3.571 \(\int (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=285 \[ \frac {5 a^{3/2} (c-15 d) (c+d)^3 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{64 d^{3/2} f}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}+\frac {a^2 (c-15 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d f \sqrt {a \sin (e+f x)+a}}+\frac {5 a^2 (c-15 d) (c+d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d f \sqrt {a \sin (e+f x)+a}}+\frac {5 a^2 (c-15 d) (c+d)^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{64 d f \sqrt {a \sin (e+f x)+a}} \]

[Out]

5/64*a^(3/2)*(c-15*d)*(c+d)^3*arctan(cos(f*x+e)*a^(1/2)*d^(1/2)/(a+a*sin(f*x+e))^(1/2)/(c+d*sin(f*x+e))^(1/2))
/d^(3/2)/f+5/96*a^2*(c-15*d)*(c+d)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/d/f/(a+a*sin(f*x+e))^(1/2)+1/24*a^2*(c-15
*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(5/2)/d/f/(a+a*sin(f*x+e))^(1/2)-1/4*a^2*cos(f*x+e)*(c+d*sin(f*x+e))^(7/2)/d/f
/(a+a*sin(f*x+e))^(1/2)+5/64*a^2*(c-15*d)*(c+d)^2*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/d/f/(a+a*sin(f*x+e))^(1/2)

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Rubi [A]  time = 0.57, antiderivative size = 285, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2763, 21, 2770, 2775, 205} \[ \frac {5 a^{3/2} (c-15 d) (c+d)^3 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}}\right )}{64 d^{3/2} f}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a \sin (e+f x)+a}}+\frac {a^2 (c-15 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d f \sqrt {a \sin (e+f x)+a}}+\frac {5 a^2 (c-15 d) (c+d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d f \sqrt {a \sin (e+f x)+a}}+\frac {5 a^2 (c-15 d) (c+d)^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{64 d f \sqrt {a \sin (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])^(5/2),x]

[Out]

(5*a^(3/2)*(c - 15*d)*(c + d)^3*ArcTan[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[a + a*Sin[e + f*x]]*Sqrt[c + d*Sin
[e + f*x]])])/(64*d^(3/2)*f) + (5*a^2*(c - 15*d)*(c + d)^2*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(64*d*f*Sqrt
[a + a*Sin[e + f*x]]) + (5*a^2*(c - 15*d)*(c + d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(3/2))/(96*d*f*Sqrt[a + a*
Sin[e + f*x]]) + (a^2*(c - 15*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2))/(24*d*f*Sqrt[a + a*Sin[e + f*x]]) -
(a^2*Cos[e + f*x]*(c + d*Sin[e + f*x])^(7/2))/(4*d*f*Sqrt[a + a*Sin[e + f*x]])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2763

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n)), x] + Dist[1/(d*
(m + n)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^n*Simp[a*b*c*(m - 2) + b^2*d*(n + 1) + a^2*d*(
m + n) - b*(b*c*(m - 1) - a*d*(3*m + 2*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] && GtQ[m, 1] &&  !LtQ[n, -1] && (IntegersQ[2*m, 2*
n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2770

