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

Optimal. Leaf size=51 \[ \frac {a+b}{2 a^2 f \left (a \cos ^2(e+f x)+b\right )}+\frac {\log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4138, 444, 43} \[ \frac {a+b}{2 a^2 f \left (a \cos ^2(e+f x)+b\right )}+\frac {\log \left (a \cos ^2(e+f x)+b\right )}{2 a^2 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.72, size = 81, normalized size = 1.59 \[ \frac {(a+2 b) \log (a \cos (2 (e+f x))+a+2 b)+a \cos (2 (e+f x)) \log (a \cos (2 (e+f x))+a+2 b)+2 (a+b)}{2 a^2 f (a \cos (2 (e+f x))+a+2 b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.54, size = 53, normalized size = 1.04 \[ \frac {{\left (a \cos \left (f x + e\right )^{2} + b\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) + a + b}{2 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} + a^{2} b f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)2/f*(-1/2/a^2*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1))+1/4/a^2*ln(((1-cos(f*x+exp(1)))/(1+cos(f*
x+exp(1))))^2*b+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*a+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b-2*(1
-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+b+a)+(-((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b-((1-cos(f*x+exp(1
)))/(1+cos(f*x+exp(1))))^2*a-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b+6*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(
1)))*a-b-a)*1/4/a^2/(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2*b+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1))))^2
*a+2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*b-2*(1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))*a+b+a))

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maple [A]  time = 1.00, size = 68, normalized size = 1.33 \[ \frac {\ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{2 a^{2} f}+\frac {1}{2 f a \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}+\frac {b}{2 a^{2} f \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.36, size = 59, normalized size = 1.16 \[ -\frac {\frac {a + b}{a^{3} \sin \left (f x + e\right )^{2} - a^{3} - a^{2} b} - \frac {\log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.59, size = 97, normalized size = 1.90 \[ -\frac {\mathrm {atanh}\left (\frac {4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{8\,b^2+\frac {8\,b^3}{a}+4\,b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2+\frac {8\,b^3\,{\mathrm {tan}\left (e+f\,x\right )}^2}{a}}\right )}{a^2\,f}-\frac {a+b}{2\,a\,b\,f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- atanh((4*b^2*tan(e + f*x)^2)/(8*b^2 + (8*b^3)/a + 4*b^2*tan(e + f*x)^2 + (8*b^3*tan(e + f*x)^2)/a))/(a^2*f)
- (a + b)/(2*a*b*f*(a + b + b*tan(e + f*x)^2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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