Optimal. Leaf size=244 \[ \frac{63 c^2 d^2 e^2}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^5}-\frac{63 c^{5/2} d^{5/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{11/2}}+\frac{21 c d e^2}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^4}+\frac{9 e}{4 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac{63 e^2}{20 (d+e x)^{5/2} \left (c d^2-a e^2\right )^3} \]
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Rubi [A] time = 0.222973, antiderivative size = 244, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.108, Rules used = {626, 51, 63, 208} \[ \frac{63 c^2 d^2 e^2}{4 \sqrt{d+e x} \left (c d^2-a e^2\right )^5}-\frac{63 c^{5/2} d^{5/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{11/2}}+\frac{21 c d e^2}{4 (d+e x)^{3/2} \left (c d^2-a e^2\right )^4}+\frac{9 e}{4 (d+e x)^{5/2} \left (c d^2-a e^2\right )^2 (a e+c d x)}-\frac{1}{2 (d+e x)^{5/2} \left (c d^2-a e^2\right ) (a e+c d x)^2}+\frac{63 e^2}{20 (d+e x)^{5/2} \left (c d^2-a e^2\right )^3} \]
Antiderivative was successfully verified.
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Rule 626
Rule 51
Rule 63
Rule 208
Rubi steps
\begin{align*} \int \frac{1}{\sqrt{d+e x} \left (a d e+\left (c d^2+a e^2\right ) x+c d e x^2\right )^3} \, dx &=\int \frac{1}{(a e+c d x)^3 (d+e x)^{7/2}} \, dx\\ &=-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}-\frac{(9 e) \int \frac{1}{(a e+c d x)^2 (d+e x)^{7/2}} \, dx}{4 \left (c d^2-a e^2\right )}\\ &=-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{\left (63 e^2\right ) \int \frac{1}{(a e+c d x) (d+e x)^{7/2}} \, dx}{8 \left (c d^2-a e^2\right )^2}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{\left (63 c d e^2\right ) \int \frac{1}{(a e+c d x) (d+e x)^{5/2}} \, dx}{8 \left (c d^2-a e^2\right )^3}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac{\left (63 c^2 d^2 e^2\right ) \int \frac{1}{(a e+c d x) (d+e x)^{3/2}} \, dx}{8 \left (c d^2-a e^2\right )^4}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac{63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt{d+e x}}+\frac{\left (63 c^3 d^3 e^2\right ) \int \frac{1}{(a e+c d x) \sqrt{d+e x}} \, dx}{8 \left (c d^2-a e^2\right )^5}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac{63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt{d+e x}}+\frac{\left (63 c^3 d^3 e\right ) \operatorname{Subst}\left (\int \frac{1}{-\frac{c d^2}{e}+a e+\frac{c d x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 \left (c d^2-a e^2\right )^5}\\ &=\frac{63 e^2}{20 \left (c d^2-a e^2\right )^3 (d+e x)^{5/2}}-\frac{1}{2 \left (c d^2-a e^2\right ) (a e+c d x)^2 (d+e x)^{5/2}}+\frac{9 e}{4 \left (c d^2-a e^2\right )^2 (a e+c d x) (d+e x)^{5/2}}+\frac{21 c d e^2}{4 \left (c d^2-a e^2\right )^4 (d+e x)^{3/2}}+\frac{63 c^2 d^2 e^2}{4 \left (c d^2-a e^2\right )^5 \sqrt{d+e x}}-\frac{63 c^{5/2} d^{5/2} e^2 \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{d} \sqrt{d+e x}}{\sqrt{c d^2-a e^2}}\right )}{4 \left (c d^2-a e^2\right )^{11/2}}\\ \end{align*}
Mathematica [C] time = 0.0207404, size = 61, normalized size = 0.25 \[ -\frac{2 e^2 \, _2F_1\left (-\frac{5}{2},3;-\frac{3}{2};-\frac{c d (d+e x)}{a e^2-c d^2}\right )}{5 (d+e x)^{5/2} \left (a e^2-c d^2\right )^3} \]
Antiderivative was successfully verified.
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Maple [A] time = 0.234, size = 294, normalized size = 1.2 \begin{align*} -{\frac{2\,{e}^{2}}{5\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{3}} \left ( ex+d \right ) ^{-{\frac{5}{2}}}}-12\,{\frac{{e}^{2}{c}^{2}{d}^{2}}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{5}\sqrt{ex+d}}}+2\,{\frac{{e}^{2}cd}{ \left ( a{e}^{2}-c{d}^{2} \right ) ^{4} \left ( ex+d \right ) ^{3/2}}}-{\frac{15\,{c}^{4}{d}^{4}{e}^{2}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{5} \left ( cdex+a{e}^{2} \right ) ^{2}} \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{17\,{e}^{4}{c}^{3}{d}^{3}a}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{5} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}+{\frac{17\,{c}^{4}{d}^{5}{e}^{2}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{5} \left ( cdex+a{e}^{2} \right ) ^{2}}\sqrt{ex+d}}-{\frac{63\,{e}^{2}{c}^{3}{d}^{3}}{4\, \left ( a{e}^{2}-c{d}^{2} \right ) ^{5}}\arctan \left ({cd\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}} \right ){\frac{1}{\sqrt{ \left ( a{e}^{2}-c{d}^{2} \right ) cd}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [B] time = 2.38589, size = 4073, normalized size = 16.69 \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 [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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