3.110 \(\int \sin ^4(e+f x) (a+b \tan ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=222 \[ \frac{3 \left (a^2-8 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 f \sqrt{a-b}}-\frac{3 (a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}+\frac{3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}+\frac{3 \sqrt{b} (a-2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac{\sin ^3(e+f x) \cos (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f} \]

[Out]

(3*(a^2 - 8*a*b + 8*b^2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*Sqrt[a - b]*f) + (3
*(a - 2*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*f) - (3*(a - 4*b)*Tan[e + f*
x]*Sqrt[a + b*Tan[e + f*x]^2])/(8*f) + (3*(a - 2*b)*Sin[e + f*x]^2*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(8
*f) - (Cos[e + f*x]*Sin[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(3/2))/(4*f)

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Rubi [A]  time = 0.321516, antiderivative size = 222, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {3663, 467, 577, 582, 523, 217, 206, 377, 203} \[ \frac{3 \left (a^2-8 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 f \sqrt{a-b}}-\frac{3 (a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}+\frac{3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}+\frac{3 \sqrt{b} (a-2 b) \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac{\sin ^3(e+f x) \cos (e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(3*(a^2 - 8*a*b + 8*b^2)*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*Sqrt[a - b]*f) + (3
*(a - 2*b)*Sqrt[b]*ArcTanh[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(2*f) - (3*(a - 4*b)*Tan[e + f*
x]*Sqrt[a + b*Tan[e + f*x]^2])/(8*f) + (3*(a - 2*b)*Sin[e + f*x]^2*Tan[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(8
*f) - (Cos[e + f*x]*Sin[e + f*x]^3*(a + b*Tan[e + f*x]^2)^(3/2))/(4*f)

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 467

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(e^(n -
1)*(e*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*n*(p + 1)), x] - Dist[e^n/(b*n*(p + 1)), Int[(e*x)^
(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(m - n + 1) + d*(m + n*(q - 1) + 1)*x^n, x], x], x] /;
FreeQ[{a, b, c, d, e}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] && GtQ[m - n + 1, 0] &
& IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 577

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> -Simp[((b*e - a*f)*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*b*g*n*(p + 1)), x] + Dist[
1/(a*b*n*(p + 1)), Int[(g*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1)*Simp[c*(b*e*n*(p + 1) + (b*e - a*f)*(m
+ 1)) + d*(b*e*n*(p + 1) + (b*e - a*f)*(m + n*q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] &&
IGtQ[n, 0] && LtQ[p, -1] && GtQ[q, 0] &&  !(EqQ[q, 1] && SimplerQ[b*c - a*d, b*e - a*f])

Rule 582

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(f*g^(n - 1)*(g*x)^(m - n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*d*(m + n*(p + q
+ 1) + 1)), x] - Dist[g^n/(b*d*(m + n*(p + q + 1) + 1)), Int[(g*x)^(m - n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*
f*c*(m - n + 1) + (a*f*d*(m + n*q + 1) + b*(f*c*(m + n*p + 1) - e*d*(m + n*(p + q + 1) + 1)))*x^n, x], x], x]
/; FreeQ[{a, b, c, d, e, f, g, p, q}, x] && IGtQ[n, 0] && GtQ[m, n - 1]

Rule 523

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/b, I
nt[1/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/((a + b*x^n)*Sqrt[c + d*x^n]), x], x] /; FreeQ[{a, b,
c, d, e, f, n}, x]

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \sin ^4(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4 \left (a+b x^2\right )^{3/2}}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac{\operatorname{Subst}\left (\int \frac{x^2 \sqrt{a+b x^2} \left (3 a+6 b x^2\right )}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 f}\\ &=\frac{3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}-\frac{\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}-\frac{\operatorname{Subst}\left (\int \frac{x^2 \left (3 a (a-6 b)+6 (a-4 b) b x^2\right )}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{3 (a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}+\frac{3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}-\frac{\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac{\operatorname{Subst}\left (\int \frac{6 a (a-4 b) b+24 (a-2 b) b^2 x^2}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{16 b f}\\ &=-\frac{3 (a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}+\frac{3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}-\frac{\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac{(3 (a-2 b) b) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{2 f}+\frac{\left (3 \left (a^2-8 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b x^2}} \, dx,x,\tan (e+f x)\right )}{8 f}\\ &=-\frac{3 (a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}+\frac{3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}-\frac{\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}+\frac{(3 (a-2 b) b) \operatorname{Subst}\left (\int \frac{1}{1-b x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}+\frac{\left (3 \left (a^2-8 a b+8 b^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-(-a+b) x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 f}\\ &=\frac{3 \left (a^2-8 a b+8 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 \sqrt{a-b} f}+\frac{3 (a-2 b) \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{2 f}-\frac{3 (a-4 b) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}+\frac{3 (a-2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f}-\frac{\cos (e+f x) \sin ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2}}{4 f}\\ \end{align*}

