3.555 \(\int \frac{1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx\)

Optimal. Leaf size=243 \[ -\frac{d^{3/2} (5 c+3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{a^{3/2} f (c-d)^3 (c+d)^{3/2}}-\frac{(c-9 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f (c-d)^3}-\frac{d (c+3 d) \cos (e+f x)}{2 a f (c-d)^2 (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))} \]

[Out]

-((c - 9*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(2*Sqrt[2]*a^(3/2)*(c - d)^3*f
) - (d^(3/2)*(5*c + 3*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(a^(3
/2)*(c - d)^3*(c + d)^(3/2)*f) - Cos[e + f*x]/(2*(c - d)*f*(a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])) -
(d*(c + 3*d)*Cos[e + f*x])/(2*a*(c - d)^2*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))

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Rubi [A]  time = 0.742985, antiderivative size = 243, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2766, 2984, 2985, 2649, 206, 2773, 208} \[ -\frac{d^{3/2} (5 c+3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a \sin (e+f x)+a}}\right )}{a^{3/2} f (c-d)^3 (c+d)^{3/2}}-\frac{(c-9 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a \sin (e+f x)+a}}\right )}{2 \sqrt{2} a^{3/2} f (c-d)^3}-\frac{d (c+3 d) \cos (e+f x)}{2 a f (c-d)^2 (c+d) \sqrt{a \sin (e+f x)+a} (c+d \sin (e+f x))}-\frac{\cos (e+f x)}{2 f (c-d) (a \sin (e+f x)+a)^{3/2} (c+d \sin (e+f x))} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])^2),x]

[Out]

-((c - 9*d)*ArcTanh[(Sqrt[a]*Cos[e + f*x])/(Sqrt[2]*Sqrt[a + a*Sin[e + f*x]])])/(2*Sqrt[2]*a^(3/2)*(c - d)^3*f
) - (d^(3/2)*(5*c + 3*d)*ArcTanh[(Sqrt[a]*Sqrt[d]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[a + a*Sin[e + f*x]])])/(a^(3
/2)*(c - d)^3*(c + d)^(3/2)*f) - Cos[e + f*x]/(2*(c - d)*f*(a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])) -
(d*(c + 3*d)*Cos[e + f*x])/(2*a*(c - d)^2*(c + d)*f*Sqrt[a + a*Sin[e + f*x]]*(c + d*Sin[e + f*x]))

Rule 2766

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

Rule 2984

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*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^(n + 1))/(f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(b*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin
[e + f*x])^(n + 1)*Simp[A*(a*d*m + b*c*(n + 1)) - B*(a*c*m + b*d*(n + 1)) + b*(B*c - A*d)*(m + n + 2)*Sin[e +
f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2
- d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || EqQ[m + 1/2, 0])

Rule 2985

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

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 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 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{1}{(a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))^2} \, dx &=-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac{\int \frac{-\frac{1}{2} a (c-6 d)-\frac{3}{2} a d \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))^2} \, dx}{2 a^2 (c-d)}\\ &=-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac{d (c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac{\int \frac{\frac{1}{2} a^2 \left (c^2-7 c d-6 d^2\right )+\frac{1}{2} a^2 d (c+3 d) \sin (e+f x)}{\sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx}{2 a^3 (c-d)^2 (c+d)}\\ &=-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac{d (c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}+\frac{(c-9 d) \int \frac{1}{\sqrt{a+a \sin (e+f x)}} \, dx}{4 a (c-d)^3}+\frac{\left (d^2 (5 c+3 d)\right ) \int \frac{\sqrt{a+a \sin (e+f x)}}{c+d \sin (e+f x)} \, dx}{2 a^2 (c-d)^3 (c+d)}\\ &=-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac{d (c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}-\frac{(c-9 d) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (e+f x)}{\sqrt{a+a \sin (e+f x)}}\right )}{2 a (c-d)^3 f}-\frac{\left (d^2 (5 c+3 d)\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 )}{a (c-d)^3 (c+d) f}\\ &=-\frac{(c-9 d) \tanh ^{-1}\left (\frac{\sqrt{a} \cos (e+f x)}{\sqrt{2} \sqrt{a+a \sin (e+f x)}}\right )}{2 \sqrt{2} a^{3/2} (c-d)^3 f}-\frac{d^{3/2} (5 c+3 d) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{d} \cos (e+f x)}{\sqrt{c+d} \sqrt{a+a \sin (e+f x)}}\right )}{a^{3/2} (c-d)^3 (c+d)^{3/2} f}-\frac{\cos (e+f x)}{2 (c-d) f (a+a \sin (e+f x))^{3/2} (c+d \sin (e+f x))}-\frac{d (c+3 d) \cos (e+f x)}{2 a (c-d)^2 (c+d) f \sqrt{a+a \sin (e+f x)} (c+d \sin (e+f x))}\\ \end{align*}

