Optimal. Leaf size=146 \[ \frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac {(b c (1+p)+a d (1+q)) \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (1+p,-q;2+p;-\frac {d \left (a+b x^2\right )}{b c-a d}\right )}{2 b^2 d (1+p) (2+p+q)} \]
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Rubi [A]
time = 0.09, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps
used = 4, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {457, 81, 72, 71}
\begin {gather*} \frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^{q+1}}{2 b d (p+q+2)}-\frac {\left (a+b x^2\right )^{p+1} \left (c+d x^2\right )^q (a d (q+1)+b c (p+1)) \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (p+1,-q;p+2;-\frac {d \left (b x^2+a\right )}{b c-a d}\right )}{2 b^2 d (p+1) (p+q+2)} \end {gather*}
Antiderivative was successfully verified.
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Rule 71
Rule 72
Rule 81
Rule 457
Rubi steps
\begin {align*} \int x^3 \left (a+b x^2\right )^p \left (c+d x^2\right )^q \, dx &=\frac {1}{2} \text {Subst}\left (\int x (a+b x)^p (c+d x)^q \, dx,x,x^2\right )\\ &=\frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac {(b c (1+p)+a d (1+q)) \text {Subst}\left (\int (a+b x)^p (c+d x)^q \, dx,x,x^2\right )}{2 b d (2+p+q)}\\ &=\frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac {\left ((b c (1+p)+a d (1+q)) \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q}\right ) \text {Subst}\left (\int (a+b x)^p \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^q \, dx,x,x^2\right )}{2 b d (2+p+q)}\\ &=\frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^{1+q}}{2 b d (2+p+q)}-\frac {(b c (1+p)+a d (1+q)) \left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (1+p,-q;2+p;-\frac {d \left (a+b x^2\right )}{b c-a d}\right )}{2 b^2 d (1+p) (2+p+q)}\\ \end {align*}
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Mathematica [A]
time = 0.14, size = 118, normalized size = 0.81 \begin {gather*} \frac {\left (a+b x^2\right )^{1+p} \left (c+d x^2\right )^q \left (b \left (c+d x^2\right )-\frac {(b c (1+p)+a d (1+q)) \left (\frac {b \left (c+d x^2\right )}{b c-a d}\right )^{-q} \, _2F_1\left (1+p,-q;2+p;\frac {d \left (a+b x^2\right )}{-b c+a d}\right )}{1+p}\right )}{2 b^2 d (2+p+q)} \end {gather*}
Antiderivative was successfully verified.
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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int x^{3} \left (b \,x^{2}+a \right )^{p} \left (d \,x^{2}+c \right )^{q}\, dx\]
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \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 x^3\,{\left (b\,x^2+a\right )}^p\,{\left (d\,x^2+c\right )}^q \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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