3.299 \(\int \frac{(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=221 \[ \frac{a^2 \left (A d (c-5 d)+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{4 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{a^{3/2} \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{4 d^{5/2} f (c+d)^{5/2}}+\frac{a (B c-A d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2} \]

[Out]

-(a^(3/2)*(A*d*(c + 7*d) + 3*B*(c^2 + 3*c*d + 4*d^2))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt
[a + a*Sin[e + f*x]])])/(4*d^(5/2)*(c + d)^(5/2)*f) + (a*(B*c - A*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(2
*d*(c + d)*f*(c + d*Sin[e + f*x])^2) + (a^2*(A*(c - 5*d)*d + B*(3*c^2 + 5*c*d - 4*d^2))*Cos[e + f*x])/(4*d^2*(
c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))

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Rubi [A]  time = 0.60951, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {2975, 2980, 2773, 208} \[ \frac{a^2 \left (A d (c-5 d)+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{4 d^2 f (c+d)^2 \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{a^{3/2} \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{4 d^{5/2} f (c+d)^{5/2}}+\frac{a (B c-A d) \cos (e+f x) \sqrt{a \sin (e+f x)+a}}{2 d f (c+d) (c+d \sin (e+f x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^(3/2)*(A*d*(c + 7*d) + 3*B*(c^2 + 3*c*d + 4*d^2))*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt
[a + a*Sin[e + f*x]])])/(4*d^(5/2)*(c + d)^(5/2)*f) + (a*(B*c - A*d)*Cos[e + f*x]*Sqrt[a + a*Sin[e + f*x]])/(2
*d*(c + d)*f*(c + d*Sin[e + f*x])^2) + (a^2*(A*(c - 5*d)*d + B*(3*c^2 + 5*c*d - 4*d^2))*Cos[e + f*x])/(4*d^2*(
c + d)^2*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))

Rule 2975

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((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]*(a + b*Sin[e + f*x])^(m - 1)*(c + d*S
in[e + f*x])^(n + 1))/(d*f*(n + 1)*(b*c + a*d)), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])
^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n +
1) - B*(b*c*m - a*d*(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] && GtQ[m, 1/2] && LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n]
|| EqQ[c, 0])

Rule 2980

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]