Mathematica [C]  time = 4.54375, size = 491, normalized size = 2.02 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right ) \left (-\frac{4 d^2 \left (\cos \left (\frac{1}{2} (e+f x)\right )-\sin \left (\frac{1}{2} (e+f x)\right )\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}{(c-d)^2 (c+d) (c+d \sin (e+f x))}+\frac{d^{3/2} (5 c+3 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (\sqrt{c+d}-\sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )+\sqrt{d} \cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(d-c)^3 (c+d)^{3/2}}+\frac{d^{3/2} (5 c+3 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \left (2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right ) \left (\sqrt{c+d}+\sqrt{d} \sin \left (\frac{1}{2} (e+f x)\right )-\sqrt{d} \cos \left (\frac{1}{2} (e+f x)\right )\right )\right )-2 \log \left (\sec ^2\left (\frac{1}{4} (e+f x)\right )\right )+e+f x\right )}{(c-d)^3 (c+d)^{3/2}}+\frac{4 \sin \left (\frac{1}{2} (e+f x)\right )}{(c-d)^2}-\frac{2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}{(c-d)^2}+\frac{(2+2 i) (-1)^{3/4} (c-9 d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (e+f x)\right )-1\right )\right )}{(c-d)^3}\right )}{4 f (a (\sin (e+f x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^(3/2)*(c + d*Sin[e + f*x])^2),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])*((4*Sin[(e + f*x)/2])/(c - d)^2 - (2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2]))/(c - d)^2 + ((2 + 2*I)*(-1)^(3/4)*(c - 9*d)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*(-1 + Tan[(e + f*x)/4])]*(Cos[
(e + f*x)/2] + Sin[(e + f*x)/2])^2)/(c - d)^3 + (d^(3/2)*(5*c + 3*d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2] + 2*
Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d] + Sqrt[d]*Cos[(e + f*x)/2] - Sqrt[d]*Sin[(e + f*x)/2])])*(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])^2)/((-c + d)^3*(c + d)^(3/2)) + (d^(3/2)*(5*c + 3*d)*(e + f*x - 2*Log[Sec[(e + f*x)/4]^2]
+ 2*Log[Sec[(e + f*x)/4]^2*(Sqrt[c + d] - Sqrt[d]*Cos[(e + f*x)/2] + Sqrt[d]*Sin[(e + f*x)/2])])*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2])^2)/((c - d)^3*(c + d)^(3/2)) - (4*d^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*(Cos[(e
+ f*x)/2] + Sin[(e + f*x)/2])^2)/((c - d)^2*(c + d)*(c + d*Sin[e + f*x]))))/(4*f*(a*(1 + Sin[e + f*x]))^(3/2))

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Maple [B]  time = 1.296, size = 978, normalized size = 4. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^2,x)

[Out]