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

Rule 2775

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*b)/f, Subst[Int[1/(b + d*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*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 (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^{5/2} \, dx &=-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a+a \sin (e+f x)}}+\frac {\int \frac {\left (-\frac {1}{2} a^2 (c-15 d)-\frac {1}{2} a^2 (c-15 d) \sin (e+f x)\right ) (c+d \sin (e+f x))^{5/2}}{\sqrt {a+a \sin (e+f x)}} \, dx}{4 d}\\ &=-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a+a \sin (e+f x)}}-\frac {(a (c-15 d)) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2} \, dx}{8 d}\\ &=\frac {a^2 (c-15 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a+a \sin (e+f x)}}-\frac {(5 a (c-15 d) (c+d)) \int \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2} \, dx}{48 d}\\ &=\frac {5 a^2 (c-15 d) (c+d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-15 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a+a \sin (e+f x)}}-\frac {\left (5 a (c-15 d) (c+d)^2\right ) \int \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)} \, dx}{64 d}\\ &=\frac {5 a^2 (c-15 d) (c+d)^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{64 d f \sqrt {a+a \sin (e+f x)}}+\frac {5 a^2 (c-15 d) (c+d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-15 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a+a \sin (e+f x)}}-\frac {\left (5 a (c-15 d) (c+d)^3\right ) \int \frac {\sqrt {a+a \sin (e+f x)}}{\sqrt {c+d \sin (e+f x)}} \, dx}{128 d}\\ &=\frac {5 a^2 (c-15 d) (c+d)^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{64 d f \sqrt {a+a \sin (e+f x)}}+\frac {5 a^2 (c-15 d) (c+d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-15 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a+a \sin (e+f x)}}+\frac {\left (5 a^2 (c-15 d) (c+d)^3\right ) \operatorname {Subst}\left (\int \frac {1}{a+d x^2} \, dx,x,\frac {a \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{64 d f}\\ &=\frac {5 a^{3/2} (c-15 d) (c+d)^3 \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {d} \cos (e+f x)}{\sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\right )}{64 d^{3/2} f}+\frac {5 a^2 (c-15 d) (c+d)^2 \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{64 d f \sqrt {a+a \sin (e+f x)}}+\frac {5 a^2 (c-15 d) (c+d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d f \sqrt {a+a \sin (e+f x)}}+\frac {a^2 (c-15 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d f \sqrt {a+a \sin (e+f x)}}-\frac {a^2 \cos (e+f x) (c+d \sin (e+f x))^{7/2}}{4 d f \sqrt {a+a \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 1.37, size = 318, normalized size = 1.12 \[ \frac {(a (\sin (e+f x)+1))^{3/2} \left (-\frac {2 \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {c+d \sin (e+f x)} \left (15 c^3+2 d \left (59 c^2+190 c d+93 d^2\right ) \sin (e+f x)+455 c^2 d-4 d^2 (17 c+15 d) \cos (2 (e+f x))+653 c d^2-12 d^3 \sin (3 (e+f x))+285 d^3\right )}{3 d}-\frac {5 (c-15 d) (c+d)^3 \left (2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \sin \left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{\sqrt {c+d \sin (e+f x)}}\right )-\log \left (\sqrt {c+d \sin (e+f x)}+\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e+2 f x-\pi )\right )\right )+\tanh ^{-1}\left (\frac {\sqrt {2} \sqrt {d} \cos \left (\frac {1}{4} (2 e+2 f x-\pi )\right )}{\sqrt {c+d \sin (e+f x)}}\right )\right )}{d^{3/2}}\right )}{128 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])^(5/2),x]

[Out]

((a*(1 + Sin[e + f*x]))^(3/2)*((-5*(c - 15*d)*(c + d)^3*(2*ArcTan[(Sqrt[2]*Sqrt[d]*Sin[(2*e - Pi + 2*f*x)/4])/
Sqrt[c + d*Sin[e + f*x]]] + ArcTanh[(Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4])/Sqrt[c + d*Sin[e + f*x]]] - Lo
g[Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]]))/d^(3/2) - (2*(Cos[(e + f*x)/2] - Sin
[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]*(15*c^3 + 455*c^2*d + 653*c*d^2 + 285*d^3 - 4*d^2*(17*c + 15*d)*Cos[2*
(e + f*x)] + 2*d*(59*c^2 + 190*c*d + 93*d^2)*Sin[e + f*x] - 12*d^3*Sin[3*(e + f*x)]))/(3*d)))/(128*f*(Cos[(e +
f*x)/2] + Sin[(e + f*x)/2])^3)

