Optimal. Leaf size=247 \[ \frac {\sqrt {3} \text {ArcTan}\left (\frac {\sqrt {3} b \sqrt [3]{c}+\sqrt {3} a \sqrt [3]{c} x}{b \sqrt [3]{c}+a \sqrt [3]{c} x-2 \sqrt [3]{d} \sqrt [3]{q+p x^4}}\right )}{c^{2/3} \sqrt [3]{d}}+\frac {\log \left (a b \sqrt [3]{c}+a^2 \sqrt [3]{c} x+a \sqrt [3]{d} \sqrt [3]{q+p x^4}\right )}{c^{2/3} \sqrt [3]{d}}-\frac {\log \left (a^2 b^2 c^{2/3}+2 a^3 b c^{2/3} x+a^4 c^{2/3} x^2+\left (-a^2 b \sqrt [3]{c} \sqrt [3]{d}-a^3 \sqrt [3]{c} \sqrt [3]{d} x\right ) \sqrt [3]{q+p x^4}+a^2 d^{2/3} \left (q+p x^4\right )^{2/3}\right )}{2 c^{2/3} \sqrt [3]{d}} \]
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Rubi [F]
time = 4.12, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps
used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {}
\begin {gather*} \int \frac {(b+a x) \left (-3 a q+4 b p x^3+a p x^4\right )}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Rubi steps
\begin {align*} \int \frac {(b+a x) \left (-3 a q+4 b p x^3+a p x^4\right )}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx &=\int \left (-\frac {a \left (a^4 c-5 b d p\right )}{d^2 p \left (q+p x^4\right )^{2/3}}+\frac {a^2 x}{d \left (q+p x^4\right )^{2/3}}+\frac {a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )+a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right ) x+3 a^3 b c \left (a^4 c-6 b d p\right ) x^2+\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) x^3}{d^2 p \left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}\right ) \, dx\\ &=\frac {a^2 \int \frac {x}{\left (q+p x^4\right )^{2/3}} \, dx}{d}+\frac {\int \frac {a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )+a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right ) x+3 a^3 b c \left (a^4 c-6 b d p\right ) x^2+\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) x^3}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}-\frac {\left (a \left (a^4 c-5 b d p\right )\right ) \int \frac {1}{\left (q+p x^4\right )^{2/3}} \, dx}{d^2 p}\\ &=\frac {a^2 \text {Subst}\left (\int \frac {1}{\left (q+p x^2\right )^{2/3}} \, dx,x,x^2\right )}{2 d}+\frac {\int \left (\frac {a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}+\frac {a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right ) x}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}+\frac {3 a^3 b c \left (a^4 c-6 b d p\right ) x^2}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}+\frac {\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) x^3}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )}\right ) \, dx}{d^2 p}-\frac {\left (a \left (a^4 c-5 b d p\right ) \left (1+\frac {p x^4}{q}\right )^{2/3}\right ) \int \frac {1}{\left (1+\frac {p x^4}{q}\right )^{2/3}} \, dx}{d^2 p \left (q+p x^4\right )^{2/3}}\\ &=-\frac {a \left (a^4 c-5 b d p\right ) x \left (1+\frac {p x^4}{q}\right )^{2/3} \, _2F_1\left (\frac {1}{4},\frac {2}{3};\frac {5}{4};-\frac {p x^4}{q}\right )}{d^2 p \left (q+p x^4\right )^{2/3}}+\frac {\left (3 a^3 b c \left (a^4 c-6 b d p\right )\right ) \int \frac {x^2}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) \int \frac {x^3}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right )\right ) \int \frac {x}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )\right ) \int \frac {1}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (3 a^2 \sqrt {p x^4}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-q+x^3}} \, dx,x,\sqrt [3]{q+p x^4}\right )}{4 d p x^2}\\ &=-\frac {3^{3/4} \sqrt {2-\sqrt {3}} a^2 \left (\sqrt [3]{q}-\sqrt [3]{q+p x^4}\right ) \sqrt {\frac {q^{2/3}+\sqrt [3]{q} \sqrt [3]{q+p x^4}+\left (q+p x^4\right )^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{q}-\sqrt [3]{q+p x^4}\right )^2}} F\left (\sin ^{-1}\left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{q}-\sqrt [3]{q+p x^4}}{\left (1-\sqrt {3}\right ) \sqrt [3]{q}-\sqrt [3]{q+p x^4}}\right )|-7+4 \sqrt {3}\right )}{2 d p x^2 \sqrt {-\frac {\sqrt [3]{q} \left (\sqrt [3]{q}-\sqrt [3]{q+p x^4}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{q}-\sqrt [3]{q+p x^4}\right )^2}}}-\frac {a \left (a^4 c-5 b d p\right ) x \left (1+\frac {p x^4}{q}\right )^{2/3} \, _2F_1\left (\frac {1}{4},\frac {2}{3};\frac {5}{4};-\frac {p x^4}{q}\right )}{d^2 p \left (q+p x^4\right )^{2/3}}+\frac {\left (3 a^3 b c \left (a^4 c-6 b d p\right )\right ) \int \frac {x^2}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a^8 c^2-8 a^4 b c d p+4 b^2 d^2 p^2\right ) \int \frac {x^3}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a^2 \left (3 a^4 b^2 c^2-4 d p \left (4 b^3 c+d q\right )\right )\right ) \int \frac {x}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}+\frac {\left (a \left (a^4 c \left (b^3 c+d q\right )-b d p \left (5 b^3 c+8 d q\right )\right )\right ) \int \frac {1}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx}{d^2 p}\\ \end {align*}
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Mathematica [F]
time = 10.54, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(b+a x) \left (-3 a q+4 b p x^3+a p x^4\right )}{\left (q+p x^4\right )^{2/3} \left (b^3 c+d q+3 a b^2 c x+3 a^2 b c x^2+a^3 c x^3+d p x^4\right )} \, dx \end {gather*}
Verification is not applicable to the result.
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Maple [F]
time = 0.05, size = 0, normalized size = 0.00 \[\int \frac {\left (a x +b \right ) \left (p a \,x^{4}+4 b p \,x^{3}-3 a q \right )}{\left (p \,x^{4}+q \right )^{\frac {2}{3}} \left (a^{3} c \,x^{3}+3 a^{2} b c \,x^{2}+d p \,x^{4}+3 a \,b^{2} c x +b^{3} c +d q \right )}\, dx\]
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(-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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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(-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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (b+a\,x\right )\,\left (a\,p\,x^4+4\,b\,p\,x^3-3\,a\,q\right )}{{\left (p\,x^4+q\right )}^{2/3}\,\left (c\,a^3\,x^3+3\,c\,a^2\,b\,x^2+3\,c\,a\,b^2\,x+c\,b^3+d\,p\,x^4+d\,q\right )} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
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