Rule 2773

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 208

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

Rubi steps

\begin{align*} \int \frac{(a+a \sin (e+f x))^{3/2} (A+B \sin (e+f x))}{(c+d \sin (e+f x))^3} \, dx &=\frac{a (B c-A d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac{\int \frac{\sqrt{a+a \sin (e+f x)} \left (-\frac{1}{2} a (B c-5 A d-4 B d)+\frac{1}{2} a (3 B c+A d+4 B d) \sin (e+f x)\right )}{(c+d \sin (e+f x))^2} \, dx}{2 d (c+d)}\\ &=\frac{a (B c-A d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac{a^2 \left (A (c-5 d) d+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac{\left (a \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right )\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{8 d^2 (c+d)^2}\\ &=\frac{a (B c-A d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac{a^2 \left (A (c-5 d) d+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac{\left (a^2 \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a c+a d-d x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{4 d^2 (c+d)^2 f}\\ &=-\frac{a^{3/2} \left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a+a \sin (e+f x)}}\right )}{4 d^{5/2} (c+d)^{5/2} f}+\frac{a (B c-A d) \cos (e+f x) \sqrt{a+a \sin (e+f x)}}{2 d (c+d) f (c+d \sin (e+f x))^2}+\frac{a^2 \left (A (c-5 d) d+B \left (3 c^2+5 c d-4 d^2\right )\right ) \cos (e+f x)}{4 d^2 (c+d)^2 f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [A]  time = 5.1612, size = 416, normalized size = 1.88 \[ \frac{(a (\sin (e+f x)+1))^{3/2} \left (-\frac{4 \sqrt{d} \left (A d (c+7 d)+B \left (-5 c^2-7 c d+4 d^2\right )\right ) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{(c+d)^2 (c+d \sin (e+f x))}+\frac{\left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \left (2 \log \left (\sqrt{d} \sqrt{c+d} \left (\tan ^2\left (\frac{1}{4} (e+f x)\right )+2 \tan \left (\frac{1}{4} (e+f x)\right )-1\right )+(c+d) \sec ^2\left (\frac{1}{4} (e+f x)\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{5/2}}-\frac{\left (A d (c+7 d)+3 B \left (c^2+3 c d+4 d^2\right )\right ) \left (2 \log \left (-\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (-\sqrt{d} \sqrt{c+d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \sqrt{c+d} \cos \left (\frac{1}{2} (e+f x)\right )+c+d\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(c+d)^{5/2}}-\frac{8 \sqrt{d} (d-c) (A d-B c) \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right )}{(c+d) (c+d \sin (e+f x))^2}\right )}{16 d^{5/2} f \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[e + f*x]))^(3/2)*(-(((A*d*(c + 7*d) + 3*B*(c^2 + 3*c*d + 4*d^2))*(e + f*x - 2*Log[Sec[(e + f*x)/4
]^2] + 2*Log[-(Sec[(e + f*x)/4]^2*(c + d + Sqrt[d]*Sqrt[c + d]*Cos[(e + f*x)/2] - Sqrt[d]*Sqrt[c + d]*Sin[(e +
f*x)/2]))]))/(c + d)^(5/2)) + ((A*d*(c + 7*d) + 3*B*(c^2 + 3*c*d + 4*d^2))*(e + f*x - 2*Log[Sec[(e + f*x)/4]^
2] + 2*Log[(c + d)*Sec[(e + f*x)/4]^2 + Sqrt[d]*Sqrt[c + d]*(-1 + 2*Tan[(e + f*x)/4] + Tan[(e + f*x)/4]^2)]))/
(c + d)^(5/2) - (8*Sqrt[d]*(-c + d)*(-(B*c) + A*d)*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/((c + d)*(c + d*Sin[
e + f*x])^2) - (4*Sqrt[d]*(A*d*(c + 7*d) + B*(-5*c^2 - 7*c*d + 4*d^2))*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2]))/
((c + d)^2*(c + d*Sin[e + f*x]))))/(16*d^(5/2)*f*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)

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Maple [B]  time = 2.233, size = 895, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(-2*sin(f*x+e)*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d*(A*c*d+7*A*d^2+3*B*c^2+9*B*c*
d+12*B*d^2)+arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^2*(A*c*d+7*A*d^2+3*B*c^2+9*B*c*d+12*B*
d^2)*cos(f*x+e)^2+A*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*c*d^2+7*A*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2
)*d^3-A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^3*d-7*A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(
a*c*d+a*d^2)^(1/2))*a^2*c^2*d^2-A*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^3-7*A*arctanh(
(a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^4-5*B*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*c^2*d-7*B*(
a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*c*d^2+4*B*(a-a*sin(f*x+e))^(3/2)*(a*(c+d)*d)^(1/2)*d^3-3*a^2*arctanh((
a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*B*c^4-9*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a
^2*c^3*d-15*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c^2*d^2-9*B*arctanh((a-a*sin(f*x+e))^(
1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*c*d^3-12*B*arctanh((a-a*sin(f*x+e))^(1/2)*d/(a*c*d+a*d^2)^(1/2))*a^2*d^4+A*(a-
a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^2*d-8*A*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d^2-9*A*(a-a*si
n(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*d^3+3*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^3+12*B*(a-a*sin(f*x+e
))^(1/2)*(a*(c+d)*d)^(1/2)*a*c^2*d+5*B*(a-a*sin(f*x+e))^(1/2)*(a*(c+d)*d)^(1/2)*a*c*d^2-4*B*(a-a*sin(f*x+e))^(
1/2)*(a*(c+d)*d)^(1/2)*a*d^3)*(-a*(-1+sin(f*x+e)))^(1/2)*(1+sin(f*x+e))/(a*(c+d)*d)^(1/2)/(c+d*sin(f*x+e))^2/(
c+d)^2/d^2/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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Maxima [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))^(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 19.132, size = 5040, normalized size = 22.81 \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))^(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

