Optimal. Leaf size=71 \[ -\frac {a^2 \log \left (a+b \tan ^2(e+f x)\right )}{2 b^2 f (a-b)}-\frac {\log (\cos (e+f x))}{f (a-b)}+\frac {\tan ^2(e+f x)}{2 b f} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.10, 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 = {3670, 446, 72} \[ -\frac {a^2 \log \left (a+b \tan ^2(e+f x)\right )}{2 b^2 f (a-b)}-\frac {\log (\cos (e+f x))}{f (a-b)}+\frac {\tan ^2(e+f x)}{2 b f} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 72
Rule 446
Rule 3670
Rubi steps
\begin {align*} \int \frac {\tan ^5(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {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 {\operatorname {Subst}\left (\int \frac {x^2}{(1+x) (a+b x)} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac {\operatorname {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*}
________________________________________________________________________________________
Mathematica [A] time = 0.17, size = 64, normalized size = 0.90 \[ \frac {-\frac {a^2 \log \left (a+b \tan ^2(e+f x)\right )}{b^2 (a-b)}-\frac {2 \log (\cos (e+f x))}{a-b}+\frac {\tan ^2(e+f x)}{b}}{2 f} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
fricas [A] time = 0.48, size = 92, normalized size = 1.30 \[ -\frac {a^{2} \log \left (\frac {b \tan \left (f x + e\right )^{2} + a}{\tan \left (f x + e\right )^{2} + 1}\right ) - {\left (a b - b^{2}\right )} \tan \left (f x + e\right )^{2} - {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{\tan \left (f x + e\right )^{2} + 1}\right )}{2 \, {\left (a b^{2} - b^{3}\right )} f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
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]
[Out]
________________________________________________________________________________________
maple [A] time = 0.18, size = 72, normalized size = 1.01 \[ \frac {\tan ^{2}\left (f x +e \right )}{2 b f}-\frac {a^{2} \ln \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )}{2 \left (a -b \right ) b^{2} f}+\frac {\ln \left (1+\tan ^{2}\left (f x +e \right )\right )}{2 f \left (a -b \right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
maxima [A] time = 0.76, size = 76, normalized size = 1.07 \[ -\frac {\frac {a^{2} \log \left (-{\left (a - b\right )} \sin \left (f x + e\right )^{2} + a\right )}{a b^{2} - b^{3}} - \frac {{\left (a + b\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right )}{b^{2}} + \frac {1}{b \sin \left (f x + e\right )^{2} - b}}{2 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
mupad [B] time = 11.86, size = 74, normalized size = 1.04 \[ \frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f\,\left (a-b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,b\,f}-\frac {a^2\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}{2\,f\,\left (a\,b^2-b^3\right )} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
sympy [A] time = 16.36, size = 348, normalized size = 4.90 \[ \begin {cases} \tilde {\infty } x \tan ^{3}{\relax (e )} & \text {for}\: a = 0 \wedge b = 0 \wedge f = 0 \\- \frac {2 \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )} \tan ^{2}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} - \frac {2 \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} + \frac {\tan ^{4}{\left (e + f x \right )}}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} - \frac {2}{2 b f \tan ^{2}{\left (e + f x \right )} + 2 b f} & \text {for}\: a = b \\\frac {\frac {\log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 f} + \frac {\tan ^{4}{\left (e + f x \right )}}{4 f} - \frac {\tan ^{2}{\left (e + f x \right )}}{2 f}}{a} & \text {for}\: b = 0 \\\frac {x \tan ^{5}{\relax (e )}}{a + b \tan ^{2}{\relax (e )}} & \text {for}\: f = 0 \\- \frac {a^{2} \log {\left (- i \sqrt {a} \sqrt {\frac {1}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b^{2} f - 2 b^{3} f} - \frac {a^{2} \log {\left (i \sqrt {a} \sqrt {\frac {1}{b}} + \tan {\left (e + f x \right )} \right )}}{2 a b^{2} f - 2 b^{3} f} + \frac {a b \tan ^{2}{\left (e + f x \right )}}{2 a b^{2} f - 2 b^{3} f} + \frac {b^{2} \log {\left (\tan ^{2}{\left (e + f x \right )} + 1 \right )}}{2 a b^{2} f - 2 b^{3} f} - \frac {b^{2} \tan ^{2}{\left (e + f x \right )}}{2 a b^{2} f - 2 b^{3} f} & \text {otherwise} \end {cases} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________