\(\int (a+b \sin ^2(e+f x))^{3/2} \tan ^4(e+f x) \, dx\) [505]

Optimal result

Integrand size = 25, antiderivative size = 275 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=-\frac {(3 a+8 b) \cos (e+f x) \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f}+\frac {8 (a+2 b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {a (5 a+8 b) \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(a+2 b) \sin ^2(e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{f}+\frac {\left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^3(e+f x)}{3 f} \]

[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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 275, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.360, Rules used = {3275, 478, 591, 596, 538, 437, 435, 432, 430} \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=-\frac {a (5 a+8 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {8 (a+2 b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{3 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\tan ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}-\frac {(a+2 b) \sin ^2(e+f x) \tan (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{f}-\frac {(3 a+8 b) \sin (e+f x) \cos (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 f} \]

[In]

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

[Out]

-1/3*((3*a + 8*b)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/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 430

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

Rule 432

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

Rule 435

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

Rule 437

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

Rule 478

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 538

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 591

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] + Dis
t[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 596

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 3275

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*} \text {integral}& = \frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\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)} \operatorname {EllipticF}\left (\arcsin (\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*}

Mathematica [A] (verified)

Time = 3.34 (sec) , antiderivative size = 211, normalized size of antiderivative = 0.77 \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\frac {32 a (a+2 b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )-4 a (5 a+8 b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )-\frac {\left (32 a^2+108 a b+18 b^2+\left (64 a^2+160 a b+17 b^2\right ) \cos (2 (e+f x))-2 b (6 a+17 b) \cos (4 (e+f x))-b^2 \cos (6 (e+f x))\right ) \sec ^2(e+f x) \tan (e+f x)}{4 \sqrt {2}}}{12 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \]

[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)]])

Maple [A] (verified)

Time = 4.98 (sec) , antiderivative size = 419, normalized size of antiderivative = 1.52

method	result	size
default	\(-\frac {\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b^{2} \left (\cos ^{6}\left (f x +e \right )\right ) \sin \left (f x +e \right )+\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (3 a +7 b \right ) \left (\cos ^{4}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (4 a^{2}+13 a b +9 b^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, a \left (5 F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +8 F\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -8 E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a -16 E\left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b \right ) \left (\cos ^{2}\left (f x +e \right )\right )+\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a^{2}+2 a b +b^{2}\right ) \sin \left (f x +e \right )}{3 \left (\sin \left (f x +e \right )-1\right ) \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(419\)

[In]

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

[Out]

-1/3*((-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^2*cos(f*x+e)^6*sin(f*x+e)+(-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)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*(cos(f*x+
e)^2)^(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(s
in(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+(-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))/(sin(f*x+e)-1)/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(f*x+e))
)^(1/2)/(1+sin(f*x+e))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

Fricas [F]

\[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{4} \,d x } \]

[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)

Sympy [F(-1)]

Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{4} \,d x } \]

[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)

Giac [F]

\[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\int { {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \tan \left (f x + e\right )^{4} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan ^4(e+f x) \, dx=\int {\mathrm {tan}\left (e+f\,x\right )}^4\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2} \,d x \]

[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)