-1/4/a^(5/2)*((a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)^2*a
*c^2*d-8*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)^2*a*c*d^
2-9*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)^2*a*d^3+20*ar
ctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(3/2)*sin(f*x+e)^2*c*d^3+12*arctanh((-a*(-1+sin(f*x+e)
))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(3/2)*sin(f*x+e)^2*d^4+(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e
)))^(1/2)*2^(1/2)/a^(1/2))*sin(f*x+e)*a*c^3-7*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)
*2^(1/2)/a^(1/2))*sin(f*x+e)*a*c^2*d-17*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/
2)/a^(1/2))*sin(f*x+e)*a*c*d^2-9*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1
/2))*sin(f*x+e)*a*d^3+20*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(3/2)*sin(f*x+e)*c^2*d^2+32
*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(3/2)*sin(f*x+e)*c*d^3+12*arctanh((-a*(-1+sin(f*x+e
)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(3/2)*sin(f*x+e)*d^4+2*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*a^(1/2)*s
in(f*x+e)*c^2*d+4*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*a^(1/2)*sin(f*x+e)*c*d^2-6*(-a*(-1+sin(f*x+e)))
^(1/2)*(a*(c+d)*d)^(1/2)*a^(1/2)*sin(f*x+e)*d^3+(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/
2)*2^(1/2)/a^(1/2))*a*c^3-8*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*
a*c^2*d-9*(a*(c+d)*d)^(1/2)*2^(1/2)*arctanh(1/2*(-a*(-1+sin(f*x+e)))^(1/2)*2^(1/2)/a^(1/2))*a*c*d^2+20*arctanh
((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c+d)*d)^(1/2))*a^(3/2)*c^2*d^2+12*arctanh((-a*(-1+sin(f*x+e)))^(1/2)*d/(a*(c
+d)*d)^(1/2))*a^(3/2)*c*d^3+2*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*a^(1/2)*c^3+2*(-a*(-1+sin(f*x+e)))^
(1/2)*(a*(c+d)*d)^(1/2)*a^(1/2)*c*d^2-4*(-a*(-1+sin(f*x+e)))^(1/2)*(a*(c+d)*d)^(1/2)*a^(1/2)*d^3)*(-a*(-1+sin(
f*x+e)))^(1/2)/(a*(c+d)*d)^(1/2)/(c+d*sin(f*x+e))/(c+d)/(c-d)^3/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(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 7.39202, size = 5644, normalized size = 23.23 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

