\(\int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\) [77]

Optimal result

Integrand size = 23, antiderivative size = 183 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {163 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\cos (c+d x) \sin ^3(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {17 \cos (c+d x) \sin ^2(c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {197 \cos (c+d x)}{24 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {95 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{48 a^3 d} \]

[Out]

1/4*cos(d*x+c)*sin(d*x+c)^3/d/(a+a*sin(d*x+c))^(5/2)+17/16*cos(d*x+c)*sin(d*x+c)^2/a/d/(a+a*sin(d*x+c))^(3/2)-
163/32*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+197/24*cos(d*x+c)/a^2/
d/(a+a*sin(d*x+c))^(1/2)-95/48*cos(d*x+c)*(a+a*sin(d*x+c))^(1/2)/a^3/d

Rubi [A] (verified)

Time = 0.55 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2844, 3056, 3047, 3102, 2830, 2728, 212} \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=-\frac {163 \text {arctanh}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {95 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{48 a^3 d}+\frac {197 \cos (c+d x)}{24 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}+\frac {17 \sin ^2(c+d x) \cos (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}} \]

[In]

Int[Sin[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-163*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + (Cos[c + d*
x]*Sin[c + d*x]^3)/(4*d*(a + a*Sin[c + d*x])^(5/2)) + (17*Cos[c + d*x]*Sin[c + d*x]^2)/(16*a*d*(a + a*Sin[c +
d*x])^(3/2)) + (197*Cos[c + d*x])/(24*a^2*d*Sqrt[a + a*Sin[c + d*x]]) - (95*Cos[c + d*x]*Sqrt[a + a*Sin[c + d*
x]])/(48*a^3*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 2728

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 2830

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

Rule 2844

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

Rule 3047

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

Rule 3056

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

Rule 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(
b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {\cos (c+d x) \sin ^3(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}-\frac {\int \frac {\sin ^2(c+d x) \left (3 a-\frac {11}{2} a \sin (c+d x)\right )}{(a+a \sin (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = \frac {\cos (c+d x) \sin ^3(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {17 \cos (c+d x) \sin ^2(c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {\int \frac {\sin (c+d x) \left (17 a^2-\frac {95}{4} a^2 \sin (c+d x)\right )}{\sqrt {a+a \sin (c+d x)}} \, dx}{8 a^4} \\ & = \frac {\cos (c+d x) \sin ^3(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {17 \cos (c+d x) \sin ^2(c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {\int \frac {17 a^2 \sin (c+d x)-\frac {95}{4} a^2 \sin ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{8 a^4} \\ & = \frac {\cos (c+d x) \sin ^3(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {17 \cos (c+d x) \sin ^2(c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}-\frac {95 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{48 a^3 d}-\frac {\int \frac {-\frac {95 a^3}{8}+\frac {197}{4} a^3 \sin (c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{12 a^5} \\ & = \frac {\cos (c+d x) \sin ^3(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {17 \cos (c+d x) \sin ^2(c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {197 \cos (c+d x)}{24 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {95 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{48 a^3 d}+\frac {163 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{32 a^2} \\ & = \frac {\cos (c+d x) \sin ^3(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {17 \cos (c+d x) \sin ^2(c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {197 \cos (c+d x)}{24 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {95 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{48 a^3 d}-\frac {163 \text {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 {163 \text {arctanh}\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 {\cos (c+d x) \sin ^3(c+d x)}{4 d (a+a \sin (c+d x))^{5/2}}+\frac {17 \cos (c+d x) \sin ^2(c+d x)}{16 a d (a+a \sin (c+d x))^{3/2}}+\frac {197 \cos (c+d x)}{24 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {95 \cos (c+d x) \sqrt {a+a \sin (c+d x)}}{48 a^3 d} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.68 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.08 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (279 \cos \left (\frac {1}{2} (c+d x)\right )+399 \cos \left (\frac {3}{2} (c+d x)\right )-88 \cos \left (\frac {5}{2} (c+d x)\right )+8 \cos \left (\frac {7}{2} (c+d x)\right )-279 \sin \left (\frac {1}{2} (c+d x)\right )+(978+978 i) (-1)^{3/4} \text {arctanh}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (-1+\tan \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^4+399 \sin \left (\frac {3}{2} (c+d x)\right )+88 \sin \left (\frac {5}{2} (c+d x)\right )+8 \sin \left (\frac {7}{2} (c+d x)\right )\right )}{96 d (a (1+\sin (c+d x)))^{5/2}} \]

[In]

