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

Optimal. Leaf size=275 \[ \frac {\tan ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}-\frac {(a+2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}-\frac {(3 a+8 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {a (5 a+8 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {8 (a+2 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}} \]

[Out]

-1/3*(3*a+8*b)*cos(f*x+e)*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/f+8/3*(a+2*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)/f/(1+b*sin(f*x+e)^2/a)^(1/2)-1/3*a*(5*a+8*b)*Ellipti
cF(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)/f/(a+b*sin(f*x+e)^2)^(1
/2)-(a+2*b)*sin(f*x+e)^2*(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/f+1/3*(a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e)^3/f

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Rubi [A]  time = 0.37, antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3196, 467, 577, 582, 524, 426, 424, 421, 419} \[ \frac {\tan ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}-\frac {(a+2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}-\frac {(3 a+8 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {a (5 a+8 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {8 (a+2 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right )}{3 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((3*a + 8*b)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*f) + (8*(a + 2*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*f*Sqrt[1 + (b*Sin[e + f*x]
^2)/a]) - (a*(5*a + 8*b)*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*f*Sqrt[a + b*Sin[e + f*x]^2]) - ((a + 2*b)*Sin[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2]*T
an[e + f*x])/f + ((a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x]^3)/(3*f)

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 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 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 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 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 \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^4 \left (a+b x^2\right )^{3/2}}{\left (1-x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \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 )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 f}\\ &=-\frac {(a+2 b) \sin ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {x^2 \left (-6 a (a+3 b)-3 b (3 a+8 b) x^2\right )}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 f}\\ &=-\frac {(3 a+8 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {(a+2 b) \sin ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname {Subst}\left (\int \frac {-3 a b (3 a+8 b)-24 b^2 (a+2 b) x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{9 b f}\\ &=-\frac {(3 a+8 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {(a+2 b) \sin ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}+\frac {\left (8 (a+2 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 f}-\frac {\left (a (5 a+8 b) \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 f}\\ &=-\frac {(3 a+8 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}-\frac {(a+2 b) \sin ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}+\frac {\left (8 (a+2 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 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {\left (a (5 a+8 b) \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 f \sqrt {a+b \sin ^2(e+f x)}}\\ &=-\frac {(3 a+8 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {8 (a+2 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 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {a (5 a+8 b) \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 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(a+2 b) \sin ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f}\\ \end {align*}

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Mathematica [A]  time = 2.74, size = 211, normalized size = 0.77 \[ \frac {-\frac {\tan (e+f x) \sec ^2(e+f x) \left (\left (64 a^2+160 a b+17 b^2\right ) \cos (2 (e+f x))+32 a^2-2 b (6 a+17 b) \cos (4 (e+f x))+108 a b-b^2 \cos (6 (e+f x))+18 b^2\right )}{4 \sqrt {2}}-4 a (5 a+8 b) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )+32 a (a+2 b) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{12 f \sqrt {2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*a*(a + 2*b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] - 4*a*(5*a + 8*b)*Sqrt[(2*a
+ b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] - ((32*a^2 + 108*a*b + 18*b^2 + (64*a^2 + 160*a*b + 17
*b^2)*Cos[2*(e + f*x)] - 2*b*(6*a + 17*b)*Cos[4*(e + f*x)] - b^2*Cos[6*(e + f*x)])*Sec[e + f*x]^2*Tan[e + f*x]
)/(4*Sqrt[2]))/(12*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \tan \left (f x + e\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 2.84, size = 419, normalized size = 1.52 \[ -\frac {\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{2} \sin \left (f x +e \right ) \left (\cos ^{6}\left (f x +e \right )\right )+\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (3 a +7 b \right ) \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (4 a^{2}+13 a b +9 b^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a^{2}+2 a b +b^{2}\right ) \sin \left (f x +e \right )-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, a \left (5 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +8 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -8 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a -16 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{3 \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*((-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^2*sin(f*x+e)*cos(f*x+e)^6+(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)
^2)^(1/2)*b*(3*a+7*b)*cos(f*x+e)^4*sin(f*x+e)-(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(4*a^2+13*a*b+9*b^2)*
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)*sin(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*(5*EllipticF(sin(f*x+e),(-
1/a*b)^(1/2))*a+8*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-8*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a-16*EllipticE
(sin(f*x+e),(-1/a*b)^(1/2))*b)*cos(f*x+e)^2)/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)/(sin(f*
x+e)-1)/(1+sin(f*x+e))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

[Out]

Timed out

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