[1/8*(sqrt(2)*((c^2*d - 8*c*d^2 - 9*d^3)*cos(f*x + e)^3 - 2*c^3 + 14*c^2*d + 34*c*d^2 + 18*d^3 + (c^3 - 6*c^2*
d - 25*c*d^2 - 18*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d - 17*c*d^2 - 9*d^3)*cos(f*x + e) - (2*c^3 - 14*c^2*d -
34*c*d^2 - 18*d^3 - (c^2*d - 8*c*d^2 - 9*d^3)*cos(f*x + e)^2 + (c^3 - 7*c^2*d - 17*c*d^2 - 9*d^3)*cos(f*x + e)
)*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*(cos(f*x + e) - si
n(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e)^2 - (cos(f*x + e
) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 2*(10*a*c^2*d + 16*a*c*d^2 + 6*a*d^3 - (5*a*c*d^2 + 3*a*d^3)*cos(f*
x + e)^3 - (5*a*c^2*d + 13*a*c*d^2 + 6*a*d^3)*cos(f*x + e)^2 + (5*a*c^2*d + 8*a*c*d^2 + 3*a*d^3)*cos(f*x + e)
+ (10*a*c^2*d + 16*a*c*d^2 + 6*a*d^3 - (5*a*c*d^2 + 3*a*d^3)*cos(f*x + e)^2 + (5*a*c^2*d + 8*a*c*d^2 + 3*a*d^3
)*cos(f*x + e))*sin(f*x + e))*sqrt(d/(a*c + a*d))*log((d^2*cos(f*x + e)^3 - (6*c*d + 7*d^2)*cos(f*x + e)^2 - c
^2 - 2*c*d - d^2 + 4*((c*d + d^2)*cos(f*x + e)^2 - c^2 - 4*c*d - 3*d^2 - (c^2 + 3*c*d + 2*d^2)*cos(f*x + e) +
(c^2 + 4*c*d + 3*d^2 + (c*d + d^2)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d/(a*c + a*d)) -
(c^2 + 8*c*d + 9*d^2)*cos(f*x + e) + (d^2*cos(f*x + e)^2 - c^2 - 2*c*d - d^2 + 2*(3*c*d + 4*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*(c^3 - c^2*d - c*d^2 + d
^3 + (c^2*d + 2*c*d^2 - 3*d^3)*cos(f*x + e)^2 + (c^3 + c*d^2 - 2*d^3)*cos(f*x + e) - (c^3 - c^2*d - c*d^2 + d^
3 - (c^2*d + 2*c*d^2 - 3*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a^2*c^4*d - 2*a^2*c^3*d^
2 + 2*a^2*c*d^4 - a^2*d^5)*f*cos(f*x + e)^3 + (a^2*c^5 - 4*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + 3*a^2*c*d^4 - 2*a^2*d
^5)*f*cos(f*x + e)^2 - (a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f*cos(f*x +
e) - 2*(a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f + ((a^2*c^4*d - 2*a^2*c^
3*d^2 + 2*a^2*c*d^4 - a^2*d^5)*f*cos(f*x + e)^2 - (a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c
*d^4 - a^2*d^5)*f*cos(f*x + e) - 2*(a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)
*f)*sin(f*x + e)), 1/8*(sqrt(2)*((c^2*d - 8*c*d^2 - 9*d^3)*cos(f*x + e)^3 - 2*c^3 + 14*c^2*d + 34*c*d^2 + 18*d
^3 + (c^3 - 6*c^2*d - 25*c*d^2 - 18*d^3)*cos(f*x + e)^2 - (c^3 - 7*c^2*d - 17*c*d^2 - 9*d^3)*cos(f*x + e) - (2
*c^3 - 14*c^2*d - 34*c*d^2 - 18*d^3 - (c^2*d - 8*c*d^2 - 9*d^3)*cos(f*x + e)^2 + (c^3 - 7*c^2*d - 17*c*d^2 - 9
*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a)*log(-(a*cos(f*x + e)^2 - 2*sqrt(2)*sqrt(a*sin(f*x + e) + a)*sqrt(a)*
(cos(f*x + e) - sin(f*x + e) + 1) + 3*a*cos(f*x + e) - (a*cos(f*x + e) - 2*a)*sin(f*x + e) + 2*a)/(cos(f*x + e
)^2 - (cos(f*x + e) + 2)*sin(f*x + e) - cos(f*x + e) - 2)) + 4*(10*a*c^2*d + 16*a*c*d^2 + 6*a*d^3 - (5*a*c*d^2
+ 3*a*d^3)*cos(f*x + e)^3 - (5*a*c^2*d + 13*a*c*d^2 + 6*a*d^3)*cos(f*x + e)^2 + (5*a*c^2*d + 8*a*c*d^2 + 3*a*
d^3)*cos(f*x + e) + (10*a*c^2*d + 16*a*c*d^2 + 6*a*d^3 - (5*a*c*d^2 + 3*a*d^3)*cos(f*x + e)^2 + (5*a*c^2*d + 8
*a*c*d^2 + 3*a*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(-d/(a*c + a*d))*arctan(1/2*sqrt(a*sin(f*x + e) + a)*(d*si
n(f*x + e) - c - 2*d)*sqrt(-d/(a*c + a*d))/(d*cos(f*x + e))) + 4*(c^3 - c^2*d - c*d^2 + d^3 + (c^2*d + 2*c*d^2
- 3*d^3)*cos(f*x + e)^2 + (c^3 + c*d^2 - 2*d^3)*cos(f*x + e) - (c^3 - c^2*d - c*d^2 + d^3 - (c^2*d + 2*c*d^2
- 3*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a))/((a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c*d^4 - a^2
*d^5)*f*cos(f*x + e)^3 + (a^2*c^5 - 4*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + 3*a^2*c*d^4 - 2*a^2*d^5)*f*cos(f*x + e)^2
- (a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f*cos(f*x + e) - 2*(a^2*c^5 - a^
2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f + ((a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c*d^4 -
a^2*d^5)*f*cos(f*x + e)^2 - (a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*f*cos
(f*x + e) - 2*(a^2*c^5 - a^2*c^4*d - 2*a^2*c^3*d^2 + 2*a^2*c^2*d^3 + a^2*c*d^4 - a^2*d^5)*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(1/(a+a*sin(f*x+e))**(3/2)/(c+d*sin(f*x+e))**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(1/(a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^2,x, algorithm="giac")

[Out]

Exception raised: TypeError