Optimal. Leaf size=116 \[ \frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b f}-\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}+\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.14503, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3670, 446, 80, 50, 63, 208} \[ \frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b f}-\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}-\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}+\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 3670
Rule 446
Rule 80
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \tan ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3 \left (a+b x^2\right )^{3/2}}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{x (a+b x)^{3/2}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=\frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b f}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b f}-\frac{(a-b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1+x} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b f}-\frac{(a-b)^2 \operatorname{Subst}\left (\int \frac{1}{(1+x) \sqrt{a+b x}} \, dx,x,\tan ^2(e+f x)\right )}{2 f}\\ &=-\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b f}-\frac{(a-b)^2 \operatorname{Subst}\left (\int \frac{1}{1-\frac{a}{b}+\frac{x^2}{b}} \, dx,x,\sqrt{a+b \tan ^2(e+f x)}\right )}{b f}\\ &=\frac{(a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{f}-\frac{(a-b) \sqrt{a+b \tan ^2(e+f x)}}{f}-\frac{\left (a+b \tan ^2(e+f x)\right )^{3/2}}{3 f}+\frac{\left (a+b \tan ^2(e+f x)\right )^{5/2}}{5 b f}\\ \end{align*}
Mathematica [A] time = 0.906409, size = 112, normalized size = 0.97 \[ \frac{\sqrt{a+b \tan ^2(e+f x)} \left (3 a^2+b (6 a-5 b) \tan ^2(e+f x)-20 a b+3 b^2 \tan ^4(e+f x)+15 b^2\right )+15 b (a-b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \tan ^2(e+f x)}}{\sqrt{a-b}}\right )}{15 b f} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.017, size = 204, normalized size = 1.8 \begin{align*}{\frac{1}{5\,fb} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{{\frac{5}{2}}}}-{\frac{b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{3\,f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}-{\frac{4\,a}{3\,f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}+{\frac{b}{f}\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}-{\frac{{b}^{2}}{f}\arctan \left ({\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}}+2\,{\frac{ab}{f\sqrt{-a+b}}\arctan \left ({\frac{\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}}{\sqrt{-a+b}}} \right ) }-{\frac{{a}^{2}}{f}\arctan \left ({\sqrt{a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2}}{\frac{1}{\sqrt{-a+b}}}} \right ){\frac{1}{\sqrt{-a+b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \tan \left (f x + e\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 2.26934, size = 797, normalized size = 6.87 \begin{align*} \left [-\frac{15 \,{\left (a b - b^{2}\right )} \sqrt{a - b} \log \left (-\frac{b^{2} \tan \left (f x + e\right )^{4} + 2 \,{\left (4 \, a b - 3 \, b^{2}\right )} \tan \left (f x + e\right )^{2} - 4 \,{\left (b \tan \left (f x + e\right )^{2} + 2 \, a - b\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{a - b} + 8 \, a^{2} - 8 \, a b + b^{2}}{\tan \left (f x + e\right )^{4} + 2 \, \tan \left (f x + e\right )^{2} + 1}\right ) - 4 \,{\left (3 \, b^{2} \tan \left (f x + e\right )^{4} +{\left (6 \, a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} - 20 \, a b + 15 \, b^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{60 \, b f}, -\frac{15 \,{\left (a b - b^{2}\right )} \sqrt{-a + b} \arctan \left (\frac{2 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} \sqrt{-a + b}}{b \tan \left (f x + e\right )^{2} + 2 \, a - b}\right ) - 2 \,{\left (3 \, b^{2} \tan \left (f x + e\right )^{4} +{\left (6 \, a b - 5 \, b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} - 20 \, a b + 15 \, b^{2}\right )} \sqrt{b \tan \left (f x + e\right )^{2} + a}}{30 \, b f}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b \tan ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}} \tan ^{3}{\left (e + f x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [A] time = 1.26429, size = 203, normalized size = 1.75 \begin{align*} -\frac{{\left (a^{2} - 2 \, a b + b^{2}\right )} \arctan \left (\frac{\sqrt{b \tan \left (f x + e\right )^{2} + a}}{\sqrt{-a + b}}\right )}{\sqrt{-a + b} f} + \frac{3 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}} b^{4} f^{4} - 5 \,{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} b^{5} f^{4} - 15 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} a b^{5} f^{4} + 15 \, \sqrt{b \tan \left (f x + e\right )^{2} + a} b^{6} f^{4}}{15 \, b^{5} f^{5}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]