\(\int \frac {\tan ^2(e+f x)}{(a-a \sin ^2(e+f x))^{3/2}} \, dx\) [483]
Optimal result
Integrand size = 26, antiderivative size = 106 \[
\int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\text {arctanh}(\sin (e+f x)) \cos (e+f x)}{8 a f \sqrt {a \cos ^2(e+f x)}}-\frac {\tan (e+f x)}{8 a f \sqrt {a \cos ^2(e+f x)}}+\frac {\sec ^2(e+f x) \tan (e+f x)}{4 a f \sqrt {a \cos ^2(e+f x)}}
\]
[Out]
-1/8*arctanh(sin(f*x+e))*cos(f*x+e)/a/f/(a*cos(f*x+e)^2)^(1/2)-1/8*tan(f*x+e)/a/f/(a*cos(f*x+e)^2)^(1/2)+1/4*s
ec(f*x+e)^2*tan(f*x+e)/a/f/(a*cos(f*x+e)^2)^(1/2)
Rubi [A] (verified)
Time = 0.28 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of
steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3255, 3286, 2691, 3853, 3855}
\[
\int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {\cos (e+f x) \text {arctanh}(\sin (e+f x))}{8 a f \sqrt {a \cos ^2(e+f x)}}-\frac {\tan (e+f x)}{8 a f \sqrt {a \cos ^2(e+f x)}}+\frac {\tan (e+f x) \sec ^2(e+f x)}{4 a f \sqrt {a \cos ^2(e+f x)}}
\]
[In]
Int[Tan[e + f*x]^2/(a - a*Sin[e + f*x]^2)^(3/2),x]
[Out]
-1/8*(ArcTanh[Sin[e + f*x]]*Cos[e + f*x])/(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 2691
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 3255
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 3286
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 3853
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 3855
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps \begin{align*}
\text {integral}& = \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 {\text {arctanh}(\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] (verified)
Time = 0.10 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.56
\[
\int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\frac {-\text {arctanh}(\sin (e+f x)) \cos (e+f x)+\left (-1+2 \sec ^2(e+f x)\right ) \tan (e+f x)}{8 a f \sqrt {a \cos ^2(e+f x)}}
\]
[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] (verified)
Time = 0.82 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.98
| | |
| method | result | size |
| | |
| default |
\(\frac {\left (\ln \left (1+\sin \left (f x +e \right )\right )-\ln \left (\sin \left (f x +e \right )-1\right )\right ) \left (\cos ^{4}\left (f x +e \right )\right )+2 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-4 \sin \left (f x +e \right )}{16 a \left (1+\sin \left (f x +e \right )\right ) \left (\sin \left (f x +e \right )-1\right ) \cos \left (f x +e \right ) \sqrt {a \left (\cos ^{2}\left (f x +e \right )\right )}\, f}\) |
\(104\) |
| risch |
\(\frac {i \left ({\mathrm e}^{6 i \left (f x +e \right )}-7 \,{\mathrm e}^{4 i \left (f x +e \right )}+7 \,{\mathrm e}^{2 i \left (f x +e \right )}-1\right )}{4 a \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{3} \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, f}-\frac {\ln \left ({\mathrm e}^{i f x}+i {\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{4 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, a}+\frac {\ln \left ({\mathrm e}^{i f x}-i {\mathrm e}^{-i e}\right ) \cos \left (f x +e \right )}{4 f \sqrt {\left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2} a \,{\mathrm e}^{-2 i \left (f x +e \right )}}\, a}\) |
\(195\) |
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[In]
int(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x,method=_RETURNVERBOSE)
[Out]
1/16/a*((ln(1+sin(f*x+e))-ln(sin(f*x+e)-1))*cos(f*x+e)^4+2*cos(f*x+e)^2*sin(f*x+e)-4*sin(f*x+e))/(1+sin(f*x+e)
)/(sin(f*x+e)-1)/cos(f*x+e)/(a*cos(f*x+e)^2)^(1/2)/f
Fricas [A] (verification not implemented)
none
Time = 0.29 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73
\[
\int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (\cos \left (f x + e\right )^{4} \log \left (-\frac {\sin \left (f x + e\right ) + 1}{\sin \left (f x + e\right ) - 1}\right ) + 2 \, {\left (\cos \left (f x + e\right )^{2} - 2\right )} \sin \left (f x + e\right )\right )} \sqrt {a \cos \left (f x + e\right )^{2}}}{16 \, a^{2} f \cos \left (f x + e\right )^{5}}
\]
[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]
\[
\int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {\tan ^{2}{\left (e + f x \right )}}{\left (- a \left (\sin {\left (e + f x \right )} - 1\right ) \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx
\]
[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)
Maxima [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 1532 vs. \(2 (94) = 188\).
Time = 0.53 (sec) , antiderivative size = 1532, normalized size of antiderivative = 14.45
\[
\int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\text {Too large to display}
\]
[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)
Giac [A] (verification not implemented)
none
Time = 0.83 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.93
\[
\int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=-\frac {{\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{3} + \frac {4}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + 4 \, \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{4 \, {\left ({\left (\frac {1}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}^{2} - 4\right )}^{2} a^{\frac {3}{2}} f \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 1\right )}
\]
[In]
integrate(tan(f*x+e)^2/(a-a*sin(f*x+e)^2)^(3/2),x, algorithm="giac")
[Out]
-1/4*((1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e))^3 + 4/tan(1/2*f*x + 1/2*e) + 4*tan(1/2*f*x + 1/2*e))/(((
1/tan(1/2*f*x + 1/2*e) + tan(1/2*f*x + 1/2*e))^2 - 4)^2*a^(3/2)*f*sgn(tan(1/2*f*x + 1/2*e)^4 - 1))
Mupad [F(-1)]
Timed out. \[
\int \frac {\tan ^2(e+f x)}{\left (a-a \sin ^2(e+f x)\right )^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{{\left (a-a\,{\sin \left (e+f\,x\right )}^2\right )}^{3/2}} \,d x
\]
[In]
int(tan(e + f*x)^2/(a - a*sin(e + f*x)^2)^(3/2),x)
[Out]
int(tan(e + f*x)^2/(a - a*sin(e + f*x)^2)^(3/2), x)