Mathematica [C]  time = 4.74926, size = 278, normalized size = 1.25 \[ \frac{\sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\frac{3 a \sin (2 (e+f x)) \csc ^2(e+f x) \left (\left (a^2-5 a b+4 b^2\right ) \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )-\left (a^2-8 a b+8 b^2\right ) \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )}{\sqrt{2} b (a-b) \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}+\frac{1}{4} ((18 b-8 a) \sin (2 (e+f x))+(a-b) \sin (4 (e+f x))+16 b \tan (e+f x))\right )}{8 \sqrt{2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4*(a + b*Tan[e + f*x]^2)^(3/2),x]

[Out]

(Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2]*((3*a*Csc[e + f*x]^2*((a^2 - 5*a*b + 4*b^2)*EllipticF
[ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] - (a^2 - 8*a*b + 8*b^2)*Ellip
ticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1])*Sin[2*(e
+ f*x)])/(Sqrt[2]*(a - b)*b*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]) + ((-8*a + 18*b)*Sin[
2*(e + f*x)] + (a - b)*Sin[4*(e + f*x)] + 16*b*Tan[e + f*x])/4))/(8*Sqrt[2]*f)

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Maple [C]  time = 0.416, size = 2630, normalized size = 11.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x)

[Out]

1/8/f/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*(-5*cos(f*x+e)^5*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2
-4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^2-12*cos(f*x+e)^5*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^2
+2*b^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^7-2*b^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*
cos(f*x+e)^6+2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^7*a^2-2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a
)^(1/2)*cos(f*x+e)^6*a^2-48*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(
a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2
)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)
^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1
/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a*b-48*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/
2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^
(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*EllipticPi((cos(f*x
+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)
*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*b^2-3*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*
(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2
)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(
1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4
*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*a^2+6*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*(1/a*(I*cos(f*x+e)
*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x
+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*EllipticPi((c
os(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*
b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a^2+48*cos(f*x+e)^2*sin(f*x+e)*
2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+
1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+
e)+1))^(1/2)*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2)*(a
-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*b^2+5
*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2+12*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a
)^(1/2)*b^2+6*cos(f*x+e)^3*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^2-6*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1
/2)+a-2*b)/a)^(1/2)*b^2+4*cos(f*x+e)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^2+12*EllipticF((cos(f*x+e)-1)
*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-
8*a*b+8*b^2)/a^2)^(1/2))*cos(f*x+e)^2*sin(f*x+e)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+
cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/
2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*2^(1/2)*a*b+24*EllipticPi((cos(f*x+e)-1)*((2*I*b^(1/2)*(
a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(a-b)^(1/2)-a+2*b)/a)
^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*cos(f*x+e)^2*sin(f*x+e)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/
2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^
(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*2^(1/2)*a*b-4*((2*I*b^(1/2)*(a-
b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^7*a*b+4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*cos(f*x+e)^6*a*b+17*cos(
f*x+e)^5*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b-17*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/
2)*a*b-cos(f*x+e)^3*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b+cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b
)/a)^(1/2)*a*b)*cos(f*x+e)*((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(3/2)*sin(f*x+e)/(cos(f*x+e)-1)/(a
*cos(f*x+e)^2-cos(f*x+e)^2*b+b)^2

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sin \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*sin(f*x + e)^4, x)

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Fricas [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(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

Timed out

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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(sin(f*x+e)**4*(a+b*tan(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sin \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4*(a+b*tan(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^(3/2)*sin(f*x + e)^4, x)