3.3.24 \(\int \frac {\tan ^5(e+f x)}{(a+b \tan ^2(e+f x))^2} \, dx\) [224]

Optimal. Leaf size=90 \[ -\frac {\log (\cos (e+f x))}{(a-b)^2 f}+\frac {a (a-2 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 (a-b)^2 b^2 f}+\frac {a^2}{2 (a-b) b^2 f \left (a+b \tan ^2(e+f x)\right )} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 90, 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, 90} \begin {gather*} \frac {a^2}{2 b^2 f (a-b) \left (a+b \tan ^2(e+f x)\right )}+\frac {a (a-2 b) \log \left (a+b \tan ^2(e+f x)\right )}{2 b^2 f (a-b)^2}-\frac {\log (\cos (e+f x))}{f (a-b)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 90

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

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Mathematica [A]
time = 0.49, size = 73, normalized size = 0.81 \begin {gather*} \frac {-2 \log (\cos (e+f x))+\frac {a (a-2 b) \log \left (a+b \tan ^2(e+f x)\right )}{b^2}+\frac {a^2 (a-b)}{b^2 \left (a+b \tan ^2(e+f x)\right )}}{2 (a-b)^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.14, size = 83, normalized size = 0.92

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 \left (a -b \right )^{2}}+\frac {a \left (\frac {\left (a -2 b \right ) \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{b^{2}}+\frac {a \left (a -b \right )}{b^{2} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}\right )}{2 \left (a -b \right )^{2}}}{f}\)	\(83\)
default	\(\frac {\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 \left (a -b \right )^{2}}+\frac {a \left (\frac {\left (a -2 b \right ) \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{b^{2}}+\frac {a \left (a -b \right )}{b^{2} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}\right )}{2 \left (a -b \right )^{2}}}{f}\)	\(83\)
norman	\(\frac {a^{2}}{2 \left (a -b \right ) b^{2} f \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \left (a^{2}-2 a b +b^{2}\right )}+\frac {a \left (a -2 b \right ) \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 b^{2} f \left (a^{2}-2 a b +b^{2}\right )}\)	\(103\)
risch	\(-\frac {i x}{a^{2}-2 a b +b^{2}}+\frac {2 i x}{b^{2}}+\frac {2 i e}{b^{2} f}-\frac {2 i a^{2} x}{b^{2} \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 i a^{2} e}{b^{2} f \left (a^{2}-2 a b +b^{2}\right )}+\frac {4 i a x}{b \left (a^{2}-2 a b +b^{2}\right )}+\frac {4 i a e}{b f \left (a^{2}-2 a b +b^{2}\right )}-\frac {2 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{b f \left (a -b \right )^{2} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}-b \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a -b \right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{b^{2} f}+\frac {a^{2} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 b^{2} f \left (a^{2}-2 a b +b^{2}\right )}-\frac {a \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{b f \left (a^{2}-2 a b +b^{2}\right )}\)	\(343\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.28, size = 131, normalized size = 1.46 \begin {gather*} -\frac {\frac {a^{2}}{a^{3} b - 2 \, a^{2} b^{2} + a b^{3} - {\left (a^{3} b - 3 \, a^{2} b^{2} + 3 \, a b^{3} - b^{4}\right )} \sin \left (f x + e\right )^{2}} - \frac {{\left (a^{2} - 2 \, a b\right )} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a^{2} b^{2} - 2 \, a b^{3} + b^{4}} + \frac {\log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b^{2}}}{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)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 193 vs. \(2 (89) = 178\).
time = 2.72, size = 193, normalized size = 2.14 \begin {gather*} -\frac {a^{2} b \tan \left (f x + e\right )^{2} + a^{2} b - {\left (a^{3} - 2 \, a^{2} b + {\left (a^{2} b - 2 \, a b^{2}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) + {\left (a^{3} - 2 \, a^{2} b + a b^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{2}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left ({\left (a^{2} b^{3} - 2 \, a b^{4} + b^{5}\right )} f \tan \left (f x + e\right )^{2} + {\left (a^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}\right )} f\right )}} \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)^2,x, algorithm="fricas")

