3.267 \(\int \tan ^2(x) (a+a \tan ^2(x))^{3/2} \, dx\)
Optimal. Leaf size=59 \[ \frac{1}{4} a \tan (x) \sec ^2(x) \sqrt{a \sec ^2(x)}-\frac{1}{8} a \tan (x) \sqrt{a \sec ^2(x)}-\frac{1}{8} a \cos (x) \sqrt{a \sec ^2(x)} \tanh ^{-1}(\sin (x)) \]
[Out]
-(a*ArcTanh[Sin[x]]*Cos[x]*Sqrt[a*Sec[x]^2])/8 - (a*Sqrt[a*Sec[x]^2]*Tan[x])/8 + (a*Sec[x]^2*Sqrt[a*Sec[x]^2]*
Tan[x])/4
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Rubi [A] time = 0.119628, antiderivative size = 59, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used =
{3657, 4125, 2611, 3768, 3770} \[ \frac{1}{4} a \tan (x) \sec ^2(x) \sqrt{a \sec ^2(x)}-\frac{1}{8} a \tan (x) \sqrt{a \sec ^2(x)}-\frac{1}{8} a \cos (x) \sqrt{a \sec ^2(x)} \tanh ^{-1}(\sin (x)) \]
Antiderivative was successfully verified.
[In]
Int[Tan[x]^2*(a + a*Tan[x]^2)^(3/2),x]
[Out]
-(a*ArcTanh[Sin[x]]*Cos[x]*Sqrt[a*Sec[x]^2])/8 - (a*Sqrt[a*Sec[x]^2]*Tan[x])/8 + (a*Sec[x]^2*Sqrt[a*Sec[x]^2]*
Tan[x])/4
Rule 3657
Int[(u_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*sec[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a, b]
Rule 4125
Int[(u_.)*((b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sec[e + f*x]^n)^FracPart[p])/(Sec[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sec[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 \tan ^2(x) \left (a+a \tan ^2(x)\right )^{3/2} \, dx &=\int \left (a \sec ^2(x)\right )^{3/2} \tan ^2(x) \, dx\\ &=\left (a \cos (x) \sqrt{a \sec ^2(x)}\right ) \int \sec ^3(x) \tan ^2(x) \, dx\\ &=\frac{1}{4} a \sec ^2(x) \sqrt{a \sec ^2(x)} \tan (x)-\frac{1}{4} \left (a \cos (x) \sqrt{a \sec ^2(x)}\right ) \int \sec ^3(x) \, dx\\ &=-\frac{1}{8} a \sqrt{a \sec ^2(x)} \tan (x)+\frac{1}{4} a \sec ^2(x) \sqrt{a \sec ^2(x)} \tan (x)-\frac{1}{8} \left (a \cos (x) \sqrt{a \sec ^2(x)}\right ) \int \sec (x) \, dx\\ &=-\frac{1}{8} a \tanh ^{-1}(\sin (x)) \cos (x) \sqrt{a \sec ^2(x)}-\frac{1}{8} a \sqrt{a \sec ^2(x)} \tan (x)+\frac{1}{4} a \sec ^2(x) \sqrt{a \sec ^2(x)} \tan (x)\\ \end{align*}
Mathematica [A] time = 0.0634659, size = 34, normalized size = 0.58 \[ \frac{1}{8} \left (a \sec ^2(x)\right )^{3/2} \left (2 \tan (x)-\sin (x) \cos (x)+\cos ^3(x) \left (-\tanh ^{-1}(\sin (x))\right )\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[Tan[x]^2*(a + a*Tan[x]^2)^(3/2),x]
[Out]
((a*Sec[x]^2)^(3/2)*(-(ArcTanh[Sin[x]]*Cos[x]^3) - Cos[x]*Sin[x] + 2*Tan[x]))/8
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Maple [A] time = 0.016, size = 54, normalized size = 0.9 \begin{align*}{\frac{\tan \left ( x \right ) }{4} \left ( a+a \left ( \tan \left ( x \right ) \right ) ^{2} \right ) ^{{\frac{3}{2}}}}-{\frac{a\tan \left ( x \right ) }{8}\sqrt{a+a \left ( \tan \left ( x \right ) \right ) ^{2}}}-{\frac{1}{8}{a}^{{\frac{3}{2}}}\ln \left ( \sqrt{a}\tan \left ( x \right ) +\sqrt{a+a \left ( \tan \left ( x \right ) \right ) ^{2}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(x)^2*(a+a*tan(x)^2)^(3/2),x)
[Out]
1/4*tan(x)*(a+a*tan(x)^2)^(3/2)-1/8*a*tan(x)*(a+a*tan(x)^2)^(1/2)-1/8*a^(3/2)*ln(a^(1/2)*tan(x)+(a+a*tan(x)^2)
^(1/2))
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Maxima [B] time = 2.86221, size = 1261, normalized size = 21.37 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(x)^2*(a+a*tan(x)^2)^(3/2),x, algorithm="maxima")
[Out]
1/16*(112*a*cos(3*x)*sin(2*x) - 16*a*cos(x)*sin(2*x) + 16*a*cos(2*x)*sin(x) - 4*(a*sin(7*x) - 7*a*sin(5*x) + 7
