(d*x + c)^2 + d*cos(d*x + c) + (d*cos(d*x + c)^4 - 2*d*co
s(d*x + c)^2 + d)*sin(d*x + c) + d)

Sympy [F(-1)]

Timed out. \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int { {\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \csc \left (d x + c\right )^{5} \,d x } \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.62 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.25 \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=-\frac {\sqrt {2} {\left (489 \, \sqrt {2} a^{2} \log \left (\frac {{\left | -2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}{{\left | 2 \, \sqrt {2} + 4 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) \right |}}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) + \frac {4 \, {\left (3912 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 7172 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4606 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1047 \, a^{2} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (2 \, \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}\right )} \sqrt {a}}{768 \, d} \]

[In]

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

[Out]

-1/768*sqrt(2)*(489*sqrt(2)*a^2*log(abs(-2*sqrt(2) + 4*sin(-1/4*pi + 1/2*d*x + 1/2*c))/abs(2*sqrt(2) + 4*sin(-
1/4*pi + 1/2*d*x + 1/2*c)))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) + 4*(3912*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*
c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^7 - 7172*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/
2*c)^5 + 4606*a^2*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 1047*a^2*sgn(cos(-1/4
*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(2*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^4)*sqrt(a)/d

Mupad [F(-1)]

Timed out. \[ \int \csc ^5(c+d x) (a+a \sin (c+d x))^{5/2} \, dx=\int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\sin \left (c+d\,x\right )}^5} \,d x \]

[In]

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

[Out]

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