Optimal. Leaf size=165 \[ -\frac{\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)}-\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{(a+b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{5 f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.355179, antiderivative size = 165, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {4141, 1975, 474, 583, 12, 377, 203} \[ -\frac{\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f (a+b)}-\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)+b}}\right )}{f}-\frac{(a+b) \cot ^5(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{5 f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)+b}}{15 f} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 4141
Rule 1975
Rule 474
Rule 583
Rule 12
Rule 377
Rule 203
Rubi steps
\begin{align*} \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b \left (1+x^2\right )\right )^{3/2}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+b+b x^2\right )^{3/2}}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{(a+b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 f}+\frac{\operatorname{Subst}\left (\int \frac{-(5 a-b) (a+b)-(4 a-b) b x^2}{x^4 \left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{5 f}\\ &=\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 f}-\frac{(a+b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 f}-\frac{\operatorname{Subst}\left (\int \frac{-(a+b) \left (15 a^2+10 a b-2 b^2\right )-2 (5 a-b) b (a+b) x^2}{x^2 \left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b) f}\\ &=-\frac{\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 f}-\frac{(a+b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 f}+\frac{\operatorname{Subst}\left (\int -\frac{15 a^2 (a+b)^2}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{15 (a+b)^2 f}\\ &=-\frac{\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 f}-\frac{(a+b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 f}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{\left (1+x^2\right ) \sqrt{a+b+b x^2}} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 f}-\frac{(a+b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 f}-\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{1+a x^2} \, dx,x,\frac{\tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{f}\\ &=-\frac{a^{3/2} \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+b+b \tan ^2(e+f x)}}\right )}{f}-\frac{\left (15 a^2+10 a b-2 b^2\right ) \cot (e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 (a+b) f}+\frac{(5 a-b) \cot ^3(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{15 f}-\frac{(a+b) \cot ^5(e+f x) \sqrt{a+b+b \tan ^2(e+f x)}}{5 f}\\ \end{align*}
Mathematica [C] time = 1.50507, size = 139, normalized size = 0.84 \[ \frac{2 \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \left (\frac{5 (a+b)^2 \text{Hypergeometric2F1}\left (-\frac{3}{2},-\frac{3}{2},-\frac{1}{2},\frac{a \sin ^2(e+f x)}{a+b}\right )}{\sqrt{\frac{-a \sin ^2(e+f x)+a+b}{a+b}}}-\frac{3}{4} \csc ^2(e+f x) (a \cos (2 (e+f x))+a+2 b)^2\right )}{15 f (a+b) (a \cos (2 (e+f x))+a+2 b)} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [C] time = 0.494, size = 5850, normalized size = 35.5 \begin{align*} \text{output too large to display} \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 \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \cot \left (f x + e\right )^{6}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [B] time = 10.3436, size = 1882, normalized size = 11.41 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \cot \left (f x + e\right )^{6}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]