\(\int \frac {\tan ^2(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [539]

Optimal result

Integrand size = 25, antiderivative size = 292 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {4 b \cos (e+f x) \sin (e+f x)}{3 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {(7 a-b) b \cos (e+f x) \sin (e+f x)}{3 a (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {(7 a-b) \sqrt {\cos ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 a (a+b)^3 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {4 \sqrt {\cos ^2(e+f x)} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

-4/3*b*cos(f*x+e)*sin(f*x+e)/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(3/2)-1/3*(7*a-b)*b*cos(f*x+e)*sin(f*x+e)/a/(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/(a+b)^3/f/(1+b*sin(f*x+e)^2/a)^(1/2)+4/3*EllipticF(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)+tan(f*x+e)/(a+b)/f/(a+b*sin
(f*x+e)^2)^(3/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 292, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3275, 482, 541, 538, 437, 435, 432, 430} \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {4 \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} \operatorname {EllipticF}\left (\arcsin (\sin (e+f x)),-\frac {b}{a}\right )}{3 f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}-\frac {(7 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\arcsin (\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{3 a f (a+b)^3 \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\tan (e+f x)}{f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {b (7 a-b) \sin (e+f x) \cos (e+f x)}{3 a f (a+b)^3 \sqrt {a+b \sin ^2(e+f x)}}-\frac {4 b \sin (e+f x) \cos (e+f x)}{3 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[In]

Int[Tan[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

(-4*b*Cos[e + f*x]*Sin[e + f*x])/(3*(a + b)^2*f*(a + b*Sin[e + f*x]^2)^(3/2)) - ((7*a - b)*b*Cos[e + f*x]*Sin[
e + f*x])/(3*a*(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*(a + b)^3*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) +
(4*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]) + Tan[e + f*x]/((a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2))

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 482

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 + 1)/(n*(b*c - a*d)*(p + 1))), x] - Dist[e^n/(n*(b*c -
a*d)*(p + 1)), Int[(e*x)^(m - n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(m - n + 1) + d*(m + n*(p + q + 1)
+ 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] && GeQ[n
, m - n + 1] && 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 541

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

Mathematica [A] (verified)

Time = 3.22 (sec) , antiderivative size = 199, normalized size of antiderivative = 0.68 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {-2 a^2 (7 a-b) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} E\left (e+f x\left |-\frac {b}{a}\right .\right )+8 a^2 (a+b) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} \operatorname {EllipticF}\left (e+f x,-\frac {b}{a}\right )+\frac {\left (24 a^3+4 a^2 b+5 a b^2+b^3-4 a b (11 a+3 b) \cos (2 (e+f x))+(7 a-b) b^2 \cos (4 (e+f x))\right ) \tan (e+f x)}{\sqrt {2}}}{6 a (a+b)^3 f (2 a+b-b \cos (2 (e+f x)))^{3/2}} \]

[In]

Integrate[Tan[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

(-2*a^2*(7*a - b)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticE[e + f*x, -(b/a)] + 8*a^2*(a + b)*((2*a +
b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticF[e + f*x, -(b/a)] + ((24*a^3 + 4*a^2*b + 5*a*b^2 + b^3 - 4*a*b*(11*a
+ 3*b)*Cos[2*(e + f*x)] + (7*a - b)*b^2*Cos[4*(e + f*x)])*Tan[e + f*x])/Sqrt[2])/(6*a*(a + b)^3*f*(2*a + b -
b*Cos[2*(e + f*x)])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(850\) vs. \(2(268)=536\).

Time = 5.92 (sec) , antiderivative size = 851, normalized size of antiderivative = 2.91

method	result	size
default	\(\text {Expression too large to display}\)	\(851\)

[In]

int(tan(f*x+e)^2/(a+b*sin(f*x+e)^2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3*((-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^2*(7*a-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)*b*(11*a^2+10*a*b-b^2)*cos(f*x+e)^2*sin(f*x+e)-(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*(-b*cos(f*x+e
)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(cos(f*x+e)^2)^(1/2)*a*b*(4*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a+4*EllipticF(s
in(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)*c
os(f*x+e)^2+3*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a*(a^2+2*a*b+b^2)*sin(f*x+e)+4*(-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)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2
))*a^3+8*(-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)*Ell
ipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b+4*(-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)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2-7*(-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)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^3-6*(-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)*EllipticE(sin(f*
x+e),(-1/a*b)^(1/2))*a^2*b+(-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)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2)/(a+b*sin(f*x+e)^2)^(3/2)/(-(a+b*sin(f*x+e)^2)*(sin(
f*x+e)-1)*(1+sin(f*x+e)))^(1/2)/(a+b)^3/a/cos(f*x+e)/f

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.26 (sec) , antiderivative size = 1632, normalized size of antiderivative = 5.59 \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(tan(f*x+e)^2/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