*a*sin(3*x) - a*sin(x))*cos(8*x) + 8*(2*a*sin(6*x) + 3*a*sin(4*x) + 2*a*sin(2*x))*cos(7*x) + 16*(7*a*sin(5*x)
- 7*a*sin(3*x) + a*sin(x))*cos(6*x) - 56*(3*a*sin(4*x) + 2*a*sin(2*x))*cos(5*x) - 24*(7*a*sin(3*x) - a*sin(x))
*cos(4*x) - (a*cos(8*x)^2 + 16*a*cos(6*x)^2 + 36*a*cos(4*x)^2 + 16*a*cos(2*x)^2 + a*sin(8*x)^2 + 16*a*sin(6*x)
^2 + 36*a*sin(4*x)^2 + 48*a*sin(4*x)*sin(2*x) + 16*a*sin(2*x)^2 + 2*(4*a*cos(6*x) + 6*a*cos(4*x) + 4*a*cos(2*x
) + a)*cos(8*x) + 8*(6*a*cos(4*x) + 4*a*cos(2*x) + a)*cos(6*x) + 12*(4*a*cos(2*x) + a)*cos(4*x) + 8*a*cos(2*x)
+ 4*(2*a*sin(6*x) + 3*a*sin(4*x) + 2*a*sin(2*x))*sin(8*x) + 16*(3*a*sin(4*x) + 2*a*sin(2*x))*sin(6*x) + a)*lo
g(cos(x)^2 + sin(x)^2 + 2*sin(x) + 1) + (a*cos(8*x)^2 + 16*a*cos(6*x)^2 + 36*a*cos(4*x)^2 + 16*a*cos(2*x)^2 +
a*sin(8*x)^2 + 16*a*sin(6*x)^2 + 36*a*sin(4*x)^2 + 48*a*sin(4*x)*sin(2*x) + 16*a*sin(2*x)^2 + 2*(4*a*cos(6*x)
+ 6*a*cos(4*x) + 4*a*cos(2*x) + a)*cos(8*x) + 8*(6*a*cos(4*x) + 4*a*cos(2*x) + a)*cos(6*x) + 12*(4*a*cos(2*x)
+ a)*cos(4*x) + 8*a*cos(2*x) + 4*(2*a*sin(6*x) + 3*a*sin(4*x) + 2*a*sin(2*x))*sin(8*x) + 16*(3*a*sin(4*x) + 2*
a*sin(2*x))*sin(6*x) + a)*log(cos(x)^2 + sin(x)^2 - 2*sin(x) + 1) + 4*(a*cos(7*x) - 7*a*cos(5*x) + 7*a*cos(3*x
) - a*cos(x))*sin(8*x) - 4*(4*a*cos(6*x) + 6*a*cos(4*x) + 4*a*cos(2*x) + a)*sin(7*x) - 16*(7*a*cos(5*x) - 7*a*
cos(3*x) + a*cos(x))*sin(6*x) + 28*(6*a*cos(4*x) + 4*a*cos(2*x) + a)*sin(5*x) + 24*(7*a*cos(3*x) - a*cos(x))*s
in(4*x) - 28*(4*a*cos(2*x) + a)*sin(3*x) + 4*a*sin(x))*sqrt(a)/(2*(4*cos(6*x) + 6*cos(4*x) + 4*cos(2*x) + 1)*c
os(8*x) + cos(8*x)^2 + 8*(6*cos(4*x) + 4*cos(2*x) + 1)*cos(6*x) + 16*cos(6*x)^2 + 12*(4*cos(2*x) + 1)*cos(4*x)
+ 36*cos(4*x)^2 + 16*cos(2*x)^2 + 4*(2*sin(6*x) + 3*sin(4*x) + 2*sin(2*x))*sin(8*x) + sin(8*x)^2 + 16*(3*sin(
4*x) + 2*sin(2*x))*sin(6*x) + 16*sin(6*x)^2 + 36*sin(4*x)^2 + 48*sin(4*x)*sin(2*x) + 16*sin(2*x)^2 + 8*cos(2*x
) + 1)
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Fricas [A] time = 1.37776, size = 174, normalized size = 2.95 \begin{align*} \frac{1}{16} \, a^{\frac{3}{2}} \log \left (2 \, a \tan \left (x\right )^{2} - 2 \, \sqrt{a \tan \left (x\right )^{2} + a} \sqrt{a} \tan \left (x\right ) + a\right ) + \frac{1}{8} \,{\left (2 \, a \tan \left (x\right )^{3} + a \tan \left (x\right )\right )} \sqrt{a \tan \left (x\right )^{2} + a} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(x)^2*(a+a*tan(x)^2)^(3/2),x, algorithm="fricas")
[Out]
1/16*a^(3/2)*log(2*a*tan(x)^2 - 2*sqrt(a*tan(x)^2 + a)*sqrt(a)*tan(x) + a) + 1/8*(2*a*tan(x)^3 + a*tan(x))*sqr
t(a*tan(x)^2 + a)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\tan ^{2}{\left (x \right )} + 1\right )\right )^{\frac{3}{2}} \tan ^{2}{\left (x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(x)**2*(a+a*tan(x)**2)**(3/2),x)
[Out]
Integral((a*(tan(x)**2 + 1))**(3/2)*tan(x)**2, x)
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Giac [A] time = 1.11016, size = 66, normalized size = 1.12 \begin{align*} \frac{1}{8} \,{\left (\sqrt{a \tan \left (x\right )^{2} + a}{\left (2 \, \tan \left (x\right )^{2} + 1\right )} \tan \left (x\right ) + \sqrt{a} \log \left ({\left | -\sqrt{a} \tan \left (x\right ) + \sqrt{a \tan \left (x\right )^{2} + a} \right |}\right )\right )} a \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(tan(x)^2*(a+a*tan(x)^2)^(3/2),x, algorithm="giac")
[Out]
1/8*(sqrt(a*tan(x)^2 + a)*(2*tan(x)^2 + 1)*tan(x) + sqrt(a)*log(abs(-sqrt(a)*tan(x) + sqrt(a*tan(x)^2 + a))))*
a