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

Optimal. Leaf size=74 \[ \frac{b^2}{4 a^3 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac{b}{a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac{\log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f} \]

[Out]

b^2/(4*a^3*f*(b + a*Cos[e + f*x]^2)^2) - 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 = 0.07414, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4138, 266, 43} \[ \frac{b^2}{4 a^3 f \left (a \cos ^2(e+f x)+b\right )^2}-\frac{b}{a^3 f \left (a \cos ^2(e+f x)+b\right )}-\frac{\log \left (a \cos ^2(e+f x)+b\right )}{2 a^3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

b^2/(4*a^3*f*(b + a*Cos[e + f*x]^2)^2) - 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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

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 (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x^5}{\left (b+a x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{x^2}{(b+a x)^3} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (\frac{b^2}{a^2 (b+a x)^3}-\frac{2 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{b^2}{4 a^3 f \left (b+a \cos ^2(e+f x)\right )^2}-\frac{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 = 1.5085, size = 129, normalized size = 1.74 \[ -\frac{a^2 \cos ^2(2 (e+f x)) \log (a \cos (2 (e+f x))+a+2 b)+(a+2 b)^2 \log (a \cos (2 (e+f x))+a+2 b)+2 a \cos (2 (e+f x)) ((a+2 b) \log (a \cos (2 (e+f x))+a+2 b)+2 b)+2 b (2 a+3 b)}{2 a^3 f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(2*b*(2*a + 3*b) + (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*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 = 0.038, size = 81, normalized size = 1.1 \begin{align*} -{\frac{\ln \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) }{2\,f{a}^{3}}}+{\frac{1}{2\,f{a}^{2} \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) }}+{\frac{1}{4\,fa \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}+{\frac{\ln \left ( \sec \left ( fx+e \right ) \right ) }{f{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.993384, size = 138, normalized size = 1.86 \begin{align*} \frac{\frac{4 \, a b \sin \left (f x + e\right )^{2} - 4 \, 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)/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 0.573321, size = 242, normalized size = 3.27 \begin{align*} -\frac{4 \, a b \cos \left (f x + e\right )^{2} + 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)/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.79737, size = 1133, normalized size = 15.31 \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)/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")

[Out]

1/4*((3*a^4 + 12*a^3*b + 18*a^2*b^2 + 12*a*b^3 + 3*b^4 + 12*a^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 40*a^3
*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 24*a^2*b^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 16*a*b^3*(cos(f*
x + e) - 1)/(cos(f*x + e) + 1) - 12*b^4*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 18*a^4*(cos(f*x + e) - 1)^2/(c
os(f*x + e) + 1)^2 + 56*a^3*b*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 12*a^2*b^2*(cos(f*x + e) - 1)^2/(cos
(f*x + e) + 1)^2 + 8*a*b^3*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 18*b^4*(cos(f*x + e) - 1)^2/(cos(f*x +
e) + 1)^2 + 12*a^4*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 40*a^3*b*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1
)^3 + 24*a^2*b^2*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 - 16*a*b^3*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^
3 - 12*b^4*(cos(f*x + e) - 1)^3/(cos(f*x + e) + 1)^3 + 3*a^4*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 12*a^
3*b*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 18*a^2*b^2*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 12*a*b^
3*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4 + 3*b^4*(cos(f*x + e) - 1)^4/(cos(f*x + e) + 1)^4)/((a^5 + 2*a^4*b
+ a^3*b^2)*(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) - 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)/f