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 1542 vs. \(2 (71) = 142\).
time = 32.77, size = 1542, normalized size = 17.13 \begin {gather*} \begin {cases} \tilde {\infty } x \tan {\left (e \right )} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\\frac {2 \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{4}{\left (e + f x \right )}}{4 b^{2} f \tan ^{4}{\left (e + f x \right )} + 8 b^{2} f \tan ^{2}{\left (e + f x \right )} + 4 b^{2} f} + \frac {4 \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{2}{\left (e + f x \right )}}{4 b^{2} f \tan ^{4}{\left (e + f x \right )} + 8 b^{2} f \tan ^{2}{\left (e + f x \right )} + 4 b^{2} f} + \frac {2 \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{4 b^{2} f \tan ^{4}{\left (e + f x \right )} + 8 b^{2} f \tan ^{2}{\left (e + f x \right )} + 4 b^{2} f} + \frac {4 \tan ^{2}{\left (e + f x \right )}}{4 b^{2} f \tan ^{4}{\left (e + f x \right )} + 8 b^{2} f \tan ^{2}{\left (e + f x \right )} + 4 b^{2} f} + \frac {3}{4 b^{2} f \tan ^{4}{\left (e + f x \right )} + 8 b^{2} f \tan ^{2}{\left (e + f x \right )} + 4 b^{2} f} & \text {for}\: a = b \\\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^{2}} & \text {for}\: b = 0 \\\frac {x \tan ^{5}{\left (e \right )}}{\left (a + b \tan ^{2}{\left (e \right )}\right )^{2}} & \text {for}\: f = 0 \\\frac {a^{3} \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} + \frac {a^{3} \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} + \frac {a^{3}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} + \frac {a^{2} b \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} - \frac {2 a^{2} b \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} + \frac {a^{2} b \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} - \frac {2 a^{2} b \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} - \frac {a^{2} b}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} - \frac {2 a b^{2} \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} - \frac {2 a b^{2} \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} + \frac {a b^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} + \frac {b^{3} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{2}{\left (e + f x \right )}}{2 a^{3} b^{2} f + 2 a^{2} b^{3} f \tan ^{2}{\left (e + f x \right )} - 4 a^{2} b^{3} f - 4 a b^{4} f \tan ^{2}{\left (e + f x \right )} + 2 a b^{4} f + 2 b^{5} f \tan ^{2}{\left (e + f x \right )}} & \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)**2,x)

[Out]

