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

Optimal. Leaf size=312 \[ -\frac{a^3 \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d^2 f \sqrt{a \sin (e+f x)+a}}-\frac{a^3 (c+d) \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{64 d^2 f \sqrt{a \sin (e+f x)+a}}-\frac{a^{5/2} (c+d)^2 \left (3 c^2-26 c d+163 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{64 d^{5/2} f}+\frac{a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d^2 f \sqrt{a \sin (e+f x)+a}}-\frac{a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f} \]

[Out]

-(a^(5/2)*(c + d)^2*(3*c^2 - 26*c*d + 163*d^2)*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^(5/2)*f) - (a^3*(c + d)*(3*c^2 - 26*c*d + 163*d^2)*Cos[e + f*x]*Sqrt[c + d*
Sin[e + f*x]])/(64*d^2*f*Sqrt[a + a*Sin[e + f*x]]) - (a^3*(3*c^2 - 26*c*d + 163*d^2)*Cos[e + f*x]*(c + d*Sin[e
+ f*x])^(3/2))/(96*d^2*f*Sqrt[a + a*Sin[e + f*x]]) + (a^3*(3*c - 17*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2
))/(24*d^2*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)
)/(4*d*f)

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Rubi [A]  time = 0.726103, antiderivative size = 312, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2763, 2981, 2770, 2775, 205} \[ -\frac{a^3 \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d^2 f \sqrt{a \sin (e+f x)+a}}-\frac{a^3 (c+d) \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{64 d^2 f \sqrt{a \sin (e+f x)+a}}-\frac{a^{5/2} (c+d)^2 \left (3 c^2-26 c d+163 d^2\right ) \tan ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{a \sin (e+f x)+a} \sqrt{c+d \sin (e+f x)}}\right )}{64 d^{5/2} f}+\frac{a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d^2 f \sqrt{a \sin (e+f x)+a}}-\frac{a^2 \cos (e+f x) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))^{5/2}}{4 d f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^(5/2)*(c + d)^2*(3*c^2 - 26*c*d + 163*d^2)*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^(5/2)*f) - (a^3*(c + d)*(3*c^2 - 26*c*d + 163*d^2)*Cos[e + f*x]*Sqrt[c + d*
Sin[e + f*x]])/(64*d^2*f*Sqrt[a + a*Sin[e + f*x]]) - (a^3*(3*c^2 - 26*c*d + 163*d^2)*Cos[e + f*x]*(c + d*Sin[e
+ f*x])^(3/2))/(96*d^2*f*Sqrt[a + a*Sin[e + f*x]]) + (a^3*(3*c - 17*d)*Cos[e + f*x]*(c + d*Sin[e + f*x])^(5/2
))/(24*d^2*f*Sqrt[a + a*Sin[e + f*x]]) - (a^2*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x])^(5/2)
)/(4*d*f)

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 2981

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

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]

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]

Rubi steps

\begin{align*} \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x))^{3/2} \, dx &=-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}{4 d f}+\frac{\int \sqrt{a+a \sin (e+f x)} \left (\frac{1}{2} a^2 (c+13 d)-\frac{1}{2} a^2 (3 c-17 d) \sin (e+f x)\right ) (c+d \sin (e+f x))^{3/2} \, dx}{4 d}\\ &=\frac{a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}{4 d f}+\frac{\left (a^2 \left (3 c^2-26 c d+163 d^2\right )\right ) \int \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2} \, dx}{48 d^2}\\ &=-\frac{a^3 \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d^2 f \sqrt{a+a \sin (e+f x)}}+\frac{a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}{4 d f}+\frac{\left (a^2 (c+d) \left (3 c^2-26 c d+163 d^2\right )\right ) \int \sqrt{a+a \sin (e+f x)} \sqrt{c+d \sin (e+f x)} \, dx}{64 d^2}\\ &=-\frac{a^3 (c+d) \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{64 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{a^3 \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d^2 f \sqrt{a+a \sin (e+f x)}}+\frac{a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}{4 d f}+\frac{\left (a^2 (c+d)^2 \left (3 c^2-26 c d+163 d^2\right )\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{\sqrt{c+d \sin (e+f x)}} \, dx}{128 d^2}\\ &=-\frac{a^3 (c+d) \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{64 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{a^3 \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d^2 f \sqrt{a+a \sin (e+f x)}}+\frac{a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}{4 d f}-\frac{\left (a^3 (c+d)^2 \left (3 c^2-26 c d+163 d^2\right )\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^2 f}\\ &=-\frac{a^{5/2} (c+d)^2 \left (3 c^2-26 c d+163 d^2\right ) \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^{5/2} f}-\frac{a^3 (c+d) \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{64 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{a^3 \left (3 c^2-26 c d+163 d^2\right ) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{96 d^2 f \sqrt{a+a \sin (e+f x)}}+\frac{a^3 (3 c-17 d) \cos (e+f x) (c+d \sin (e+f x))^{5/2}}{24 d^2 f \sqrt{a+a \sin (e+f x)}}-\frac{a^2 \cos (e+f x) \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^{5/2}}{4 d f}\\ \end{align*}

