Optimal. Leaf size=107 \[ \frac {\left (8 a^2 d^2-8 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{5/2}}-\frac {3 b x \sqrt {c+d x^2} (b c-2 a d)}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d} \]
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Rubi [A] time = 0.06, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {416, 388, 217, 206} \begin {gather*} \frac {\left (8 a^2 d^2-8 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{5/2}}-\frac {3 b x \sqrt {c+d x^2} (b c-2 a d)}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d} \end {gather*}
Antiderivative was successfully verified.
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Rule 206
Rule 217
Rule 388
Rule 416
Rubi steps
\begin {align*} \int \frac {\left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx &=\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d}+\frac {\int \frac {-a (b c-4 a d)-3 b (b c-2 a d) x^2}{\sqrt {c+d x^2}} \, dx}{4 d}\\ &=-\frac {3 b (b c-2 a d) x \sqrt {c+d x^2}}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d}-\frac {(2 a d (b c-4 a d)-3 b c (b c-2 a d)) \int \frac {1}{\sqrt {c+d x^2}} \, dx}{8 d^2}\\ &=-\frac {3 b (b c-2 a d) x \sqrt {c+d x^2}}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d}-\frac {(2 a d (b c-4 a d)-3 b c (b c-2 a d)) \operatorname {Subst}\left (\int \frac {1}{1-d x^2} \, dx,x,\frac {x}{\sqrt {c+d x^2}}\right )}{8 d^2}\\ &=-\frac {3 b (b c-2 a d) x \sqrt {c+d x^2}}{8 d^2}+\frac {b x \left (a+b x^2\right ) \sqrt {c+d x^2}}{4 d}+\frac {\left (3 b^2 c^2-8 a b c d+8 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{8 d^{5/2}}\\ \end {align*}
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Mathematica [A] time = 0.06, size = 91, normalized size = 0.85 \begin {gather*} \frac {\left (8 a^2 d^2-8 a b c d+3 b^2 c^2\right ) \log \left (\sqrt {d} \sqrt {c+d x^2}+d x\right )+b \sqrt {d} x \sqrt {c+d x^2} \left (8 a d-3 b c+2 b d x^2\right )}{8 d^{5/2}} \end {gather*}
Antiderivative was successfully verified.
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IntegrateAlgebraic [A] time = 0.11, size = 95, normalized size = 0.89 \begin {gather*} \frac {\left (-8 a^2 d^2+8 a b c d-3 b^2 c^2\right ) \log \left (\sqrt {c+d x^2}-\sqrt {d} x\right )}{8 d^{5/2}}+\frac {\sqrt {c+d x^2} \left (8 a b d x-3 b^2 c x+2 b^2 d x^3\right )}{8 d^2} \end {gather*}
Antiderivative was successfully verified.
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fricas [A] time = 1.43, size = 194, normalized size = 1.81 \begin {gather*} \left [\frac {{\left (3 \, b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (2 \, b^{2} d^{2} x^{3} - {\left (3 \, b^{2} c d - 8 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{16 \, d^{3}}, -\frac {{\left (3 \, b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (2 \, b^{2} d^{2} x^{3} - {\left (3 \, b^{2} c d - 8 \, a b d^{2}\right )} x\right )} \sqrt {d x^{2} + c}}{8 \, d^{3}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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giac [A] time = 0.49, size = 91, normalized size = 0.85 \begin {gather*} \frac {1}{8} \, {\left (\frac {2 \, b^{2} x^{2}}{d} - \frac {3 \, b^{2} c d - 8 \, a b d^{2}}{d^{3}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (3 \, b^{2} c^{2} - 8 \, a b c d + 8 \, a^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{8 \, d^{\frac {5}{2}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maple [A] time = 0.01, size = 131, normalized size = 1.22 \begin {gather*} \frac {\sqrt {d \,x^{2}+c}\, b^{2} x^{3}}{4 d}+\frac {a^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{\sqrt {d}}-\frac {a b c \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{d^{\frac {3}{2}}}+\frac {3 b^{2} c^{2} \ln \left (\sqrt {d}\, x +\sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {5}{2}}}+\frac {\sqrt {d \,x^{2}+c}\, a b x}{d}-\frac {3 \sqrt {d \,x^{2}+c}\, b^{2} c x}{8 d^{2}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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maxima [A] time = 0.89, size = 109, normalized size = 1.02 \begin {gather*} \frac {\sqrt {d x^{2} + c} b^{2} x^{3}}{4 \, d} - \frac {3 \, \sqrt {d x^{2} + c} b^{2} c x}{8 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a b x}{d} + \frac {3 \, b^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {5}{2}}} - \frac {a b c \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{d^{\frac {3}{2}}} + \frac {a^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{\sqrt {d}} \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 (b\,x^2+a\right )}^2}{\sqrt {d\,x^2+c}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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sympy [A] time = 14.27, size = 238, normalized size = 2.22 \begin {gather*} a^{2} \left (\begin {cases} \frac {\sqrt {- \frac {c}{d}} \operatorname {asin}{\left (x \sqrt {- \frac {d}{c}} \right )}}{\sqrt {c}} & \text {for}\: c > 0 \wedge d < 0 \\\frac {\sqrt {\frac {c}{d}} \operatorname {asinh}{\left (x \sqrt {\frac {d}{c}} \right )}}{\sqrt {c}} & \text {for}\: c > 0 \wedge d > 0 \\\frac {\sqrt {- \frac {c}{d}} \operatorname {acosh}{\left (x \sqrt {- \frac {d}{c}} \right )}}{\sqrt {- c}} & \text {for}\: d > 0 \wedge c < 0 \end {cases}\right ) + \frac {a b \sqrt {c} x \sqrt {1 + \frac {d x^{2}}{c}}}{d} - \frac {a b c \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{d^{\frac {3}{2}}} - \frac {3 b^{2} c^{\frac {3}{2}} x}{8 d^{2} \sqrt {1 + \frac {d x^{2}}{c}}} - \frac {b^{2} \sqrt {c} x^{3}}{8 d \sqrt {1 + \frac {d x^{2}}{c}}} + \frac {3 b^{2} c^{2} \operatorname {asinh}{\left (\frac {\sqrt {d} x}{\sqrt {c}} \right )}}{8 d^{\frac {5}{2}}} + \frac {b^{2} x^{5}}{4 \sqrt {c} \sqrt {1 + \frac {d x^{2}}{c}}} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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