3.3.12 \(\int \frac {\tan ^3(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) [212]

Optimal. Leaf size=50 \[ \frac {\log (\cos (e+f x))}{(a-b) f}+\frac {a \log \left (a+b \tan ^2(e+f x)\right )}{2 (a-b) b f} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 50, 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, 78} \begin {gather*} \frac {a \log \left (a+b \tan ^2(e+f x)\right )}{2 b f (a-b)}+\frac {\log (\cos (e+f x))}{f (a-b)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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

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Mathematica [A]
time = 0.02, size = 41, normalized size = 0.82 \begin {gather*} \frac {2 b \log (\cos (e+f x))+a \log \left (a+b \tan ^2(e+f x)\right )}{2 a b f-2 b^2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.09, size = 52, normalized size = 1.04

method	result	size
derivativedivides	\(\frac {-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 \left (a -b \right )}+\frac {a \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 \left (a -b \right ) b}}{f}\)	\(52\)
default	\(\frac {-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 \left (a -b \right )}+\frac {a \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 \left (a -b \right ) b}}{f}\)	\(52\)
norman	\(-\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \left (a -b \right )}+\frac {a \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 \left (a -b \right ) b f}\)	\(54\)
risch	\(\frac {i x}{a -b}+\frac {2 i x}{b}+\frac {2 i e}{b f}-\frac {2 i a x}{b \left (a -b \right )}-\frac {2 i a e}{f b \left (a -b \right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{b f}+\frac {a \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a -b}+1\right )}{2 f b \left (a -b \right )}\)	\(132\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 55, normalized size = 1.10 \begin {gather*} \frac {\frac {a \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a b - b^{2}} - \frac {\log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.34, size = 68, normalized size = 1.36 \begin {gather*} \frac {a \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (a - b\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (a b - b^{2}\right )} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 230 vs. \(2 (36) = 72\).
time = 1.93, size = 230, normalized size = 4.60 \begin {gather*} \begin {cases} \tilde {\infty } x \tan {\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 ^{2}{\left (e + f x \right )}}{2 f}}{a} & \text {for}\: b = 0 \\\frac {\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 {\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 {1}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} & \text {for}\: a = b \\\frac {x \tan ^{3}{\left (e \right )}}{a + b \tan ^{2}{\left (e \right )}} & \text {for}\: f = 0 \\\frac {a \log {\left (- \sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b f - 2 b^{2} f} + \frac {a \log {\left (\sqrt {- \frac {a}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b f - 2 b^{2} f} - \frac {b \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a b f - 2 b^{2} f} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 189 vs. \(2 (50) = 100\).
time = 0.84, size = 189, normalized size = 3.78 \begin {gather*} \frac {\frac {a^{2} \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 - a b^{2}} - \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 {\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 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 11.78, size = 54, normalized size = 1.08 \begin {gather*} \frac {a\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}{2\,f\,\left (a\,b-b^2\right )}-\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,\left (a-b\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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