1/6*((2*((-7*I*a*b^4 + I*b^5)*cos(f*x + e)^5 - 2*(-7*I*a^2*b^3 - 6*I*a*b^4 + I*b^5)*cos(f*x + e)^3 + (-7*I*a^3
*b^2 - 13*I*a^2*b^3 - 5*I*a*b^4 + I*b^5)*cos(f*x + e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((14*I*a^2*b^3 + 5*I*a
*b^4 - I*b^5)*cos(f*x + e)^5 + 2*(-14*I*a^3*b^2 - 19*I*a^2*b^3 - 4*I*a*b^4 + I*b^5)*cos(f*x + e)^3 + (14*I*a^4
*b + 33*I*a^3*b^2 + 23*I*a^2*b^3 + 3*I*a*b^4 - I*b^5)*cos(f*x + e))*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2)
+ 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e)))
, (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*((7*I*a*b^4 - I*b^5)*cos(f*x + e)^5
- 2*(7*I*a^2*b^3 + 6*I*a*b^4 - I*b^5)*cos(f*x + e)^3 + (7*I*a^3*b^2 + 13*I*a^2*b^3 + 5*I*a*b^4 - I*b^5)*cos(f*
x + e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((-14*I*a^2*b^3 - 5*I*a*b^4 + I*b^5)*cos(f*x + e)^5 + 2*(14*I*a^3*b^2
+ 19*I*a^2*b^3 + 4*I*a*b^4 - I*b^5)*cos(f*x + e)^3 + (-14*I*a^4*b - 33*I*a^3*b^2 - 23*I*a^2*b^3 - 3*I*a*b^4 +
I*b^5)*cos(f*x + e))*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt
((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt(
(a^2 + a*b)/b^2))/b^2) - 2*(2*((-3*I*a^2*b^3 - 2*I*a*b^4 + I*b^5)*cos(f*x + e)^5 + 2*(3*I*a^3*b^2 + 5*I*a^2*b^
3 + I*a*b^4 - I*b^5)*cos(f*x + e)^3 + (-3*I*a^4*b - 8*I*a^3*b^2 - 6*I*a^2*b^3 + I*b^5)*cos(f*x + e))*sqrt(-b)*
sqrt((a^2 + a*b)/b^2) + ((6*I*a^3*b^2 - 7*I*a^2*b^3 - 5*I*a*b^4)*cos(f*x + e)^5 + 2*(-6*I*a^4*b + I*a^3*b^2 +
12*I*a^2*b^3 + 5*I*a*b^4)*cos(f*x + e)^3 + (6*I*a^5 + 5*I*a^4*b - 13*I*a^3*b^2 - 17*I*a^2*b^3 - 5*I*a*b^4)*cos
(f*x + e))*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b
)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)
/b^2))/b^2) - 2*(2*((3*I*a^2*b^3 + 2*I*a*b^4 - I*b^5)*cos(f*x + e)^5 + 2*(-3*I*a^3*b^2 - 5*I*a^2*b^3 - I*a*b^4
+ I*b^5)*cos(f*x + e)^3 + (3*I*a^4*b + 8*I*a^3*b^2 + 6*I*a^2*b^3 - I*b^5)*cos(f*x + e))*sqrt(-b)*sqrt((a^2 +
a*b)/b^2) + ((-6*I*a^3*b^2 + 7*I*a^2*b^3 + 5*I*a*b^4)*cos(f*x + e)^5 + 2*(6*I*a^4*b - I*a^3*b^2 - 12*I*a^2*b^3
- 5*I*a*b^4)*cos(f*x + e)^3 + (-6*I*a^5 - 5*I*a^4*b + 13*I*a^3*b^2 + 17*I*a^2*b^3 + 5*I*a*b^4)*cos(f*x + e))*
sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*
a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2)
+ 2*(3*a^3*b^2 + 6*a^2*b^3 + 3*a*b^4 + (7*a*b^4 - b^5)*cos(f*x + e)^4 - (11*a^2*b^3 + 10*a*b^4 - b^5)*cos(f*x
+ e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^4*b^4 + 3*a^3*b^5 + 3*a^2*b^6 + a*b^7)*f*cos(f*x +
e)^5 - 2*(a^5*b^3 + 4*a^4*b^4 + 6*a^3*b^5 + 4*a^2*b^6 + a*b^7)*f*cos(f*x + e)^3 + (a^6*b^2 + 5*a^5*b^3 + 10*a^
4*b^4 + 10*a^3*b^5 + 5*a^2*b^6 + a*b^7)*f*cos(f*x + e))

Sympy [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate(tan(f*x+e)**2/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Integral(tan(e + f*x)**2/(a + b*sin(e + f*x)**2)**(5/2), x)

Maxima [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tan(f*x+e)^2/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

integrate(tan(f*x + e)^2/(b*sin(f*x + e)^2 + a)^(5/2), x)

Giac [F]

\[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate(tan(f*x+e)^2/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \]

[In]

int(tan(e + f*x)^2/(a + b*sin(e + f*x)^2)^(5/2),x)

[Out]

int(tan(e + f*x)^2/(a + b*sin(e + f*x)^2)^(5/2), x)