Mathematica [A]  time = 1.76093, size = 327, normalized size = 1.05 \[ \frac{(a (\sin (e+f x)+1))^{5/2} \left (\frac{2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \sqrt{c+d \sin (e+f x)} \left (-2 d \left (3 c^2+158 c d+181 d^2\right ) \sin (e+f x)-63 c^2 d+9 c^3+4 d^2 (9 c+23 d) \cos (2 (e+f x))-773 c d^2+12 d^3 \sin (3 (e+f x))-581 d^3\right )}{3 d^2}+\frac{\left (3 c^2-26 c d+163 d^2\right ) (c+d)^2 \left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \sin \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )-\log \left (\sqrt{c+d \sin (e+f x)}+\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )\right )+\tanh ^{-1}\left (\frac{\sqrt{2} \sqrt{d} \cos \left (\frac{1}{4} (2 e+2 f x-\pi )\right )}{\sqrt{c+d \sin (e+f x)}}\right )\right )}{d^{5/2}}\right )}{128 f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^(5/2)*(((c + d)^2*(3*c^2 - 26*c*d + 163*d^2)*(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]]] - Log[Sqrt[2]*Sqrt[d]*Cos[(2*e - Pi + 2*f*x)/4] + Sqrt[c + d*Sin[e + f*x]]]))/d^(5/2) + (2*(Cos[(e +
f*x)/2] - Sin[(e + f*x)/2])*Sqrt[c + d*Sin[e + f*x]]*(9*c^3 - 63*c^2*d - 773*c*d^2 - 581*d^3 + 4*d^2*(9*c + 23
*d)*Cos[2*(e + f*x)] - 2*d*(3*c^2 + 158*c*d + 181*d^2)*Sin[e + f*x] + 12*d^3*Sin[3*(e + f*x)]))/(3*d^2)))/(128
*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5)

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Maple [F]  time = 180., size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sin \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}} \left ( c+d\sin \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sin \left (f x + e\right ) + a\right )}^{\frac{5}{2}}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 14.7684, size = 4084, normalized size = 13.09 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/1536*(3*(3*a^2*c^4 - 20*a^2*c^3*d + 114*a^2*c^2*d^2 + 300*a^2*c*d^3 + 163*a^2*d^4 + (3*a^2*c^4 - 20*a^2*c^3
*d + 114*a^2*c^2*d^2 + 300*a^2*c*d^3 + 163*a^2*d^4)*cos(f*x + e) + (3*a^2*c^4 - 20*a^2*c^3*d + 114*a^2*c^2*d^2
+ 300*a^2*c*d^3 + 163*a^2*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 + 31*c*d^3 - 25*d^4)*cos(f*x + e) + (16*d^4*cos(f*x + e)^3 + c^3*d - 1
7*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^2*d^3*cos(f*x + e)^4 + 9*a^2*c^3 - 57*a^2*c^2*d - 493*a^2*c*d^2 - 299*a^2*d^3 + 8*(9*a^2
*c*d^2 + 23*a^2*d^3)*cos(f*x + e)^3 - 2*(3*a^2*c^2*d + 122*a^2*c*d^2 + 119*a^2*d^3)*cos(f*x + e)^2 + (9*a^2*c^
3 - 63*a^2*c^2*d - 809*a^2*c*d^2 - 673*a^2*d^3)*cos(f*x + e) + (48*a^2*d^3*cos(f*x + e)^3 - 9*a^2*c^3 + 57*a^2
*c^2*d + 493*a^2*c*d^2 + 299*a^2*d^3 - 8*(9*a^2*c*d^2 + 17*a^2*d^3)*cos(f*x + e)^2 - 2*(3*a^2*c^2*d + 158*a^2*
c*d^2 + 187*a^2*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^2*f*cos
(f*x + e) + d^2*f*sin(f*x + e) + d^2*f), 1/768*(3*(3*a^2*c^4 - 20*a^2*c^3*d + 114*a^2*c^2*d^2 + 300*a^2*c*d^3
+ 163*a^2*d^4 + (3*a^2*c^4 - 20*a^2*c^3*d + 114*a^2*c^2*d^2 + 300*a^2*c*d^3 + 163*a^2*d^4)*cos(f*x + e) + (3*a
^2*c^4 - 20*a^2*c^3*d + 114*a^2*c^2*d^2 + 300*a^2*c*d^3 + 163*a^2*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^2*d^3*cos(f*x + e)^4 + 9*a^2*c^3 - 57*a^2*c^2*d - 493*a^2*c*d^2 - 299*a^2*d^3
+ 8*(9*a^2*c*d^2 + 23*a^2*d^3)*cos(f*x + e)^3 - 2*(3*a^2*c^2*d + 122*a^2*c*d^2 + 119*a^2*d^3)*cos(f*x + e)^2
+ (9*a^2*c^3 - 63*a^2*c^2*d - 809*a^2*c*d^2 - 673*a^2*d^3)*cos(f*x + e) + (48*a^2*d^3*cos(f*x + e)^3 - 9*a^2*c
^3 + 57*a^2*c^2*d + 493*a^2*c*d^2 + 299*a^2*d^3 - 8*(9*a^2*c*d^2 + 17*a^2*d^3)*cos(f*x + e)^2 - 2*(3*a^2*c^2*d
+ 158*a^2*c*d^2 + 187*a^2*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^2*f*cos(f*x + e) + d^2*f*sin(f*x + e) + d^2*f)]

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

[Out]

Timed out

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError