Optimal. Leaf size=251 \[ \frac{2 b (a+b x) (A b-a B)}{\sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}+\frac{2 (a+b x) (A b-a B)}{3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac{2 (a+b x) (B d-A e)}{5 e \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac{2 b^{3/2} (a+b x) (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]
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Rubi [A] time = 0.155796, antiderivative size = 251, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {770, 78, 51, 63, 208} \[ \frac{2 b (a+b x) (A b-a B)}{\sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}+\frac{2 (a+b x) (A b-a B)}{3 \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{3/2} (b d-a e)^2}-\frac{2 (a+b x) (B d-A e)}{5 e \sqrt{a^2+2 a b x+b^2 x^2} (d+e x)^{5/2} (b d-a e)}-\frac{2 b^{3/2} (a+b x) (A b-a B) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]
Antiderivative was successfully verified.
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Rule 770
Rule 78
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{A+B x}{(d+e x)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{A+B x}{\left (a b+b^2 x\right ) (d+e x)^{7/2}} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left ((A b-a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) (d+e x)^{5/2}} \, dx}{(b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (b (A b-a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{(b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 b (A b-a B) (a+b x)}{(b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (b^2 (A b-a B) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{(b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 b (A b-a B) (a+b x)}{(b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (2 b^2 (A b-a B) \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 )}{e (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{2 (B d-A e) (a+b x)}{5 e (b d-a e) (d+e x)^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 (A b-a B) (a+b x)}{3 (b d-a e)^2 (d+e x)^{3/2} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 b (A b-a B) (a+b x)}{(b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 b^{3/2} (A b-a B) (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{(b d-a e)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}
Mathematica [C] time = 0.0604375, size = 102, normalized size = 0.41 \[ \frac{2 (a+b x) \left (5 e (d+e x) (A b-a B) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )-3 (b d-a e) (B d-A e)\right )}{15 e \sqrt{(a+b x)^2} (d+e x)^{5/2} (b d-a e)^2} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.016, size = 386, normalized size = 1.5 \begin{align*} -{\frac{2\,bx+2\,a}{15\,e \left ( ae-bd \right ) ^{3}} \left ( 15\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{5/2}{b}^{3}e-15\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \left ( ex+d \right ) ^{5/2}a{b}^{2}e+15\,A\sqrt{ \left ( ae-bd \right ) b}{x}^{2}{b}^{2}{e}^{3}-15\,B\sqrt{ \left ( ae-bd \right ) b}{x}^{2}ab{e}^{3}-5\,A\sqrt{ \left ( ae-bd \right ) b}xab{e}^{3}+35\,A\sqrt{ \left ( ae-bd \right ) b}x{b}^{2}d{e}^{2}+5\,B\sqrt{ \left ( ae-bd \right ) b}x{a}^{2}{e}^{3}-35\,B\sqrt{ \left ( ae-bd \right ) b}xabd{e}^{2}+3\,A\sqrt{ \left ( ae-bd \right ) b}{a}^{2}{e}^{3}-11\,A\sqrt{ \left ( ae-bd \right ) b}abd{e}^{2}+23\,A\sqrt{ \left ( ae-bd \right ) b}{b}^{2}{d}^{2}e+2\,B\sqrt{ \left ( ae-bd \right ) b}{a}^{2}d{e}^{2}-14\,B\sqrt{ \left ( ae-bd \right ) b}ab{d}^{2}e-3\,B\sqrt{ \left ( ae-bd \right ) b}{b}^{2}{d}^{3} \right ){\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B x + A}{\sqrt{{\left (b x + a\right )}^{2}}{\left (e x + d\right )}^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.66683, size = 1823, normalized size = 7.26 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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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.
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Giac [A] time = 1.17966, size = 497, normalized size = 1.98 \begin{align*} -\frac{2 \,{\left (B a b^{2} \mathrm{sgn}\left (b x + a\right ) - A b^{3} \mathrm{sgn}\left (b x + a\right )\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \sqrt{-b^{2} d + a b e}} - \frac{2 \,{\left (3 \, B b^{2} d^{3} \mathrm{sgn}\left (b x + a\right ) + 15 \,{\left (x e + d\right )}^{2} B a b e \mathrm{sgn}\left (b x + a\right ) - 15 \,{\left (x e + d\right )}^{2} A b^{2} e \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )} B a b d e \mathrm{sgn}\left (b x + a\right ) - 5 \,{\left (x e + d\right )} A b^{2} d e \mathrm{sgn}\left (b x + a\right ) - 6 \, B a b d^{2} e \mathrm{sgn}\left (b x + a\right ) - 3 \, A b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) - 5 \,{\left (x e + d\right )} B a^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 5 \,{\left (x e + d\right )} A a b e^{2} \mathrm{sgn}\left (b x + a\right ) + 3 \, B a^{2} d e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, A a b d e^{2} \mathrm{sgn}\left (b x + a\right ) - 3 \, A a^{2} e^{3} \mathrm{sgn}\left (b x + a\right )\right )}}{15 \,{\left (b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}\right )}{\left (x e + d\right )}^{\frac{5}{2}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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