3.483 \(\int \frac{\tan ^2(e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\)
Optimal. Leaf size=106 \[ -\frac{\tan (e+f x)}{8 a f \sqrt{a \cos ^2(e+f x)}}-\frac{\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a f \sqrt{a \cos ^2(e+f x)}}+\frac{\tan (e+f x) \sec ^2(e+f x)}{4 a f \sqrt{a \cos ^2(e+f x)}} \]
[Out]
-(ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(8*a*f*Sqrt[a*Cos[e + f*x]^2]) - Tan[e + f*x]/(8*a*f*Sqrt[a*Cos[e + f*x]
^2]) + (Sec[e + f*x]^2*Tan[e + f*x])/(4*a*f*Sqrt[a*Cos[e + f*x]^2])
________________________________________________________________________________________
Rubi [A] time = 0.157268, antiderivative size = 106, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used =
{3176, 3207, 2611, 3768, 3770} \[ -\frac{\tan (e+f x)}{8 a f \sqrt{a \cos ^2(e+f x)}}-\frac{\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a f \sqrt{a \cos ^2(e+f x)}}+\frac{\tan (e+f x) \sec ^2(e+f x)}{4 a f \sqrt{a \cos ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Tan[e + f*x]^2/(a - a*Sin[e + f*x]^2)^(3/2),x]
[Out]
-(ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(8*a*f*Sqrt[a*Cos[e + f*x]^2]) - Tan[e + f*x]/(8*a*f*Sqrt[a*Cos[e + f*x]
^2]) + (Sec[e + f*x]^2*Tan[e + f*x])/(4*a*f*Sqrt[a*Cos[e + f*x]^2])
Rule 3176
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]
Rule 3207
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])
Rule 2611
Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]
Rule 3768
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int \frac{\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx &=\int \frac{\tan ^2(e+f x)}{\left (a \cos ^2(e+f x)\right )^{3/2}} \, dx\\ &=\frac{\cos (e+f x) \int \sec ^3(e+f x) \tan ^2(e+f x) \, dx}{a \sqrt{a \cos ^2(e+f x)}}\\ &=\frac{\sec ^2(e+f x) \tan (e+f x)}{4 a f \sqrt{a \cos ^2(e+f x)}}-\frac{\cos (e+f x) \int \sec ^3(e+f x) \, dx}{4 a \sqrt{a \cos ^2(e+f x)}}\\ &=-\frac{\tan (e+f x)}{8 a f \sqrt{a \cos ^2(e+f x)}}+\frac{\sec ^2(e+f x) \tan (e+f x)}{4 a f \sqrt{a \cos ^2(e+f x)}}-\frac{\cos (e+f x) \int \sec (e+f x) \, dx}{8 a \sqrt{a \cos ^2(e+f x)}}\\ &=-\frac{\tanh ^{-1}(\sin (e+f x)) \cos (e+f x)}{8 a f \sqrt{a \cos ^2(e+f x)}}-\frac{\tan (e+f x)}{8 a f \sqrt{a \cos ^2(e+f x)}}+\frac{\sec ^2(e+f x) \tan (e+f x)}{4 a f \sqrt{a \cos ^2(e+f x)}}\\ \end{align*}
Mathematica [A] time = 0.0951606, size = 59, normalized size = 0.56 \[ \frac{\tan (e+f x) \left (2 \sec ^2(e+f x)-1\right )-\cos (e+f x) \tanh ^{-1}(\sin (e+f x))}{8 a f \sqrt{a \cos ^2(e+f x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[e + f*x]^2/(a - a*Sin[e + f*x]^2)^(3/2),x]
[Out]
(-(ArcTanh[Sin[e + f*x]]*Cos[e + f*x]) + (-1 + 2*Sec[e + f*x]^2)*Tan[e + f*x])/(8*a*f*Sqrt[a*Cos[e + f*x]^2])
________________________________________________________________________________________
Maple [A] time = 1.285, size = 104, normalized size = 1. \begin{align*} -{\frac{-2\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) +4\,\sin \left ( fx+e \right ) + \left ( \ln \left ( -1+\sin \left ( fx+e \right ) \right ) -\ln \left ( 1+\sin \left ( fx+e \right ) \right ) \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}}{16\,a \left ( 1+\sin \left ( fx+e \right ) \right ) \left ( -1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f}{\frac{1}{\sqrt{a \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x)
[Out]
-1/16/a*(-2*cos(f*x+e)^2*sin(f*x+e)+4*sin(f*x+e)+(ln(-1+sin(f*x+e))-ln(1+sin(f*x+e)))*cos(f*x+e)^4)/(1+sin(f*x
+e))/(-1+sin(f*x+e))/cos(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f
________________________________________________________________________________________
Maxima [B] time = 2.65676, size = 2068, normalized size = 19.51 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")
[Out]
-1/16*(4*(sin(7*f*x + 7*e) - 7*sin(5*f*x + 5*e) + 7*sin(3*f*x + 3*e) - sin(f*x + e))*cos(8*f*x + 8*e) - 8*(2*s
in(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*cos(7*f*x + 7*e) - 16*(7*sin(5*f*x + 5*e) - 7*sin(3
