Optimal. Leaf size=250 \[ -\frac{(d+e x)^{7/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) (d+e x)^{3/2} (b d-a e)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) \sqrt{d+e x} (b d-a e)^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 e (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]
[Out]
________________________________________________________________________________________
Rubi [A] time = 0.172754, antiderivative size = 250, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {768, 646, 50, 63, 208} \[ -\frac{(d+e x)^{7/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) (d+e x)^{3/2} (b d-a e)}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) \sqrt{d+e x} (b d-a e)^2}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 e (a+b x) (b d-a e)^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]
Antiderivative was successfully verified.
[In]
[Out]
Rule 768
Rule 646
Rule 50
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{(a+b x) (d+e x)^{7/2}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=-\frac{(d+e x)^{7/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(7 e) \int \frac{(d+e x)^{5/2}}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx}{2 b}\\ &=-\frac{(d+e x)^{7/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (7 e \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{5/2}}{a b+b^2 x} \, dx}{2 b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{7/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (7 e \left (b^2 d-a b e\right ) \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{a b+b^2 x} \, dx}{2 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{7/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (7 e \left (b^2 d-a b e\right )^2 \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{a b+b^2 x} \, dx}{2 b^5 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e (b d-a e)^2 (a+b x) \sqrt{d+e x}}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{7/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (7 e \left (b^2 d-a b e\right )^3 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{2 b^7 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e (b d-a e)^2 (a+b x) \sqrt{d+e x}}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{7/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (7 \left (b^2 d-a b e\right )^3 \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b-\frac{b^2 d}{e}+\frac{b^2 x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b^7 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{7 e (b d-a e)^2 (a+b x) \sqrt{d+e x}}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (b d-a e) (a+b x) (d+e x)^{3/2}}{3 b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{7 e (a+b x) (d+e x)^{5/2}}{5 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{7/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{7 e (b d-a e)^{5/2} (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{9/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}
Mathematica [C] time = 0.0359011, size = 66, normalized size = 0.26 \[ \frac{2 e (a+b x) (d+e x)^{9/2} \, _2F_1\left (2,\frac{9}{2};\frac{11}{2};-\frac{b (d+e x)}{a e-b d}\right )}{9 \sqrt{(a+b x)^2} (a e-b d)^2} \]
Antiderivative was successfully verified.
[In]
[Out]
________________________________________________________________________________________
Maple [B] time = 0.016, size = 662, normalized size = 2.7 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2}}{15\,{b}^{4}} \left ( 6\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}x{b}^{3}e+6\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{5/2}a{b}^{2}e-20\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}xa{b}^{2}{e}^{2}+20\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}x{b}^{3}de-105\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{3}b{e}^{4}+315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{a}^{2}{b}^{2}d{e}^{3}-315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xa{b}^{3}{d}^{2}{e}^{2}+105\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{b}^{4}{d}^{3}e-20\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}{a}^{2}b{e}^{2}+20\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}a{b}^{2}de+90\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}x{a}^{2}b{e}^{3}-180\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}xa{b}^{2}d{e}^{2}+90\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}x{b}^{3}{d}^{2}e-105\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{4}{e}^{4}+315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{3}bd{e}^{3}-315\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}{b}^{2}{d}^{2}{e}^{2}+105\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) a{b}^{3}{d}^{3}e+105\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{3}{e}^{3}-225\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{a}^{2}bd{e}^{2}+135\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}a{b}^{2}{d}^{2}e-15\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}{b}^{3}{d}^{3} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \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 \frac{{\left (b x + a\right )}{\left (e x + d\right )}^{\frac{7}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]
________________________________________________________________________________________
Fricas [A] time = 1.04401, size = 1049, normalized size = 4.2 \begin{align*} \left [\frac{105 \,{\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} +{\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) + 2 \,{\left (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \,{\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \,{\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{30 \,{\left (b^{5} x + a b^{4}\right )}}, -\frac{105 \,{\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} +{\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (6 \, b^{3} e^{3} x^{3} - 15 \, b^{3} d^{3} + 161 \, a b^{2} d^{2} e - 245 \, a^{2} b d e^{2} + 105 \, a^{3} e^{3} + 2 \,{\left (16 \, b^{3} d e^{2} - 7 \, a b^{2} e^{3}\right )} x^{2} + 2 \,{\left (58 \, b^{3} d^{2} e - 84 \, a b^{2} d e^{2} + 35 \, a^{2} b e^{3}\right )} x\right )} \sqrt{e x + d}}{15 \,{\left (b^{5} x + a b^{4}\right )}}\right ] \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 [B] time = 1.22411, size = 485, normalized size = 1.94 \begin{align*} \frac{7 \,{\left (b^{3} d^{3} e^{2} - 3 \, a b^{2} d^{2} e^{3} + 3 \, a^{2} b d e^{4} - a^{3} e^{5}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{\left (-1\right )}}{\sqrt{-b^{2} d + a b e} b^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{{\left (\sqrt{x e + d} b^{3} d^{3} e^{2} - 3 \, \sqrt{x e + d} a b^{2} d^{2} e^{3} + 3 \, \sqrt{x e + d} a^{2} b d e^{4} - \sqrt{x e + d} a^{3} e^{5}\right )} e^{\left (-1\right )}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} + \frac{2 \,{\left (3 \,{\left (x e + d\right )}^{\frac{5}{2}} b^{8} e^{6} + 10 \,{\left (x e + d\right )}^{\frac{3}{2}} b^{8} d e^{6} + 45 \, \sqrt{x e + d} b^{8} d^{2} e^{6} - 10 \,{\left (x e + d\right )}^{\frac{3}{2}} a b^{7} e^{7} - 90 \, \sqrt{x e + d} a b^{7} d e^{7} + 45 \, \sqrt{x e + d} a^{2} b^{6} e^{8}\right )} e^{\left (-5\right )}}{15 \, b^{10} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
[Out]