Optimal. Leaf size=215 \[ \frac{2 (d+e x) (d g+e f)^2 \left (36 d^2 g^2-8 d e f g+e^2 f^2\right )}{15 d^3 e^5 \sqrt{d^2-e^2 x^2}}-\frac{g^3 (3 d g+4 e f) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5}+\frac{2 (d+e x)^2 (e f-9 d g) (d g+e f)^3}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{(d+e x)^3 (d g+e f)^4}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{g^4 \sqrt{d^2-e^2 x^2}}{e^5} \]
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Rubi [A] time = 0.66619, antiderivative size = 215, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.129, Rules used = {1635, 641, 217, 203} \[ \frac{2 (d+e x) (d g+e f)^2 \left (36 d^2 g^2-8 d e f g+e^2 f^2\right )}{15 d^3 e^5 \sqrt{d^2-e^2 x^2}}-\frac{g^3 (3 d g+4 e f) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5}+\frac{2 (d+e x)^2 (e f-9 d g) (d g+e f)^3}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{(d+e x)^3 (d g+e f)^4}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{g^4 \sqrt{d^2-e^2 x^2}}{e^5} \]
Antiderivative was successfully verified.
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Rule 1635
Rule 641
Rule 217
Rule 203
Rubi steps
\begin{align*} \int \frac{(d+e x)^3 (f+g x)^4}{\left (d^2-e^2 x^2\right )^{7/2}} \, dx &=\frac{(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}-\frac{\int \frac{(d+e x)^2 \left (-\frac{2 e^4 f^4-12 d e^3 f^3 g-18 d^2 e^2 f^2 g^2-12 d^3 e f g^3-3 d^4 g^4}{e^4}+\frac{5 d g^2 \left (6 e^2 f^2+4 d e f g+d^2 g^2\right ) x}{e^3}+\frac{5 d g^3 (4 e f+d g) x^2}{e^2}+\frac{5 d g^4 x^3}{e}\right )}{\left (d^2-e^2 x^2\right )^{5/2}} \, dx}{5 d}\\ &=\frac{(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{\int \frac{(d+e x) \left (\frac{2 e^4 f^4-12 d e^3 f^3 g+42 d^2 e^2 f^2 g^2+68 d^3 e f g^3+27 d^4 g^4}{e^4}+\frac{30 d^2 g^3 (2 e f+d g) x}{e^3}+\frac{15 d^2 g^4 x^2}{e^2}\right )}{\left (d^2-e^2 x^2\right )^{3/2}} \, dx}{15 d^2}\\ &=\frac{(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 (e f+d g)^2 \left (e^2 f^2-8 d e f g+36 d^2 g^2\right ) (d+e x)}{15 d^3 e^5 \sqrt{d^2-e^2 x^2}}-\frac{\int \frac{\frac{15 d^3 g^3 (4 e f+3 d g)}{e^4}+\frac{15 d^3 g^4 x}{e^3}}{\sqrt{d^2-e^2 x^2}} \, dx}{15 d^3}\\ &=\frac{(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 (e f+d g)^2 \left (e^2 f^2-8 d e f g+36 d^2 g^2\right ) (d+e x)}{15 d^3 e^5 \sqrt{d^2-e^2 x^2}}+\frac{g^4 \sqrt{d^2-e^2 x^2}}{e^5}-\frac{\left (g^3 (4 e f+3 d g)\right ) \int \frac{1}{\sqrt{d^2-e^2 x^2}} \, dx}{e^4}\\ &=\frac{(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 (e f+d g)^2 \left (e^2 f^2-8 d e f g+36 d^2 g^2\right ) (d+e x)}{15 d^3 e^5 \sqrt{d^2-e^2 x^2}}+\frac{g^4 \sqrt{d^2-e^2 x^2}}{e^5}-\frac{\left (g^3 (4 e f+3 d g)\right ) \operatorname{Subst}\left (\int \frac{1}{1+e^2 x^2} \, dx,x,\frac{x}{\sqrt{d^2-e^2 x^2}}\right )}{e^4}\\ &=\frac{(e f+d g)^4 (d+e x)^3}{5 d e^5 \left (d^2-e^2 x^2\right )^{5/2}}+\frac{2 (e f-9 d g) (e f+d g)^3 (d+e x)^2}{15 d^2 e^5 \left (d^2-e^2 x^2\right )^{3/2}}+\frac{2 (e f+d g)^2 \left (e^2 f^2-8 d e f g+36 d^2 g^2\right ) (d+e x)}{15 d^3 e^5 \sqrt{d^2-e^2 x^2}}+\frac{g^4 \sqrt{d^2-e^2 x^2}}{e^5}-\frac{g^3 (4 e f+3 d g) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{e^5}\\ \end{align*}
Mathematica [A] time = 0.759525, size = 168, normalized size = 0.78 \[ \frac{\frac{\sqrt{d^2-e^2 x^2} \left (2 (d-e x)^2 (d g+e f)^2 \left (36 d^2 g^2-8 d e f g+e^2 f^2\right )+3 d^2 (d g+e f)^4+15 d^3 g^4 (d-e x)^3+2 d (d-e x) (e f-9 d g) (d g+e f)^3\right )}{d^3 (d-e x)^3}-15 g^3 (3 d g+4 e f) \tan ^{-1}\left (\frac{e x}{\sqrt{d^2-e^2 x^2}}\right )}{15 e^5} \]
Antiderivative was successfully verified.