Piecewise((zoo*x*tan(e), Eq(a, 0) & Eq(b, 0) & Eq(f, 0)), (2*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**4/(4*b**2*
f*tan(e + f*x)**4 + 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f) + 4*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(4*b**2*
f*tan(e + f*x)**4 + 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f) + 2*log(tan(e + f*x)**2 + 1)/(4*b**2*f*tan(e + f*x)**
4 + 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f) + 4*tan(e + f*x)**2/(4*b**2*f*tan(e + f*x)**4 + 8*b**2*f*tan(e + f*x)
**2 + 4*b**2*f) + 3/(4*b**2*f*tan(e + f*x)**4 + 8*b**2*f*tan(e + f*x)**2 + 4*b**2*f), Eq(a, b)), ((log(tan(e +
f*x)**2 + 1)/(2*f) + tan(e + f*x)**4/(4*f) - tan(e + f*x)**2/(2*f))/a**2, Eq(b, 0)), (x*tan(e)**5/(a + b*tan(
e)**2)**2, Eq(f, 0)), (a**3*log(-sqrt(-a/b) + tan(e + f*x))/(2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4
*a**2*b**3*f - 4*a*b**4*f*tan(e + f*x)**2 + 2*a*b**4*f + 2*b**5*f*tan(e + f*x)**2) + a**3*log(sqrt(-a/b) + tan
(e + f*x))/(2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f - 4*a*b**4*f*tan(e + f*x)**2 + 2*a*b
**4*f + 2*b**5*f*tan(e + f*x)**2) + a**3/(2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f - 4*a*
b**4*f*tan(e + f*x)**2 + 2*a*b**4*f + 2*b**5*f*tan(e + f*x)**2) + a**2*b*log(-sqrt(-a/b) + tan(e + f*x))*tan(e
+ f*x)**2/(2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f - 4*a*b**4*f*tan(e + f*x)**2 + 2*a*b
**4*f + 2*b**5*f*tan(e + f*x)**2) - 2*a**2*b*log(-sqrt(-a/b) + tan(e + f*x))/(2*a**3*b**2*f + 2*a**2*b**3*f*ta
n(e + f*x)**2 - 4*a**2*b**3*f - 4*a*b**4*f*tan(e + f*x)**2 + 2*a*b**4*f + 2*b**5*f*tan(e + f*x)**2) + a**2*b*l
og(sqrt(-a/b) + tan(e + f*x))*tan(e + f*x)**2/(2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f -
4*a*b**4*f*tan(e + f*x)**2 + 2*a*b**4*f + 2*b**5*f*tan(e + f*x)**2) - 2*a**2*b*log(sqrt(-a/b) + tan(e + f*x))
/(2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f - 4*a*b**4*f*tan(e + f*x)**2 + 2*a*b**4*f + 2*
b**5*f*tan(e + f*x)**2) - a**2*b/(2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f - 4*a*b**4*f*t
an(e + f*x)**2 + 2*a*b**4*f + 2*b**5*f*tan(e + f*x)**2) - 2*a*b**2*log(-sqrt(-a/b) + tan(e + f*x))*tan(e + f*x
)**2/(2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f - 4*a*b**4*f*tan(e + f*x)**2 + 2*a*b**4*f
+ 2*b**5*f*tan(e + f*x)**2) - 2*a*b**2*log(sqrt(-a/b) + tan(e + f*x))*tan(e + f*x)**2/(2*a**3*b**2*f + 2*a**2*
b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f - 4*a*b**4*f*tan(e + f*x)**2 + 2*a*b**4*f + 2*b**5*f*tan(e + f*x)**2) +
a*b**2*log(tan(e + f*x)**2 + 1)/(2*a**3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f - 4*a*b**4*f*t
an(e + f*x)**2 + 2*a*b**4*f + 2*b**5*f*tan(e + f*x)**2) + b**3*log(tan(e + f*x)**2 + 1)*tan(e + f*x)**2/(2*a**
3*b**2*f + 2*a**2*b**3*f*tan(e + f*x)**2 - 4*a**2*b**3*f - 4*a*b**4*f*tan(e + f*x)**2 + 2*a*b**4*f + 2*b**5*f*
tan(e + f*x)**2), True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 399 vs. \(2 (89) = 178\).
time = 1.75, size = 399, normalized size = 4.43 \begin {gather*} \frac {\frac {{\left (a^{3} - 2 \, a^{2} b\right )} \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^{3} b^{2} - 2 \, a^{2} b^{3} + a b^{4}} + \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^{2} - 2 \, a b + b^{2}} - \frac {a^{3} {\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^{2} 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^{3} - 12 \, a^{2} b + 12 \, a b^{2}}{{\left (a^{2} b^{2} - 2 \, a b^{3} + b^{4}\right )} {\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 )}} - \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 )}{b^{2}}}{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)^2,x, algorithm="giac")

[Out]

1/2*((a^3 - 2*a^2*b)*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^3*b^2 - 2*a^2*b^3 + a*b^4) + log(abs(-(cos(f*x + e) + 1)/(cos(f*x + e) - 1) - (cos(f*x + e)
- 1)/(cos(f*x + e) + 1) + 2))/(a^2 - 2*a*b + b^2) - (a^3*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x +
e) - 1)/(cos(f*x + e) + 1)) - 2*a^2*b*((cos(f*x + e) + 1)/(cos(f*x + e) - 1) + (cos(f*x + e) - 1)/(cos(f*x + e
) + 1)) + 2*a^3 - 12*a^2*b + 12*a*b^2)/((a^2*b^2 - 2*a*b^3 + b^4)*(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)) - 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)/f

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Mupad [B]
time = 11.59, size = 90, normalized size = 1.00 \begin {gather*} \frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,{\left (a-b\right )}^2}+\frac {a^2}{2\,b^2\,f\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (a-b\right )}+\frac {a\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )\,\left (a-2\,b\right )}{2\,b^2\,f\,{\left (a-b\right )}^2} \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)^2,x)

[Out]

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

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