*f*x + 3*e) + sin(f*x + e))*cos(6*f*x + 6*e) + 56*(3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*cos(5*f*x + 5*e) +
24*(7*sin(3*f*x + 3*e) - sin(f*x + e))*cos(4*f*x + 4*e) + (2*(4*cos(6*f*x + 6*e) + 6*cos(4*f*x + 4*e) + 4*cos
(2*f*x + 2*e) + 1)*cos(8*f*x + 8*e) + cos(8*f*x + 8*e)^2 + 8*(6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) + 1)*cos
(6*f*x + 6*e) + 16*cos(6*f*x + 6*e)^2 + 12*(4*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 36*cos(4*f*x + 4*e)^2 +
16*cos(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(8*f*x + 8*e) + s
in(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(6*f*x + 6*e) + 16*sin(6*f*x + 6*e)^2 + 36
*sin(4*f*x + 4*e)^2 + 48*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 16*sin(2*f*x + 2*e)^2 + 8*cos(2*f*x + 2*e) + 1)*l
og(cos(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) - (2*(4*cos(6*f*x + 6*e) + 6*cos(4*f*x + 4*e) + 4*cos
(2*f*x + 2*e) + 1)*cos(8*f*x + 8*e) + cos(8*f*x + 8*e)^2 + 8*(6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) + 1)*cos
(6*f*x + 6*e) + 16*cos(6*f*x + 6*e)^2 + 12*(4*cos(2*f*x + 2*e) + 1)*cos(4*f*x + 4*e) + 36*cos(4*f*x + 4*e)^2 +
16*cos(2*f*x + 2*e)^2 + 4*(2*sin(6*f*x + 6*e) + 3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(8*f*x + 8*e) + s
in(8*f*x + 8*e)^2 + 16*(3*sin(4*f*x + 4*e) + 2*sin(2*f*x + 2*e))*sin(6*f*x + 6*e) + 16*sin(6*f*x + 6*e)^2 + 36
*sin(4*f*x + 4*e)^2 + 48*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 16*sin(2*f*x + 2*e)^2 + 8*cos(2*f*x + 2*e) + 1)*l
og(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1) - 4*(cos(7*f*x + 7*e) - 7*cos(5*f*x + 5*e) + 7*cos(3*
f*x + 3*e) - cos(f*x + e))*sin(8*f*x + 8*e) + 4*(4*cos(6*f*x + 6*e) + 6*cos(4*f*x + 4*e) + 4*cos(2*f*x + 2*e)
+ 1)*sin(7*f*x + 7*e) + 16*(7*cos(5*f*x + 5*e) - 7*cos(3*f*x + 3*e) + cos(f*x + e))*sin(6*f*x + 6*e) - 28*(6*c
os(4*f*x + 4*e) + 4*cos(2*f*x + 2*e) + 1)*sin(5*f*x + 5*e) - 24*(7*cos(3*f*x + 3*e) - cos(f*x + e))*sin(4*f*x
+ 4*e) + 28*(4*cos(2*f*x + 2*e) + 1)*sin(3*f*x + 3*e) - 112*cos(3*f*x + 3*e)*sin(2*f*x + 2*e) + 16*cos(f*x + e
)*sin(2*f*x + 2*e) - 16*cos(2*f*x + 2*e)*sin(f*x + e) - 4*sin(f*x + e))/((a*cos(8*f*x + 8*e)^2 + 16*a*cos(6*f*
x + 6*e)^2 + 36*a*cos(4*f*x + 4*e)^2 + 16*a*cos(2*f*x + 2*e)^2 + a*sin(8*f*x + 8*e)^2 + 16*a*sin(6*f*x + 6*e)^
2 + 36*a*sin(4*f*x + 4*e)^2 + 48*a*sin(4*f*x + 4*e)*sin(2*f*x + 2*e) + 16*a*sin(2*f*x + 2*e)^2 + 2*(4*a*cos(6*
f*x + 6*e) + 6*a*cos(4*f*x + 4*e) + 4*a*cos(2*f*x + 2*e) + a)*cos(8*f*x + 8*e) + 8*(6*a*cos(4*f*x + 4*e) + 4*a
*cos(2*f*x + 2*e) + a)*cos(6*f*x + 6*e) + 12*(4*a*cos(2*f*x + 2*e) + a)*cos(4*f*x + 4*e) + 8*a*cos(2*f*x + 2*e
) + 4*(2*a*sin(6*f*x + 6*e) + 3*a*sin(4*f*x + 4*e) + 2*a*sin(2*f*x + 2*e))*sin(8*f*x + 8*e) + 16*(3*a*sin(4*f*
x + 4*e) + 2*a*sin(2*f*x + 2*e))*sin(6*f*x + 6*e) + a)*sqrt(a)*f)
________________________________________________________________________________________
Fricas [A] time = 1.71558, size = 205, normalized size = 1.93 \begin{align*} -\frac{{\left (\cos \left (f x + e\right )^{4} \log \left (-\frac{\sin \left (f x + e\right ) + 1}{\sin \left (f x + e\right ) - 1}\right ) + 2 \,{\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right )\right )} \sqrt{a \cos \left (f x + e\right )^{2}}}{16 \, a^{2} f \cos \left (f x + e\right )^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")
[Out]
-1/16*(cos(f*x + e)^4*log(-(sin(f*x + e) + 1)/(sin(f*x + e) - 1)) + 2*(cos(f*x + e)^2 - 2)*sin(f*x + e))*sqrt(
a*cos(f*x + e)^2)/(a^2*f*cos(f*x + e)^5)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{2}{\left (e + f x \right )}}{\left (- a \left (\sin{\left (e + f x \right )} - 1\right ) \left (\sin{\left (e + f x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)**2/(a-a*sin(f*x+e)**2)**(3/2),x)
[Out]
Integral(tan(e + f*x)**2/(-a*(sin(e + f*x) - 1)*(sin(e + f*x) + 1))**(3/2), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{2}}{{\left (-a \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
integrate(tan(f*x + e)^2/(-a*sin(f*x + e)^2 + a)^(3/2), x)