3.363 \(\int \frac{\tan ^5(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=78 \[ \frac{(a+b)^2}{4 a^3 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac{a+b}{a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac{\log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f} \]

[Out]

(a + b)^2/(4*a^3*f*(b + a*Cos[e + f*x]^2)^2) - (a + b)/(a^3*f*(b + a*Cos[e + f*x]^2)) - Log[b + a*Cos[e + f*x]
^2]/(2*a^3*f)

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Rubi [A]  time = 0.107703, antiderivative size = 78, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4138, 444, 43} \[ \frac{(a+b)^2}{4 a^3 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac{a+b}{a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac{\log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

(a + b)^2/(4*a^3*f*(b + a*Cos[e + f*x]^2)^2) - (a + b)/(a^3*f*(b + a*Cos[e + f*x]^2)) - Log[b + a*Cos[e + f*x]
^2]/(2*a^3*f)

Rule 4138

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> Module[{ff =
FreeFactors[Cos[e + f*x], x]}, -Dist[(f*ff^(m + n*p - 1))^(-1), Subst[Int[((1 - ff^2*x^2)^((m - 1)/2)*(b + a*
(ff*x)^n)^p)/x^(m + n*p), x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, n}, x] && IntegerQ[(m - 1)/2] &&
IntegerQ[n] && IntegerQ[p]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 2.19357, size = 136, normalized size = 1.74 \[ -\frac{2 \left (a^2+4 a b+3 b^2\right )+a^2 \cos ^2(2 (e+f x)) \log (a \cos (2 (e+f x))+a+2 b)+(a+2 b)^2 \log (a \cos (2 (e+f x))+a+2 b)+2 a \cos (2 (e+f x)) ((a+2 b) \log (a \cos (2 (e+f x))+a+2 b)+2 (a+b))}{2 a^3 f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^5/(a + b*Sec[e + f*x]^2)^3,x]

[Out]

-(2*(a^2 + 4*a*b + 3*b^2) + (a + 2*b)^2*Log[a + 2*b + a*Cos[2*(e + f*x)]] + a^2*Cos[2*(e + f*x)]^2*Log[a + 2*b
+ a*Cos[2*(e + f*x)]] + 2*a*Cos[2*(e + f*x)]*(2*(a + b) + (a + 2*b)*Log[a + 2*b + a*Cos[2*(e + f*x)]]))/(2*a^
3*f*(a + 2*b + a*Cos[2*(e + f*x)])^2)

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Maple [A]  time = 0.089, size = 138, normalized size = 1.8 \begin{align*} -{\frac{\ln \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,{a}^{3}f}}-{\frac{1}{f{a}^{2} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{b}{{a}^{3}f \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{1}{4\,fa \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{b}{2\,f{a}^{2} \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{{b}^{2}}{4\,{a}^{3}f \left ( b+a \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x)

[Out]

-1/2*ln(b+a*cos(f*x+e)^2)/a^3/f-1/f/a^2/(b+a*cos(f*x+e)^2)-b/a^3/f/(b+a*cos(f*x+e)^2)+1/4/f/a/(b+a*cos(f*x+e)^
2)^2+1/2/f/a^2/(b+a*cos(f*x+e)^2)^2*b+1/4*b^2/a^3/f/(b+a*cos(f*x+e)^2)^2

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Maxima [A]  time = 1.02008, size = 151, normalized size = 1.94 \begin{align*} \frac{\frac{4 \,{\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{2} - 3 \, a^{2} - 6 \, a b - 3 \, b^{2}}{a^{5} \sin \left (f x + e\right )^{4} + a^{5} + 2 \, a^{4} b + a^{3} b^{2} - 2 \,{\left (a^{5} + a^{4} b\right )} \sin \left (f x + e\right )^{2}} - \frac{2 \, \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{3}}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

1/4*((4*(a^2 + a*b)*sin(f*x + e)^2 - 3*a^2 - 6*a*b - 3*b^2)/(a^5*sin(f*x + e)^4 + a^5 + 2*a^4*b + a^3*b^2 - 2*
(a^5 + a^4*b)*sin(f*x + e)^2) - 2*log(a*sin(f*x + e)^2 - a - b)/a^3)/f

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Fricas [A]  time = 0.574916, size = 271, normalized size = 3.47 \begin{align*} -\frac{4 \,{\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} - a^{2} + 2 \, a b + 3 \, b^{2} + 2 \,{\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right )}{4 \,{\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

-1/4*(4*(a^2 + a*b)*cos(f*x + e)^2 - a^2 + 2*a*b + 3*b^2 + 2*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)
*log(a*cos(f*x + e)^2 + b))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(f*x+e)**5/(a+b*sec(f*x+e)**2)**3,x)

[Out]

Timed out

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Giac [B]  time = 2.91474, size = 772, normalized size = 9.9 \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)^5/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

-1/4*(2*log(a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*
(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)/a^3 - 4*log(-(cos(f*x
+ e) - 1)/(cos(f*x + e) + 1) + 1)/a^3 - (3*a^2 + 6*a*b + 3*b^2 + 20*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1)
+ 8*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 12*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 50*a^2*(cos(f*
x + e) - 1)^2/(cos(f*x + e) + 1)^2 - 28*a*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 18*b^2*(cos(f*x + e) -
1)^2/(cos(f*x + e) + 1)^2 + 20*a^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 8*a*b*(cos(f*x + e) - 1)^3/(co
s(f*x + e) + 1)^3 - 12*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 3*a^2*(cos(f*x + e) - 1)^4/(cos(f*x + e
) + 1)^4 + 6*a*b*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 3*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/
((a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + a*(cos(f*x +
e) - 1)^2/(cos(f*x + e) + 1)^2 + b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)^2*a^3))/f