Optimal. Leaf size=525 \[ \frac{(d+e x)^{m+1} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right ) \left (e g^2 (m+1) (b d-a e)+c \left (2 d^2 g^2 (p+1)-2 d e f g (m+2 p+3)+e^2 f^2 (m+2 p+3)\right )\right )}{c e^3 (m+1) (m+2 p+3)}-\frac{g (d+e x)^{m+2} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} (b e g (m+p+2)+2 c (d g (p+1)-e f (m+2 p+3))) F_1\left (m+2;-p,-p;m+3;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{c e^3 (m+2) (m+2 p+3)}+\frac{g^2 (d+e x)^{m+1} \left (a+b x+c x^2\right )^{p+1}}{c e (m+2 p+3)} \]
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Rubi [A] time = 0.701753, antiderivative size = 523, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {1653, 843, 759, 133} \[ \frac{(d+e x)^{m+1} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} F_1\left (m+1;-p,-p;m+2;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right ) \left (g^2 (b d-a e)+\frac{c \left (2 d^2 g^2 (p+1)-2 d e f g (m+2 p+3)+e^2 f^2 (m+2 p+3)\right )}{e (m+1)}\right )}{c e^2 (m+2 p+3)}-\frac{g (d+e x)^{m+2} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (d+e x)}{2 c d-e \left (b-\sqrt{b^2-4 a c}\right )}\right )^{-p} \left (1-\frac{2 c (d+e x)}{2 c d-e \left (\sqrt{b^2-4 a c}+b\right )}\right )^{-p} (b e g (m+p+2)+2 c d g (p+1)-2 c e f (m+2 p+3)) F_1\left (m+2;-p,-p;m+3;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{c e^3 (m+2) (m+2 p+3)}+\frac{g^2 (d+e x)^{m+1} \left (a+b x+c x^2\right )^{p+1}}{c e (m+2 p+3)} \]
Antiderivative was successfully verified.
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Rule 1653
Rule 843
Rule 759
Rule 133
Rubi steps
\begin{align*} \int (d+e x)^m (f+g x)^2 \left (a+b x+c x^2\right )^p \, dx &=\frac{g^2 (d+e x)^{1+m} \left (a+b x+c x^2\right )^{1+p}}{c e (3+m+2 p)}+\frac{\int (d+e x)^m \left (e \left (c e f^2 (3+m+2 p)-g^2 (a e (1+m)+b d (1+p))\right )-e g (2 c d g (1+p)+b e g (2+m+p)-2 c e f (3+m+2 p)) x\right ) \left (a+b x+c x^2\right )^p \, dx}{c e^2 (3+m+2 p)}\\ &=\frac{g^2 (d+e x)^{1+m} \left (a+b x+c x^2\right )^{1+p}}{c e (3+m+2 p)}-\frac{(g (2 c d g (1+p)+b e g (2+m+p)-2 c e f (3+m+2 p))) \int (d+e x)^{1+m} \left (a+b x+c x^2\right )^p \, dx}{c e^2 (3+m+2 p)}+\frac{\left (e (b d-a e) g^2 (1+m)+c \left (2 d^2 g^2 (1+p)+e^2 f^2 (3+m+2 p)-2 d e f g (3+m+2 p)\right )\right ) \int (d+e x)^m \left (a+b x+c x^2\right )^p \, dx}{c e^2 (3+m+2 p)}\\ &=\frac{g^2 (d+e x)^{1+m} \left (a+b x+c x^2\right )^{1+p}}{c e (3+m+2 p)}-\frac{\left (g (2 c d g (1+p)+b e g (2+m+p)-2 c e f (3+m+2 p)) \left (a+b x+c x^2\right )^p \left (1-\frac{d+e x}{d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c}}\right )^{-p} \left (1-\frac{d+e x}{d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c}}\right )^{-p}\right ) \operatorname{Subst}\left (\int x^{1+m} \left (1-\frac{2 c x}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )^p \left (1-\frac{2 c x}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^p \, dx,x,d+e x\right )}{c e^3 (3+m+2 p)}+\frac{\left (\left (e (b d-a e) g^2 (1+m)+c \left (2 d^2 g^2 (1+p)+e^2 f^2 (3+m+2 p)-2 d e f g (3+m+2 p)\right )\right ) \left (a+b x+c x^2\right )^p \left (1-\frac{d+e x}{d-\frac{\left (b-\sqrt{b^2-4 a c}\right ) e}{2 c}}\right )^{-p} \left (1-\frac{d+e x}{d-\frac{\left (b+\sqrt{b^2-4 a c}\right ) e}{2 c}}\right )^{-p}\right ) \operatorname{Subst}\left (\int x^m \left (1-\frac{2 c x}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )^p \left (1-\frac{2 c x}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^p \, dx,x,d+e x\right )}{c e^3 (3+m+2 p)}\\ &=\frac{g^2 (d+e x)^{1+m} \left (a+b x+c x^2\right )^{1+p}}{c e (3+m+2 p)}+\frac{\left (e (b d-a e) g^2 (1+m)+c \left (2 d^2 g^2 (1+p)+e^2 f^2 (3+m+2 p)-2 d e f g (3+m+2 p)\right )\right ) (d+e x)^{1+m} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^{-p} F_1\left (1+m;-p,-p;2+m;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{c e^3 (1+m) (3+m+2 p)}-\frac{g (2 c d g (1+p)+b e g (2+m+p)-2 c e f (3+m+2 p)) (d+e x)^{2+m} \left (a+b x+c x^2\right )^p \left (1-\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e}\right )^{-p} \left (1-\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )^{-p} F_1\left (2+m;-p,-p;3+m;\frac{2 c (d+e x)}{2 c d-\left (b-\sqrt{b^2-4 a c}\right ) e},\frac{2 c (d+e x)}{2 c d-\left (b+\sqrt{b^2-4 a c}\right ) e}\right )}{c e^3 (2+m) (3+m+2 p)}\\ \end{align*}
Mathematica [F] time = 2.2038, size = 0, normalized size = 0. \[ \int (d+e x)^m (f+g x)^2 \left (a+b x+c x^2\right )^p \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 1.473, size = 0, normalized size = 0. \begin{align*} \int \left ( ex+d \right ) ^{m} \left ( gx+f \right ) ^{2} \left ( c{x}^{2}+bx+a \right ) ^{p}\, 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{\left (g x + f\right )}^{2}{\left (c x^{2} + b x + a\right )}^{p}{\left (e x + d\right )}^{m}\,{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 ({\left (g^{2} x^{2} + 2 \, f g x + f^{2}\right )}{\left (c x^{2} + b x + a\right )}^{p}{\left (e x + d\right )}^{m}, 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{\left (g x + f\right )}^{2}{\left (c x^{2} + b x + a\right )}^{p}{\left (e x + d\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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