[-1/16*((3*B*a*c^4 + (A + 15*B)*a*c^3*d + 3*(3*A + 11*B)*a*c^2*d^2 + 3*(5*A + 11*B)*a*c*d^3 + (7*A + 12*B)*a*d
^4 - (3*B*a*c^2*d^2 + (A + 9*B)*a*c*d^3 + (7*A + 12*B)*a*d^4)*cos(f*x + e)^3 - (6*B*a*c^3*d + (2*A + 21*B)*a*c
^2*d^2 + 3*(5*A + 11*B)*a*c*d^3 + (7*A + 12*B)*a*d^4)*cos(f*x + e)^2 + (3*B*a*c^4 + (A + 9*B)*a*c^3*d + (7*A +
15*B)*a*c^2*d^2 + (A + 9*B)*a*c*d^3 + (7*A + 12*B)*a*d^4)*cos(f*x + e) + (3*B*a*c^4 + (A + 15*B)*a*c^3*d + 3*
(3*A + 11*B)*a*c^2*d^2 + 3*(5*A + 11*B)*a*c*d^3 + (7*A + 12*B)*a*d^4 - (3*B*a*c^2*d^2 + (A + 9*B)*a*c*d^3 + (7
*A + 12*B)*a*d^4)*cos(f*x + e)^2 + 2*(3*B*a*c^3*d + (A + 9*B)*a*c^2*d^2 + (7*A + 12*B)*a*c*d^3)*cos(f*x + e))*
sin(f*x + e))*sqrt(a/(c*d + d^2))*log((a*d^2*cos(f*x + e)^3 - a*c^2 - 2*a*c*d - a*d^2 - (6*a*c*d + 7*a*d^2)*co
s(f*x + e)^2 + 4*(c^2*d + 4*c*d^2 + 3*d^3 - (c*d^2 + d^3)*cos(f*x + e)^2 + (c^2*d + 3*c*d^2 + 2*d^3)*cos(f*x +
e) - (c^2*d + 4*c*d^2 + 3*d^3 + (c*d^2 + d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(a/(c*
d + d^2)) - (a*c^2 + 8*a*c*d + 9*a*d^2)*cos(f*x + e) + (a*d^2*cos(f*x + e)^2 - a*c^2 - 2*a*c*d - a*d^2 + 2*(3*
a*c*d + 4*a*d^2)*cos(f*x + e))*sin(f*x + e))/(d^2*cos(f*x + e)^3 + (2*c*d + d^2)*cos(f*x + e)^2 - c^2 - 2*c*d
- d^2 - (c^2 + d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - 2*c*d*cos(f*x + e) - c^2 - 2*c*d - d^2)*sin(f*x + e))
) + 4*(3*B*a*c^3 + (A + 2*B)*a*c^2*d - 3*(2*A + 3*B)*a*c*d^2 + (5*A + 4*B)*a*d^3 + (5*B*a*c^2*d - (A - 7*B)*a*
c*d^2 - (7*A + 4*B)*a*d^3)*cos(f*x + e)^2 + (3*B*a*c^3 + (A + 7*B)*a*c^2*d - (7*A + 2*B)*a*c*d^2 - 2*A*a*d^3)*
cos(f*x + e) - (3*B*a*c^3 + (A + 2*B)*a*c^2*d - 3*(2*A + 3*B)*a*c*d^2 + (5*A + 4*B)*a*d^3 - (5*B*a*c^2*d - (A
- 7*B)*a*c*d^2 - (7*A + 4*B)*a*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^4 + 2*c*d^5
+ d^6)*f*cos(f*x + e)^3 + (2*c^3*d^3 + 5*c^2*d^4 + 4*c*d^5 + d^6)*f*cos(f*x + e)^2 - (c^4*d^2 + 2*c^3*d^3 + 2*
c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f + ((c^2*d^4 + 2*
c*d^5 + d^6)*f*cos(f*x + e)^2 - 2*(c^3*d^3 + 2*c^2*d^4 + c*d^5)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*
