3.6.28 \(\int \frac {\tan ^2(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [528]

Optimal. Leaf size=224 \[ -\frac {2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{(a+b)^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.15, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3275, 482, 541, 538, 437, 435, 432, 430} \begin {gather*} \frac {\sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{f (a+b) \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{f (a+b)^2 \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {\tan (e+f x)}{f (a+b) \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 b \sin (e+f x) \cos (e+f x)}{f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*Cos[e + f*x]*Sin[e + f*x])/((a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) - (2*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])/((a + b)^2*f*Sqrt[1 + (b*Sin[e + f*x]^2
)/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])/((a + b)*f*Sqrt[a + b*Sin[e + f*x]^2]) + Tan[e + f*x]/((a + b)*f*Sqrt[a + b*Sin[e + f*x]^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*} \int \frac {\tan ^2(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 ) \text {Subst}\left (\int \frac {x^2}{\left (1-x^2\right )^{3/2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {a-b x^2}{\sqrt {1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f}\\ &=-\frac {2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {-a (a-b)-2 a b x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{a (a+b)^2 f}\\ &=-\frac {2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (2 \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 )}{(a+b)^2 f}+\frac {\left (\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 )}{(a+b) f}\\ &=-\frac {2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (2 \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 )}{(a+b)^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left (\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 )}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}\\ &=-\frac {2 b \cos (e+f x) \sin (e+f x)}{(a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {2 \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)}}{(a+b)^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\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}}}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.62, size = 145, normalized size = 0.65 \begin {gather*} \frac {-2 \sqrt {2} a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )+\sqrt {2} (a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )+2 (a-b \cos (2 (e+f x))) \tan (e+f x)}{\sqrt {2} (a+b)^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sqrt[2]*a*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + Sqrt[2]*(a + b)*Sqrt[(2*a +
b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + 2*(a - b*Cos[2*(e + f*x)])*Tan[e + f*x])/(Sqrt[2]*(a +
b)^2*f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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Maple [A]
time = 18.81, size = 278, normalized size = 1.24

method	result	size
default	\(\frac {\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )+b \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-2 a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-2 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) b +a \sin \left (f x +e \right )+b \sin \left (f x +e \right )\right )}{\left (a +b \right )^{2} \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(278\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(a*(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))+b*(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))-2*a*(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))-2*
sin(f*x+e)*cos(f*x+e)^2*b+a*sin(f*x+e)+b*sin(f*x+e))/(a+b)^2/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(f*x+e)
))^(1/2)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 0.20, size = 1048, normalized size = 4.68 \begin {gather*} \frac {{\left (2 \, {\left (-i \, b^{3} \cos \left (f x + e\right )^{3} + {\left (i \, a b^{2} + i \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left ({\left (2 i \, a b^{2} + i \, b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (-2 i \, a^{2} b - 3 i \, a b^{2} - i \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 \, {\left (i \, b^{3} \cos \left (f x + e\right )^{3} + {\left (-i \, a b^{2} - i \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left ({\left (-2 i \, a b^{2} - i \, b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (2 i \, a^{2} b + 3 i \, a b^{2} + i \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} E(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 \, {\left ({\left (i \, a b^{2} + i \, b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (-i \, a^{2} b - 2 i \, a b^{2} - i \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left ({\left (2 i \, a^{2} b - i \, a b^{2} - i \, b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (-2 i \, a^{3} - i \, a^{2} b + 2 i \, a b^{2} + i \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 \, {\left ({\left (-i \, a b^{2} - i \, b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (i \, a^{2} b + 2 i \, a b^{2} + i \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b} \sqrt {\frac {a^{2} + a b}{b^{2}}} - {\left ({\left (-2 i \, a^{2} b + i \, a b^{2} + i \, b^{3}\right )} \cos \left (f x + e\right )^{3} + {\left (2 i \, a^{3} + i \, a^{2} b - 2 i \, a b^{2} - i \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (2 \, b^{3} \cos \left (f x + e\right )^{2} - a b^{2} - b^{3}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{{\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{3} - {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((2*(-I*b^3*cos(f*x + e)^3 + (I*a*b^2 + I*b^3)*cos(f*x + e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((2*I*a*b^2 + I*
b^3)*cos(f*x + e)^3 + (-2*I*a^2*b - 3*I*a*b^2 - I*b^3)*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*(I*b^3*cos(f*x + e)^3 + (-I*a*b^2 -
I*b^3)*cos(f*x + e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((-2*I*a*b^2 - I*b^3)*cos(f*x + e)^3 + (2*I*a^2*b + 3*I
*a*b^2 + I*b^3)*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*((I*a*b^2 + I*b^3)*cos(f*x + e)^3 + (-I*a^2*b - 2*I*a*b^2 - I*b^3)*cos(f*x
+ e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((2*I*a^2*b - I*a*b^2 - I*b^3)*cos(f*x + e)^3 + (-2*I*a^3 - I*a^2*b +
2*I*a*b^2 + I*b^3)*cos(f*x + e))*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqr
t((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*((-I*a*b^2 - I*b^3)*cos(f*x + e)^3 + (I*a^2*b + 2*I*a*b^2 + I*b^3)*cos(
f*x + e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((-2*I*a^2*b + I*a*b^2 + I*b^3)*cos(f*x + e)^3 + (2*I*a^3 + I*a^2*b
- 2*I*a*b^2 - I*b^3)*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*b^3*cos(f*x + e)^2 - a*b^2 - b^3)*sqrt(-b*cos(f*x + e)^2 + a + b)*si
n(f*x + e))/((a^2*b^3 + 2*a*b^4 + b^5)*f*cos(f*x + e)^3 - (a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5)*f*cos(f*x + e)
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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