Integrate[Sin[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

((Cos[(c + d*x)/2] + Sin[(c + d*x)/2])*(279*Cos[(c + d*x)/2] + 399*Cos[(3*(c + d*x))/2] - 88*Cos[(5*(c + d*x))
/2] + 8*Cos[(7*(c + d*x))/2] - 279*Sin[(c + d*x)/2] + (978 + 978*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 + 399*Sin[(3*(c + d*x))/2] + 88*Sin[(5*(c + d
*x))/2] + 8*Sin[(7*(c + d*x))/2]))/(96*d*(a*(1 + Sin[c + d*x]))^(5/2))

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 293, normalized size of antiderivative = 1.60

method	result	size
default	\(-\frac {\left (64 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}\, \left (\cos ^{2}\left (d x +c \right )\right )+384 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right )-489 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \left (\cos ^{2}\left (d x +c \right )\right )-128 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}\, \sin \left (d x +c \right )-768 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}} \sin \left (d x +c \right )+978 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2} \sin \left (d x +c \right )+46 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}-1092 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}+978 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{96 a^{\frac {9}{2}} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d}\)	\(293\)

[In]

int(sin(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/96*(64*(a-a*sin(d*x+c))^(3/2)*a^(1/2)*cos(d*x+c)^2+384*(a-a*sin(d*x+c))^(1/2)*a^(3/2)*cos(d*x+c)^2-489*2^(1
/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2*cos(d*x+c)^2-128*(a-a*sin(d*x+c))^(3/2)*a^(1/2)*si
n(d*x+c)-768*(a-a*sin(d*x+c))^(1/2)*a^(3/2)*sin(d*x+c)+978*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/
a^(1/2))*a^2*sin(d*x+c)+46*(a-a*sin(d*x+c))^(3/2)*a^(1/2)-1092*(a-a*sin(d*x+c))^(1/2)*a^(3/2)+978*2^(1/2)*arct
anh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2)*(-a*(sin(d*x+c)-1))^(1/2)/a^(9/2)/(1+sin(d*x+c))/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. 360 vs. \(2 (156) = 312\).

Time = 0.29 (sec) , antiderivative size = 360, normalized size of antiderivative = 1.97 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {489 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (32 \, \cos \left (d x + c\right )^{4} - 160 \, \cos \left (d x + c\right )^{3} + 279 \, \cos \left (d x + c\right )^{2} + {\left (32 \, \cos \left (d x + c\right )^{3} + 192 \, \cos \left (d x + c\right )^{2} + 471 \, \cos \left (d x + c\right ) + 12\right )} \sin \left (d x + c\right ) + 459 \, \cos \left (d x + c\right ) - 12\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

[In]

integrate(sin(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/192*(489*sqrt(2)*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + (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)) - 4*(32*cos(d*x + c)^4 - 160*cos(d*x + c)^3 + 279*cos(d*x + c)^2 +
(32*cos(d*x + c)^3 + 192*cos(d*x + c)^2 + 471*cos(d*x + c) + 12)*sin(d*x + c) + 459*cos(d*x + c) - 12)*sqrt(a
*sin(d*x + c) + a))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d + (a^3*d*c
os(d*x + c)^2 - 2*a^3*d*cos(d*x + c) - 4*a^3*d)*sin(d*x + c))

Sympy [F]

\[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {\sin ^{4}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(sin(d*x+c)**4/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Integral(sin(c + d*x)**4/(a*(sin(c + d*x) + 1))**(5/2), x)

Maxima [F]

\[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int { \frac {\sin \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(sin(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(d*x + c)^4/(a*sin(d*x + c) + a)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 0.59 (sec) , antiderivative size = 143, normalized size of antiderivative = 0.78 \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\frac {\frac {3 \, \sqrt {2} {\left (29 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, \sqrt {a} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a^{3} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {128 \, \sqrt {2} {\left (a^{\frac {13}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{\frac {13}{2}} \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a^{9} \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{96 \, d} \]

[In]

integrate(sin(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/96*(3*sqrt(2)*(29*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 27*sqrt(a)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/((si
n(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2*a^3*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))) - 128*sqrt(2)*(a^(13/2)*sin(-1/
4*pi + 1/2*d*x + 1/2*c)^3 + 3*a^(13/2)*sin(-1/4*pi + 1/2*d*x + 1/2*c))/(a^9*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)
)))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^4}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

[In]

int(sin(c + d*x)^4/(a + a*sin(c + d*x))^(5/2),x)

[Out]

int(sin(c + d*x)^4/(a + a*sin(c + d*x))^(5/2), x)