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

Optimal. Leaf size=292 \[ -\frac {4 a \tan (e+f x)}{3 f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}+\frac {b (7 a-b) \sin (e+f x) \cos (e+f x)}{3 f (a+b)^3 \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x) \sec ^2(e+f x)}{3 f (a+b) \sqrt {a+b \sin ^2(e+f x)}}-\frac {4 a \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}+\frac {(7 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 f (a+b)^3 \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}} \]

[Out]

1/3*(7*a-b)*b*cos(f*x+e)*sin(f*x+e)/(a+b)^3/f/(a+b*sin(f*x+e)^2)^(1/2)+1/3*(7*a-b)*EllipticE(sin(f*x+e),(-b/a)
^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(a+b*sin(f*x+e)^2)^(1/2)/(a+b)^3/f/(1+b*sin(f*x+e)^2/a)^(1/2)-4/3*a*El
lipticF(sin(f*x+e),(-b/a)^(1/2))*sec(f*x+e)*(cos(f*x+e)^2)^(1/2)*(1+b*sin(f*x+e)^2/a)^(1/2)/(a+b)^2/f/(a+b*sin
(f*x+e)^2)^(1/2)-4/3*a*tan(f*x+e)/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)+1/3*sec(f*x+e)^2*tan(f*x+e)/(a+b)/f/(a+b*
sin(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.31, antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3196, 470, 527, 524, 426, 424, 421, 419} \[ -\frac {4 a \tan (e+f x)}{3 f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}+\frac {b (7 a-b) \sin (e+f x) \cos (e+f x)}{3 f (a+b)^3 \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x) \sec ^2(e+f x)}{3 f (a+b) \sqrt {a+b \sin ^2(e+f x)}}-\frac {4 a \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}+\frac {(7 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 f (a+b)^3 \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((7*a - b)*b*Cos[e + f*x]*Sin[e + f*x])/(3*(a + b)^3*f*Sqrt[a + b*Sin[e + f*x]^2]) + ((7*a - b)*Sqrt[Cos[e + f
*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*(a + b)^3*f*Sqrt[1
+ (b*Sin[e + f*x]^2)/a]) - (4*a*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt
[1 + (b*Sin[e + f*x]^2)/a])/(3*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) - (4*a*Tan[e + f*x])/(3*(a + b)^2*f*Sqr
t[a + b*Sin[e + f*x]^2]) + (Sec[e + f*x]^2*Tan[e + f*x])/(3*(a + b)*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 419

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Simp[(1*EllipticF[ArcSin[Rt[-(d/c),
2]*x], (b*c)/(a*d)])/(Sqrt[a]*Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &
& GtQ[a, 0] &&  !(NegQ[b/a] && SimplerSqrtQ[-(b/a), -(d/c)])

Rule 421

Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*Sqrt[(c_) + (d_.)*(x_)^2]), x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*
x^2], Int[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 524

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-(b/a), -(d/c)]))))))

Rule 527

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> -Simp[
((b*e - a*f)*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*n*(b*c - a*d)*(p + 1)), x] + Dist[1/(a*n*(b*c - a*d
)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*f)
*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 3196

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[(ff^(m + 1)*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(x^m*(a + b*ff^2*
x^2)^p)/(1 - ff^2*x^2)^((m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rubi steps

\begin {align*} \int \frac {\tan ^4(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^4}{\left (1-x^2\right )^{5/2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {a+3 a x^2}{\left (1-x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b) f}\\ &=-\frac {4 a \tan (e+f x)}{3 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {-a (3 a-b)+4 a b x^2}{\sqrt {1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^2 f}\\ &=\frac {(7 a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {4 a \tan (e+f x)}{3 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {a^2 (3 a-5 b)+a (7 a-b) b x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a (a+b)^3 f}\\ &=\frac {(7 a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {4 a \tan (e+f x)}{3 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left ((7 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^3 f}-\frac {\left (4 a \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^2 f}\\ &=\frac {(7 a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {4 a \tan (e+f x)}{3 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left ((7 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^3 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {\left (4 a \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{3 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}\\ &=\frac {(7 a-b) b \cos (e+f x) \sin (e+f x)}{3 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(7 a-b) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 (a+b)^3 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {4 a \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {4 a \tan (e+f x)}{3 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{3 (a+b) f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 2.28, size = 197, normalized size = 0.67 \[ \frac {-\frac {\tan (e+f x) \sec ^2(e+f x) \left (4 \left (4 a^2-3 a b+b^2\right ) \cos (2 (e+f x))+8 a^2+b (b-7 a) \cos (4 (e+f x))-21 a b-5 b^2\right )}{2 \sqrt {2}}-8 a (a+b) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )+2 a (7 a-b) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{6 f (a+b)^3 \sqrt {2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(7*a - b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] - 8*a*(a + b)*Sqrt[(2*a + b -
b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] - ((8*a^2 - 21*a*b - 5*b^2 + 4*(4*a^2 - 3*a*b + b^2)*Cos[2*
(e + f*x)] + b*(-7*a + b)*Cos[4*(e + f*x)])*Sec[e + f*x]^2*Tan[e + f*x])/(2*Sqrt[2]))/(6*(a + b)^3*f*Sqrt[2*a
+ b - b*Cos[2*(e + f*x)]])

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fricas [F]  time = 0.51, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )^{4}}{b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-b*cos(f*x + e)^2 + a + b)*tan(f*x + e)^4/(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a
^2 + 2*a*b + b^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan \left (f x + e\right )^{4}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 3.10, size = 368, normalized size = 1.26 \[ -\frac {\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (7 a -b \right ) \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )-4 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, a \left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a^{2}+2 a b +b^{2}\right ) \sin \left (f x +e \right )-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, a \left (4 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +4 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -7 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +\EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{3 \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-1\right ) \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \left (a +b \right )^{3} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b*(7*a-b)*sin(f*x+e)*cos(f*x+e)^4-4*(-b*cos(f*x+e)^4+(a+b)*co
s(f*x+e)^2)^(1/2)*a*(a+b)*cos(f*x+e)^2*sin(f*x+e)+(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(a^2+2*a*b+b^2)*s
in(f*x+e)-(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*a*
(4*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a+4*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-7*EllipticE(sin(f*x+e),(-1/
a*b)^(1/2))*a+EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*b)*cos(f*x+e)^2)/(1+sin(f*x+e))/(sin(f*x+e)-1)/(-(a+b*sin(f
*x+e)^2)*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)/(a+b)^3/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{4}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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