Optimal. Leaf size=116 \[ -\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {5 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e} \]
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Rubi [A]
time = 0.03, antiderivative size = 116, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {685, 655, 223,
209} \begin {gather*} \frac {5 d^3 \text {ArcTan}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}-\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e} \end {gather*}
Antiderivative was successfully verified.
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Rule 209
Rule 223
Rule 655
Rule 685
Rubi steps
\begin {align*} \int \frac {(d+e x)^3}{\sqrt {d^2-e^2 x^2}} \, dx &=-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{3} (5 d) \int \frac {(d+e x)^2}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{2} \left (5 d^2\right ) \int \frac {d+e x}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{2} \left (5 d^3\right ) \int \frac {1}{\sqrt {d^2-e^2 x^2}} \, dx\\ &=-\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {1}{2} \left (5 d^3\right ) \text {Subst}\left (\int \frac {1}{1+e^2 x^2} \, dx,x,\frac {x}{\sqrt {d^2-e^2 x^2}}\right )\\ &=-\frac {5 d^2 \sqrt {d^2-e^2 x^2}}{2 e}-\frac {5 d (d+e x) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {(d+e x)^2 \sqrt {d^2-e^2 x^2}}{3 e}+\frac {5 d^3 \tan ^{-1}\left (\frac {e x}{\sqrt {d^2-e^2 x^2}}\right )}{2 e}\\ \end {align*}
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Mathematica [A]
time = 0.22, size = 89, normalized size = 0.77 \begin {gather*} \frac {\left (-22 d^2-9 d e x-2 e^2 x^2\right ) \sqrt {d^2-e^2 x^2}}{6 e}-\frac {5 d^3 \log \left (-\sqrt {-e^2} x+\sqrt {d^2-e^2 x^2}\right )}{2 \sqrt {-e^2}} \end {gather*}
Antiderivative was successfully verified.
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Maple [A]
time = 0.48, size = 166, normalized size = 1.43
method | result | size |
risch | \(-\frac {\left (2 e^{2} x^{2}+9 d x e +22 d^{2}\right ) \sqrt {-e^{2} x^{2}+d^{2}}}{6 e}+\frac {5 d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 \sqrt {e^{2}}}\) | \(72\) |
default | \(e^{3} \left (-\frac {x^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{2}}-\frac {2 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{3 e^{4}}\right )+3 d \,e^{2} \left (-\frac {x \sqrt {-e^{2} x^{2}+d^{2}}}{2 e^{2}}+\frac {d^{2} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{2 e^{2} \sqrt {e^{2}}}\right )-\frac {3 d^{2} \sqrt {-e^{2} x^{2}+d^{2}}}{e}+\frac {d^{3} \arctan \left (\frac {\sqrt {e^{2}}\, x}{\sqrt {-e^{2} x^{2}+d^{2}}}\right )}{\sqrt {e^{2}}}\) | \(166\) |
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [A]
time = 0.56, size = 73, normalized size = 0.63 \begin {gather*} \frac {5}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} - \frac {1}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} x^{2} e - \frac {11}{3} \, \sqrt {-x^{2} e^{2} + d^{2}} d^{2} e^{\left (-1\right )} - \frac {3}{2} \, \sqrt {-x^{2} e^{2} + d^{2}} d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [A]
time = 1.98, size = 68, normalized size = 0.59 \begin {gather*} -\frac {1}{6} \, {\left (30 \, d^{3} \arctan \left (-\frac {{\left (d - \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )}}{x}\right ) + {\left (2 \, x^{2} e^{2} + 9 \, d x e + 22 \, d^{2}\right )} \sqrt {-x^{2} e^{2} + d^{2}}\right )} e^{\left (-1\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [A]
time = 2.80, size = 337, normalized size = 2.91 \begin {gather*} d^{3} \left (\begin {cases} \frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {asin}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} > 0 \\\frac {\sqrt {- \frac {d^{2}}{e^{2}}} \operatorname {asinh}{\left (x \sqrt {- \frac {e^{2}}{d^{2}}} \right )}}{\sqrt {d^{2}}} & \text {for}\: d^{2} > 0 \wedge e^{2} < 0 \\\frac {\sqrt {\frac {d^{2}}{e^{2}}} \operatorname {acosh}{\left (x \sqrt {\frac {e^{2}}{d^{2}}} \right )}}{\sqrt {- d^{2}}} & \text {for}\: d^{2} < 0 \wedge e^{2} < 0 \end {cases}\right ) + 3 d^{2} e \left (\begin {cases} \frac {x^{2}}{2 \sqrt {d^{2}}} & \text {for}\: e^{2} = 0 \\- \frac {\sqrt {d^{2} - e^{2} x^{2}}}{e^{2}} & \text {otherwise} \end {cases}\right ) + 3 d e^{2} \left (\begin {cases} - \frac {i d^{2} \operatorname {acosh}{\left (\frac {e x}{d} \right )}}{2 e^{3}} + \frac {i d x}{2 e^{2} \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} - \frac {i x^{3}}{2 d \sqrt {-1 + \frac {e^{2} x^{2}}{d^{2}}}} & \text {for}\: \left |{\frac {e^{2} x^{2}}{d^{2}}}\right | > 1 \\\frac {d^{2} \operatorname {asin}{\left (\frac {e x}{d} \right )}}{2 e^{3}} - \frac {d x \sqrt {1 - \frac {e^{2} x^{2}}{d^{2}}}}{2 e^{2}} & \text {otherwise} \end {cases}\right ) + e^{3} \left (\begin {cases} - \frac {2 d^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{4}} - \frac {x^{2} \sqrt {d^{2} - e^{2} x^{2}}}{3 e^{2}} & \text {for}\: e \neq 0 \\\frac {x^{4}}{4 \sqrt {d^{2}}} & \text {otherwise} \end {cases}\right ) \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [A]
time = 2.65, size = 52, normalized size = 0.45 \begin {gather*} \frac {5}{2} \, d^{3} \arcsin \left (\frac {x e}{d}\right ) e^{\left (-1\right )} \mathrm {sgn}\left (d\right ) - \frac {1}{6} \, \sqrt {-x^{2} e^{2} + d^{2}} {\left (22 \, d^{2} e^{\left (-1\right )} + {\left (2 \, x e + 9 \, d\right )} x\right )} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^3}{\sqrt {d^2-e^2\,x^2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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