Optimal. Leaf size=280 \[ -\frac{A b-a B}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)}-\frac{a B e-5 A b e+4 b B d}{4 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2}-\frac{3 e (a+b x) (a B e-5 A b e+4 b B d)}{4 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}+\frac{3 e (a+b x) (a B e-5 A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{b} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]
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Rubi [A] time = 0.255782, antiderivative size = 280, 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} \[ -\frac{A b-a B}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)}-\frac{a B e-5 A b e+4 b B d}{4 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^2}-\frac{3 e (a+b x) (a B e-5 A b e+4 b B d)}{4 b \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{d+e x} (b d-a e)^3}+\frac{3 e (a+b x) (a B e-5 A b e+4 b B d) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{b} \sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^{7/2}} \]
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)^{3/2} \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{A+B x}{\left (a b+b^2 x\right )^3 (d+e x)^{3/2}} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{A b-a B}{2 b (b d-a e) (a+b x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left ((4 b B d-5 A b e+a B e) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right )^2 (d+e x)^{3/2}} \, dx}{4 (b d-a e) \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A b-a B}{2 b (b d-a e) (a+b x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (3 e (4 b B d-5 A b e+a B e) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) (d+e x)^{3/2}} \, dx}{8 b (b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A b-a B}{2 b (b d-a e) (a+b x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e (4 b B d-5 A b e+a B e) (a+b x)}{4 b (b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (3 e (4 b B d-5 A b e+a B e) \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{8 (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A b-a B}{2 b (b d-a e) (a+b x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e (4 b B d-5 A b e+a B e) (a+b x)}{4 b (b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{\left (3 (4 b B d-5 A b e+a B e) \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 )}{4 (b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{4 b B d-5 A b e+a B e}{4 b (b d-a e)^2 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{A b-a B}{2 b (b d-a e) (a+b x) \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e (4 b B d-5 A b e+a B e) (a+b x)}{4 b (b d-a e)^3 \sqrt{d+e x} \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e (4 b B d-5 A b e+a B e) (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 \sqrt{b} (b d-a e)^{7/2} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}
Mathematica [C] time = 0.0803107, size = 110, normalized size = 0.39 \[ \frac{(a+b x) \left (\frac{e (a+b x)^2 (-a B e+5 A b e-4 b B d) \, _2F_1\left (-\frac{1}{2},2;\frac{1}{2};\frac{b (d+e x)}{b d-a e}\right )}{(b d-a e)^2}+a B-A b\right )}{2 b \left ((a+b x)^2\right )^{3/2} \sqrt{d+e x} (b d-a e)} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.023, size = 681, normalized size = 2.4 \begin{align*} -{\frac{bx+a}{4\, \left ( ae-bd \right ) ^{3}} \left ( 15\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}{x}^{2}{b}^{3}{e}^{2}-3\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}{x}^{2}a{b}^{2}{e}^{2}-12\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}{x}^{2}{b}^{3}de+30\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}xa{b}^{2}{e}^{2}-6\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}x{a}^{2}b{e}^{2}-24\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}xa{b}^{2}de+15\,A\sqrt{ \left ( ae-bd \right ) b}{x}^{2}{b}^{2}{e}^{2}+15\,A\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}{a}^{2}b{e}^{2}-3\,B\sqrt{ \left ( ae-bd \right ) b}{x}^{2}ab{e}^{2}-12\,B\sqrt{ \left ( ae-bd \right ) b}{x}^{2}{b}^{2}de-3\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}{a}^{3}{e}^{2}-12\,B\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) \sqrt{ex+d}{a}^{2}bde+25\,A\sqrt{ \left ( ae-bd \right ) b}xab{e}^{2}+5\,A\sqrt{ \left ( ae-bd \right ) b}x{b}^{2}de-5\,B\sqrt{ \left ( ae-bd \right ) b}x{a}^{2}{e}^{2}-21\,B\sqrt{ \left ( ae-bd \right ) b}xabde-4\,B\sqrt{ \left ( ae-bd \right ) b}x{b}^{2}{d}^{2}+8\,A\sqrt{ \left ( ae-bd \right ) b}{a}^{2}{e}^{2}+9\,A\sqrt{ \left ( ae-bd \right ) b}abde-2\,A\sqrt{ \left ( ae-bd \right ) b}{b}^{2}{d}^{2}-13\,B\sqrt{ \left ( ae-bd \right ) b}{a}^{2}de-2\,B\sqrt{ \left ( ae-bd \right ) b}ab{d}^{2} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}{\frac{1}{\sqrt{ex+d}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \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}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}{\left (e x + d\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 1.86079, size = 2884, normalized size = 10.3 \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 [B] time = 1.30776, size = 832, normalized size = 2.97 \begin{align*} -\frac{3 \,{\left (4 \, B b d e^{2} + B a e^{3} - 5 \, A b e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{4 \,{\left (b^{3} d^{3} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt{-b^{2} d + a b e}} - \frac{2 \,{\left (B d e^{2} - A e^{3}\right )}}{{\left (b^{3} d^{3} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )} \sqrt{x e + d}} - \frac{4 \,{\left (x e + d\right )}^{\frac{3}{2}} B b^{2} d e^{2} - 4 \, \sqrt{x e + d} B b^{2} d^{2} e^{2} + 3 \,{\left (x e + d\right )}^{\frac{3}{2}} B a b e^{3} - 7 \,{\left (x e + d\right )}^{\frac{3}{2}} A b^{2} e^{3} - \sqrt{x e + d} B a b d e^{3} + 9 \, \sqrt{x e + d} A b^{2} d e^{3} + 5 \, \sqrt{x e + d} B a^{2} e^{4} - 9 \, \sqrt{x e + d} A a b e^{4}}{4 \,{\left (b^{3} d^{3} e \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - 3 \, a b^{2} d^{2} e^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) + 3 \, a^{2} b d e^{3} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right ) - a^{3} e^{4} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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