Optimal. Leaf size=313 \[ \frac{\left (\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{2 c (q+1) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}+\frac{\left (-\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c (q+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{(b e+c d) \left (d+e x^2\right )^{q+1}}{2 c^2 e^2 (q+1)}+\frac{\left (d+e x^2\right )^{q+2}}{2 c e^2 (q+2)} \]
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Rubi [A] time = 0.938896, antiderivative size = 313, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.111, Rules used = {1251, 1628, 68} \[ \frac{\left (\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{2 c (q+1) \left (2 c d-e \left (b-\sqrt{b^2-4 a c}\right )\right )}+\frac{\left (-\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) \left (d+e x^2\right )^{q+1} \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c (q+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{(b e+c d) \left (d+e x^2\right )^{q+1}}{2 c^2 e^2 (q+1)}+\frac{\left (d+e x^2\right )^{q+2}}{2 c e^2 (q+2)} \]
Antiderivative was successfully verified.
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Rule 1251
Rule 1628
Rule 68
Rubi steps
\begin{align*} \int \frac{x^7 \left (d+e x^2\right )^q}{a+b x^2+c x^4} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{x^3 (d+e x)^q}{a+b x+c x^2} \, dx,x,x^2\right )\\ &=\frac{1}{2} \operatorname{Subst}\left (\int \left (\frac{(-c d-b e) (d+e x)^q}{c^2 e}+\frac{\left (\frac{b^2}{c^2}-\frac{a}{c}-\frac{b \left (b^2-3 a c\right )}{c^2 \sqrt{b^2-4 a c}}\right ) (d+e x)^q}{b-\sqrt{b^2-4 a c}+2 c x}+\frac{\left (\frac{b^2}{c^2}-\frac{a}{c}+\frac{b \left (b^2-3 a c\right )}{c^2 \sqrt{b^2-4 a c}}\right ) (d+e x)^q}{b+\sqrt{b^2-4 a c}+2 c x}+\frac{(d+e x)^{1+q}}{c e}\right ) \, dx,x,x^2\right )\\ &=-\frac{(c d+b e) \left (d+e x^2\right )^{1+q}}{2 c^2 e^2 (1+q)}+\frac{\left (d+e x^2\right )^{2+q}}{2 c e^2 (2+q)}-\frac{\left (a-\frac{b^2}{c}-\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{(d+e x)^q}{b+\sqrt{b^2-4 a c}+2 c x} \, dx,x,x^2\right )}{2 c}-\frac{\left (a-\frac{b^2}{c}+\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}\right ) \operatorname{Subst}\left (\int \frac{(d+e x)^q}{b-\sqrt{b^2-4 a c}+2 c x} \, dx,x,x^2\right )}{2 c}\\ &=-\frac{(c d+b e) \left (d+e x^2\right )^{1+q}}{2 c^2 e^2 (1+q)}+\frac{\left (d+e x^2\right )^{2+q}}{2 c e^2 (2+q)}+\frac{\left (a-\frac{b^2}{c}+\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \, _2F_1\left (1,1+q;2+q;\frac{2 c \left (d+e x^2\right )}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )}{2 c \left (2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e\right ) (1+q)}+\frac{\left (a-\frac{b^2}{c}-\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}\right ) \left (d+e x^2\right )^{1+q} \, _2F_1\left (1,1+q;2+q;\frac{2 c \left (d+e x^2\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{2 c \left (2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e\right ) (1+q)}\\ \end{align*}
Mathematica [A] time = 0.762778, size = 272, normalized size = 0.87 \[ \frac{\left (d+e x^2\right )^{q+1} \left (\frac{c \left (\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d+\left (\sqrt{b^2-4 a c}-b\right ) e}\right )}{(q+1) \left (e \left (\sqrt{b^2-4 a c}-b\right )+2 c d\right )}+\frac{c \left (-\frac{b \left (b^2-3 a c\right )}{c \sqrt{b^2-4 a c}}+a-\frac{b^2}{c}\right ) \, _2F_1\left (1,q+1;q+2;\frac{2 c \left (e x^2+d\right )}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{(q+1) \left (2 c d-e \left (\sqrt{b^2-4 a c}+b\right )\right )}-\frac{b e+c d}{e^2 (q+1)}+\frac{c \left (d+e x^2\right )}{e^2 (q+2)}\right )}{2 c^2} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.06, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{7} \left ( e{x}^{2}+d \right ) ^{q}}{c{x}^{4}+b{x}^{2}+a}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{q} x^{7}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (e x^{2} + d\right )}^{q} x^{7}}{c x^{4} + b x^{2} + a}, x\right ) \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] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x^{2} + d\right )}^{q} x^{7}}{c x^{4} + b x^{2} + a}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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