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Maple [B] time = 0.121, size = 1030, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [B] time = 1.56566, size = 1608, normalized size = 7.48 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 3.22675, size = 1277, normalized size = 5.94 \begin{align*} -\frac{7 \, d^{3} e^{4} f^{4} - 12 \, d^{4} e^{3} f^{3} g + 12 \, d^{5} e^{2} f^{2} g^{2} + 88 \, d^{6} e f g^{3} + 72 \, d^{7} g^{4} -{\left (7 \, e^{7} f^{4} - 12 \, d e^{6} f^{3} g + 12 \, d^{2} e^{5} f^{2} g^{2} + 88 \, d^{3} e^{4} f g^{3} + 72 \, d^{4} e^{3} g^{4}\right )} x^{3} + 3 \,{\left (7 \, d e^{6} f^{4} - 12 \, d^{2} e^{5} f^{3} g + 12 \, d^{3} e^{4} f^{2} g^{2} + 88 \, d^{4} e^{3} f g^{3} + 72 \, d^{5} e^{2} g^{4}\right )} x^{2} - 3 \,{\left (7 \, d^{2} e^{5} f^{4} - 12 \, d^{3} e^{4} f^{3} g + 12 \, d^{4} e^{3} f^{2} g^{2} + 88 \, d^{5} e^{2} f g^{3} + 72 \, d^{6} e g^{4}\right )} x + 30 \,{\left (4 \, d^{6} e f g^{3} + 3 \, d^{7} g^{4} -{\left (4 \, d^{3} e^{4} f g^{3} + 3 \, d^{4} e^{3} g^{4}\right )} x^{3} + 3 \,{\left (4 \, d^{4} e^{3} f g^{3} + 3 \, d^{5} e^{2} g^{4}\right )} x^{2} - 3 \,{\left (4 \, d^{5} e^{2} f g^{3} + 3 \, d^{6} e g^{4}\right )} x\right )} \arctan \left (-\frac{d - \sqrt{-e^{2} x^{2} + d^{2}}}{e x}\right ) -{\left (15 \, d^{3} e^{3} g^{4} x^{3} - 7 \, d^{2} e^{4} f^{4} + 12 \, d^{3} e^{3} f^{3} g - 12 \, d^{4} e^{2} f^{2} g^{2} - 88 \, d^{5} e f g^{3} - 72 \, d^{6} g^{4} -{\left (2 \, e^{6} f^{4} - 12 \, d e^{5} f^{3} g + 42 \, d^{2} e^{4} f^{2} g^{2} + 128 \, d^{3} e^{3} f g^{3} + 117 \, d^{4} e^{2} g^{4}\right )} x^{2} + 3 \,{\left (2 \, d e^{5} f^{4} - 12 \, d^{2} e^{4} f^{3} g + 12 \, d^{3} e^{3} f^{2} g^{2} + 68 \, d^{4} e^{2} f g^{3} + 57 \, d^{5} e g^{4}\right )} x\right )} \sqrt{-e^{2} x^{2} + d^{2}}}{15 \,{\left (d^{3} e^{8} x^{3} - 3 \, d^{4} e^{7} x^{2} + 3 \, d^{5} e^{6} x - d^{6} e^{5}\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d + e x\right )^{3} \left (f + g x\right )^{4}}{\left (- \left (- d + e x\right ) \left (d + e x\right )\right )^{\frac{7}{2}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [B] time = 1.21178, size = 555, normalized size = 2.58 \begin{align*} -{\left (3 \, d g^{4} + 4 \, f g^{3} e\right )} \arcsin \left (\frac{x e}{d}\right ) e^{\left (-5\right )} \mathrm{sgn}\left (d\right ) + \frac{\sqrt{-x^{2} e^{2} + d^{2}}{\left ({\left ({\left ({\left ({\left ({\left (15 \, g^{4} x e - \frac{2 \,{\left (36 \, d^{5} g^{4} e^{10} + 64 \, d^{4} f g^{3} e^{11} + 21 \, d^{3} f^{2} g^{2} e^{12} - 6 \, d^{2} f^{3} g e^{13} + d f^{4} e^{14}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x - \frac{45 \,{\left (3 \, d^{6} g^{4} e^{9} + 4 \, d^{5} f g^{3} e^{10} + 2 \, d^{4} f^{2} g^{2} e^{11}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x + \frac{5 \,{\left (21 \, d^{7} g^{4} e^{8} + 28 \, d^{6} f g^{3} e^{9} - 6 \, d^{5} f^{2} g^{2} e^{10} - 12 \, d^{4} f^{3} g e^{11} + d^{3} f^{4} e^{12}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x + \frac{5 \,{\left (36 \, d^{8} g^{4} e^{7} + 44 \, d^{7} f g^{3} e^{8} + 6 \, d^{6} f^{2} g^{2} e^{9} - 12 \, d^{5} f^{3} g e^{10} - d^{4} f^{4} e^{11}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x - \frac{15 \,{\left (3 \, d^{9} g^{4} e^{6} + 4 \, d^{8} f g^{3} e^{7} + d^{5} f^{4} e^{10}\right )} e^{\left (-10\right )}}{d^{4}}\right )} x - \frac{{\left (72 \, d^{10} g^{4} e^{5} + 88 \, d^{9} f g^{3} e^{6} + 12 \, d^{8} f^{2} g^{2} e^{7} - 12 \, d^{7} f^{3} g e^{8} + 7 \, d^{6} f^{4} e^{9}\right )} e^{\left (-10\right )}}{d^{4}}\right )}}{15 \,{\left (x^{2} e^{2} - d^{2}\right )}^{3}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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