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

Optimal. Leaf size=69 \[ -\frac {(a+b)^2 \log \left (a \cos ^2(e+f x)+b\right )}{2 a b^2 f}+\frac {(a+2 b) \log (\cos (e+f x))}{b^2 f}+\frac {\sec ^2(e+f x)}{2 b f} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 69, 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, 446, 88} \[ -\frac {(a+b)^2 \log \left (a \cos ^2(e+f x)+b\right )}{2 a b^2 f}+\frac {(a+2 b) \log (\cos (e+f x))}{b^2 f}+\frac {\sec ^2(e+f x)}{2 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

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

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Mathematica [A]  time = 0.27, size = 99, normalized size = 1.43 \[ \frac {\sec ^2(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (a b \sec ^2(e+f x)+(a+b)^2 \left (-\log \left (-a \sin ^2(e+f x)+a+b\right )\right )+2 a (a+2 b) \log (\cos (e+f x))\right )}{4 a b^2 f \left (a+b \sec ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.71, size = 84, normalized size = 1.22 \[ -\frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, {\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} \log \left (-\cos \left (f x + e\right )\right ) - a b}{2 \, a b^{2} f \cos \left (f x + e\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)^5/(a+b*sec(f*x+e)^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)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)2/f*(1/4/a*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+c
os(f*x+exp(1)))+2))+(-b^3-3*b^2*a-3*b*a^2-a^3)/(4*b^3*a+4*b^2*a^2)*ln(abs(((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(
1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1))))*b+((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)
))*(1+cos(f*x+exp(1))))*a+2*b-2*a))+(2*b+a)*1/4/b^2*ln(abs((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*
x+exp(1)))*(1+cos(f*x+exp(1)))-2))+(-2*((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f
*x+exp(1))))*b-((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1))))*a+8*b+2*a)*
1/4/b^2/((1-cos(f*x+exp(1)))/(1+cos(f*x+exp(1)))+1/(1-cos(f*x+exp(1)))*(1+cos(f*x+exp(1)))-2))

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maple [A]  time = 0.73, size = 112, normalized size = 1.62 \[ -\frac {a \ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{2 f \,b^{2}}-\frac {\ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{f b}-\frac {\ln \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right )}{2 a f}+\frac {\ln \left (\cos \left (f x +e \right )\right ) a}{f \,b^{2}}+\frac {2 \ln \left (\cos \left (f x +e \right )\right )}{b f}+\frac {1}{2 f b \cos \left (f x +e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 81, normalized size = 1.17 \[ \frac {\frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b^{2}} - \frac {1}{b \sin \left (f x + e\right )^{2} - b} - \frac {{\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a b^{2}}}{2 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.61, size = 103, normalized size = 1.49 \[ \frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a\,f}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{b\,f}-\frac {\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,a\,f}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,b\,f}-\frac {a\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,b^2\,f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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