3.3.11 \(\int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) [211]

Optimal. Leaf size=71 \[ -\frac {\log (\cos (e+f x))}{(a-b) f}-\frac {a^2 \log \left (a+b \tan ^2(e+f x)\right )}{2 (a-b) b^2 f}+\frac {\tan ^2(e+f x)}{2 b f} \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 71, 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 = {3751, 457, 84} \begin {gather*} -\frac {a^2 \log \left (a+b \tan ^2(e+f x)\right )}{2 b^2 f (a-b)}-\frac {\log (\cos (e+f x))}{f (a-b)}+\frac {\tan ^2(e+f x)}{2 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 457

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 3751

Int[((d_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol]
:> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[c*(ff/f), Subst[Int[(d*ff*(x/c))^m*((a + b*(ff*x)^n)^p/(c^2
+ ff^2*x^2)), x], x, c*(Tan[e + f*x]/ff)], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 64, normalized size = 0.90 \begin {gather*} \frac {-\frac {2 \log (\cos (e+f x))}{a-b}-\frac {a^2 \log \left (a+b \tan ^2(e+f x)\right )}{(a-b) b^2}+\frac {\tan ^2(e+f x)}{b}}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.11, size = 67, normalized size = 0.94

method	result	size
derivativedivides	\(\frac {\frac {\tan ^{2}\left (f x +e \right )}{2 b}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 a -2 b}-\frac {a^{2} \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 b^{2} \left (a -b \right )}}{f}\)	\(67\)
default	\(\frac {\frac {\tan ^{2}\left (f x +e \right )}{2 b}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 a -2 b}-\frac {a^{2} \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 b^{2} \left (a -b \right )}}{f}\)	\(67\)
norman	\(\frac {\tan ^{2}\left (f x +e \right )}{2 b f}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \left (a -b \right )}-\frac {a^{2} \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 \left (a -b \right ) b^{2} f}\)	\(72\)
risch	\(-\frac {i x}{a -b}-\frac {2 i a x}{b^{2}}-\frac {2 i a e}{b^{2} f}-\frac {2 i x}{b}-\frac {2 i e}{b f}+\frac {2 i a^{2} x}{\left (a -b \right ) b^{2}}+\frac {2 i a^{2} e}{\left (a -b \right ) b^{2} f}+\frac {2 \,{\mathrm e}^{2 i \left (f x +e \right )}}{f b \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{b^{2} f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{b f}-\frac {a^{2} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 \left (a -b \right ) b^{2} f}\)	\(206\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 79, normalized size = 1.11 \begin {gather*} -\frac {\frac {a^{2} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a b^{2} - b^{3}} - \frac {{\left (a + 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}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.51, size = 96, normalized size = 1.35 \begin {gather*} -\frac {a^{2} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (a b^{2} - b^{3}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (53) = 106\).
time = 9.00, size = 338, normalized size = 4.76 \begin {gather*} \begin {cases} \tilde {\infty } x \tan ^{3}{\left (e \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {\frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {\tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {\tan ^{2}{\left (e + f x \right )}}{2 f}}{a} & \text {for}\: b = 0 \\- \frac {2 \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{2}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} - \frac {2 \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac {\tan ^{4}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} - \frac {2}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} & \text {for}\: a = b \\\frac {x \tan ^{5}{\left (e \right )}}{a + b \tan ^{2}{\left (e \right )}} & \text {for}\: f = 0 \\- \frac {a^{2} \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b^{2} f - 2 b^{3} f} - \frac {a^{2} \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b^{2} f - 2 b^{3} f} + \frac {a b \tan ^{2}{\left (e + f x \right )}}{2 a b^{2} f - 2 b^{3} f} + \frac {b^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a b^{2} f - 2 b^{3} f} - \frac {b^{2} \tan ^{2}{\left (e + f x \right )}}{2 a b^{2} f - 2 b^{3} f} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*tan(e)**3, Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), ((log(tan(e + f*x)**2 + 1)/(2*f) + tan(e + f*x)**
4/(4*f) - tan(e + f*x)**2/(2*f))/a, Eq(b, 0)), (-2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(2*b*f*tan(e + f*x
)**2 + 2*b*f) - 2*log(tan(e + f*x)**2 + 1)/(2*b*f*tan(e + f*x)**2 + 2*b*f) + tan(e + f*x)**4/(2*b*f*tan(e + f*
x)**2 + 2*b*f) - 2/(2*b*f*tan(e + f*x)**2 + 2*b*f), Eq(a, b)), (x*tan(e)**5/(a + b*tan(e)**2), Eq(f, 0)), (-a*
*2*log(-sqrt(-a/b) + tan(e + f*x))/(2*a*b**2*f - 2*b**3*f) - a**2*log(sqrt(-a/b) + tan(e + f*x))/(2*a*b**2*f -
2*b**3*f) + a*b*tan(e + f*x)**2/(2*a*b**2*f - 2*b**3*f) + b**2*log(tan(e + f*x)**2 + 1)/(2*a*b**2*f - 2*b**3*
f) - b**2*tan(e + f*x)**2/(2*a*b**2*f - 2*b**3*f), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (70) = 140\).
time = 1.59, size = 341, normalized size = 4.80 \begin {gather*} -\frac {\frac {a^{3} \log \left ({\left | -a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} - 2 \, a + 4 \, b \right |}\right )}{a^{2} b^{2} - a b^{3}} - \frac {\log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2 \right |}\right )}{a - b} - \frac {{\left (a + b\right )} \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2 \right |}\right )}{b^{2}} + \frac {a {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + b {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 2 \, a + 6 \, b}{b^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2\right )}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 11.86, size = 74, normalized size = 1.04 \begin {gather*} \frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,\left (a-b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,b\,f}-\frac {a^2\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}{2\,f\,\left (a\,b^2-b^3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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