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fricas [B]  time = 1.59, size = 1579, normalized size = 5.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/1536*(15*(a*c^4 - 12*a*c^3*d - 42*a*c^2*d^2 - 44*a*c*d^3 - 15*a*d^4 + (a*c^4 - 12*a*c^3*d - 42*a*c^2*d^2 -
44*a*c*d^3 - 15*a*d^4)*cos(f*x + e) + (a*c^4 - 12*a*c^3*d - 42*a*c^2*d^2 - 44*a*c*d^3 - 15*a*d^4)*sin(f*x + e
))*sqrt(-a/d)*log((128*a*d^4*cos(f*x + e)^5 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 + 128*(2*a*c
*d^3 - a*d^4)*cos(f*x + e)^4 - 32*(5*a*c^2*d^2 - 14*a*c*d^3 + 13*a*d^4)*cos(f*x + e)^3 - 32*(a*c^3*d - 2*a*c^2
*d^2 + 9*a*c*d^3 - 4*a*d^4)*cos(f*x + e)^2 - 8*(16*d^4*cos(f*x + e)^4 - c^3*d + 17*c^2*d^2 - 59*c*d^3 + 51*d^4
+ 24*(c*d^3 - d^4)*cos(f*x + e)^3 - 2*(5*c^2*d^2 - 26*c*d^3 + 33*d^4)*cos(f*x + e)^2 - (c^3*d - 7*c^2*d^2 + 3
1*c*d^3 - 25*d^4)*cos(f*x + e) + (16*d^4*cos(f*x + e)^3 + c^3*d - 17*c^2*d^2 + 59*c*d^3 - 51*d^4 - 8*(3*c*d^3
- 5*d^4)*cos(f*x + e)^2 - 2*(5*c^2*d^2 - 14*c*d^3 + 13*d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) +
a)*sqrt(d*sin(f*x + e) + c)*sqrt(-a/d) + (a*c^4 - 28*a*c^3*d + 230*a*c^2*d^2 - 476*a*c*d^3 + 289*a*d^4)*cos(f*
x + e) + (128*a*d^4*cos(f*x + e)^4 + a*c^4 + 4*a*c^3*d + 6*a*c^2*d^2 + 4*a*c*d^3 + a*d^4 - 256*(a*c*d^3 - a*d^
4)*cos(f*x + e)^3 - 32*(5*a*c^2*d^2 - 6*a*c*d^3 + 5*a*d^4)*cos(f*x + e)^2 + 32*(a*c^3*d - 7*a*c^2*d^2 + 15*a*c
*d^3 - 9*a*d^4)*cos(f*x + e))*sin(f*x + e))/(cos(f*x + e) + sin(f*x + e) + 1)) - 8*(48*a*d^3*cos(f*x + e)^4 -
15*a*c^3 - 337*a*c^2*d - 341*a*c*d^2 - 147*a*d^3 + 8*(17*a*c*d^2 + 15*a*d^3)*cos(f*x + e)^3 - 2*(59*a*c^2*d +
122*a*c*d^2 + 63*a*d^3)*cos(f*x + e)^2 - (15*a*c^3 + 455*a*c^2*d + 721*a*c*d^2 + 345*a*d^3)*cos(f*x + e) + (48
*a*d^3*cos(f*x + e)^3 + 15*a*c^3 + 337*a*c^2*d + 341*a*c*d^2 + 147*a*d^3 - 8*(17*a*c*d^2 + 9*a*d^3)*cos(f*x +
e)^2 - 2*(59*a*c^2*d + 190*a*c*d^2 + 99*a*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin
(f*x + e) + c))/(d*f*cos(f*x + e) + d*f*sin(f*x + e) + d*f), -1/768*(15*(a*c^4 - 12*a*c^3*d - 42*a*c^2*d^2 - 4
4*a*c*d^3 - 15*a*d^4 + (a*c^4 - 12*a*c^3*d - 42*a*c^2*d^2 - 44*a*c*d^3 - 15*a*d^4)*cos(f*x + e) + (a*c^4 - 12*
a*c^3*d - 42*a*c^2*d^2 - 44*a*c*d^3 - 15*a*d^4)*sin(f*x + e))*sqrt(a/d)*arctan(1/4*(8*d^2*cos(f*x + e)^2 - c^2
+ 6*c*d - 9*d^2 - 8*(c*d - d^2)*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)*sqrt(a/d)/(2*
a*d^2*cos(f*x + e)^3 - (3*a*c*d - a*d^2)*cos(f*x + e)*sin(f*x + e) - (a*c^2 - a*c*d + 2*a*d^2)*cos(f*x + e)))
- 4*(48*a*d^3*cos(f*x + e)^4 - 15*a*c^3 - 337*a*c^2*d - 341*a*c*d^2 - 147*a*d^3 + 8*(17*a*c*d^2 + 15*a*d^3)*co
s(f*x + e)^3 - 2*(59*a*c^2*d + 122*a*c*d^2 + 63*a*d^3)*cos(f*x + e)^2 - (15*a*c^3 + 455*a*c^2*d + 721*a*c*d^2
+ 345*a*d^3)*cos(f*x + e) + (48*a*d^3*cos(f*x + e)^3 + 15*a*c^3 + 337*a*c^2*d + 341*a*c*d^2 + 147*a*d^3 - 8*(1
7*a*c*d^2 + 9*a*d^3)*cos(f*x + e)^2 - 2*(59*a*c^2*d + 190*a*c*d^2 + 99*a*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt
(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c))/(d*f*cos(f*x + e) + d*f*sin(f*x + e) + d*f)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)^(5/2), x)

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int \left (a +a \sin \left (f x +e \right )\right )^{\frac {3}{2}} \left (c +d \sin \left (f x +e \right )\right )^{\frac {5}{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(5/2),x)

[Out]

int((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(5/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(3/2)*(c+d*sin(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(3/2)*(d*sin(f*x + e) + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*x))^(5/2),x)

[Out]

int((a + a*sin(e + f*x))^(3/2)*(c + d*sin(e + f*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((a+a*sin(f*x+e))**(3/2)*(c+d*sin(f*x+e))**(5/2),x)

[Out]

Timed out

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