d^4 + 4*c*d^5 + d^6)*f)*sin(f*x + e)), 1/8*((3*B*a*c^4 + (A + 15*B)*a*c^3*d + 3*(3*A + 11*B)*a*c^2*d^2 + 3*(5*
A + 11*B)*a*c*d^3 + (7*A + 12*B)*a*d^4 - (3*B*a*c^2*d^2 + (A + 9*B)*a*c*d^3 + (7*A + 12*B)*a*d^4)*cos(f*x + e)
^3 - (6*B*a*c^3*d + (2*A + 21*B)*a*c^2*d^2 + 3*(5*A + 11*B)*a*c*d^3 + (7*A + 12*B)*a*d^4)*cos(f*x + e)^2 + (3*
B*a*c^4 + (A + 9*B)*a*c^3*d + (7*A + 15*B)*a*c^2*d^2 + (A + 9*B)*a*c*d^3 + (7*A + 12*B)*a*d^4)*cos(f*x + e) +
(3*B*a*c^4 + (A + 15*B)*a*c^3*d + 3*(3*A + 11*B)*a*c^2*d^2 + 3*(5*A + 11*B)*a*c*d^3 + (7*A + 12*B)*a*d^4 - (3*
B*a*c^2*d^2 + (A + 9*B)*a*c*d^3 + (7*A + 12*B)*a*d^4)*cos(f*x + e)^2 + 2*(3*B*a*c^3*d + (A + 9*B)*a*c^2*d^2 +
(7*A + 12*B)*a*c*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(-a/(c*d + d^2))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*
sin(f*x + e) - c - 2*d)*sqrt(-a/(c*d + d^2))/(a*cos(f*x + e))) - 2*(3*B*a*c^3 + (A + 2*B)*a*c^2*d - 3*(2*A + 3
*B)*a*c*d^2 + (5*A + 4*B)*a*d^3 + (5*B*a*c^2*d - (A - 7*B)*a*c*d^2 - (7*A + 4*B)*a*d^3)*cos(f*x + e)^2 + (3*B*
a*c^3 + (A + 7*B)*a*c^2*d - (7*A + 2*B)*a*c*d^2 - 2*A*a*d^3)*cos(f*x + e) - (3*B*a*c^3 + (A + 2*B)*a*c^2*d - 3
*(2*A + 3*B)*a*c*d^2 + (5*A + 4*B)*a*d^3 - (5*B*a*c^2*d - (A - 7*B)*a*c*d^2 - (7*A + 4*B)*a*d^3)*cos(f*x + e))
*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^3 + (2*c^3*d^3 + 5*c^2*d^4
+ 4*c*d^5 + d^6)*f*cos(f*x + e)^2 - (c^4*d^2 + 2*c^3*d^3 + 2*c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e) - (c^4*d^
2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f + ((c^2*d^4 + 2*c*d^5 + d^6)*f*cos(f*x + e)^2 - 2*(c^3*d^3 + 2*c^
2*d^4 + c*d^5)*f*cos(f*x + e) - (c^4*d^2 + 4*c^3*d^3 + 6*c^2*d^4 + 4*c*d^5 + d^6)*f)*sin(f*x + e))]

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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))**(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))**3,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))^(3/2)*(A+B*sin(f*x+e))/(c+d*sin(f*x+e))^3,x, algorithm="giac")

[Out]

Exception raised: TypeError