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

Optimal. Leaf size=183 \[ -\frac {163 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {95 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{48 a^3 d}+\frac {197 \cos (c+d x)}{24 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}+\frac {17 \sin ^2(c+d x) \cos (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}} \]

[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

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Rubi [A]  time = 0.39, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2765, 2977, 2968, 3023, 2751, 2649, 206} \[ -\frac {95 \cos (c+d x) \sqrt {a \sin (c+d x)+a}}{48 a^3 d}+\frac {197 \cos (c+d x)}{24 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {163 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {\sin ^3(c+d x) \cos (c+d x)}{4 d (a \sin (c+d x)+a)^{5/2}}+\frac {17 \sin ^2(c+d x) \cos (c+d x)}{16 a d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[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 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 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 2751

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*Sin
[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 2765

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 2968

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 2977

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 3023

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*} \int \frac {\sin ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=\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 \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 {163 \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 {\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*}

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Mathematica [C]  time = 0.48, size = 197, normalized size = 1.08 \[ \frac {\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \left (-279 \sin \left (\frac {1}{2} (c+d x)\right )+399 \sin \left (\frac {3}{2} (c+d x)\right )+88 \sin \left (\frac {5}{2} (c+d x)\right )+8 \sin \left (\frac {7}{2} (c+d x)\right )+279 \cos \left (\frac {1}{2} (c+d x)\right )+399 \cos \left (\frac {3}{2} (c+d x)\right )-88 \cos \left (\frac {5}{2} (c+d x)\right )+8 \cos \left (\frac {7}{2} (c+d x)\right )+(978+978 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )\right )}{96 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[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))

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fricas [B]  time = 0.53, size = 360, normalized size = 1.97 \[ \frac {489 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right ) - 4\right )} \sin \left (d x + c\right ) - 2 \, \cos \left (d x + c\right ) - 4\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) - 4 \, {\left (32 \, \cos \left (d x + c\right )^{4} - 160 \, \cos \left (d x + c\right )^{3} + 279 \, \cos \left (d x + c\right )^{2} + {\left (32 \, \cos \left (d x + c\right )^{3} + 192 \, \cos \left (d x + c\right )^{2} + 471 \, \cos \left (d x + c\right ) + 12\right )} \sin \left (d x + c\right ) + 459 \, \cos \left (d x + c\right ) - 12\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{192 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d + {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) - 4 \, a^{3} d\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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giac [B]  time = 8.72, size = 584, normalized size = 3.19 \[ -\frac {\frac {32 \, {\left ({\left ({\left (\frac {7 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} - \frac {9}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {9}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {7}{a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}\right )}}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a\right )}^{\frac {3}{2}}} - \frac {489 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} + \sqrt {a}\right )}}{2 \, \sqrt {-a}}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )} + \frac {6 \, {\left (67 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{7} + 341 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{6} \sqrt {a} + 233 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{5} a - 325 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{4} a^{\frac {3}{2}} + 33 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{3} a^{2} + 159 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} a^{\frac {5}{2}} - 133 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} a^{3} + 25 \, a^{\frac {7}{2}}\right )}}{{\left ({\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 2 \, {\left (\sqrt {a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )} \sqrt {a} - a\right )}^{4} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*(32*(((7*tan(1/2*d*x + 1/2*c)/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)) - 9/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)))*t
an(1/2*d*x + 1/2*c) + 9/(a*sgn(tan(1/2*d*x + 1/2*c) + 1)))*tan(1/2*d*x + 1/2*c) - 7/(a*sgn(tan(1/2*d*x + 1/2*c
) + 1)))/(a*tan(1/2*d*x + 1/2*c)^2 + a)^(3/2) - 489*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c)
- sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) + sqrt(a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 6*(67
*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^7 + 341*(sqrt(a)*tan(1/2*d*x + 1/2*c) - s
qrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*sqrt(a) + 233*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)
^2 + a))^5*a - 325*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*a^(3/2) + 33*(sqrt(a)
*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a^2 + 159*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a
*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(5/2) - 133*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))*a^3 + 25*a^(7/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 2*(sqrt(a)*tan
(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^4*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.97, size = 269, normalized size = 1.47 \[ \frac {\left (\sin \left (d x +c \right ) \left (128 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}+768 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}-978 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}\right )+\left (-64 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}-384 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}+489 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-46 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {a}+1092 \sqrt {a -a \sin \left (d x +c \right )}\, a^{\frac {3}{2}}-978 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}\right ) \sqrt {-a \left (\sin \left (d x +c \right )-1\right )}}{96 a^{\frac {9}{2}} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96/a^(9/2)*(sin(d*x+c)*(128*(a-a*sin(d*x+c))^(3/2)*a^(1/2)+768*(a-a*sin(d*x+c))^(1/2)*a^(3/2)-978*2^(1/2)*ar
ctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2)+(-64*(a-a*sin(d*x+c))^(3/2)*a^(1/2)-384*(a-a*sin(d*x+c)
)^(1/2)*a^(3/2)+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)^2-46*(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)*arctanh(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)/(1+sin(d*x+c))/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (d x + c\right )^{4}}{{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (c+d\,x\right )}^4}{{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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