Optimal. Leaf size=393 \[ \frac{(b g-a h)^2 (a+b x)^{m+1} (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;-n,-p;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^3 (m+1)}+\frac{2 h (b g-a h) (a+b x)^{m+2} (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+2;-n,-p;m+3;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^3 (m+2)}+\frac{h^2 (a+b x)^{m+3} (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+3;-n,-p;m+4;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^3 (m+3)} \]
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Rubi [A] time = 0.492588, antiderivative size = 393, normalized size of antiderivative = 1., number of steps used = 15, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {181, 159, 140, 139, 138} \[ \frac{(b g-a h)^2 (a+b x)^{m+1} (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+1;-n,-p;m+2;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^3 (m+1)}+\frac{2 h (b g-a h) (a+b x)^{m+2} (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+2;-n,-p;m+3;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^3 (m+2)}+\frac{h^2 (a+b x)^{m+3} (c+d x)^n (e+f x)^p \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (m+3;-n,-p;m+4;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^3 (m+3)} \]
Antiderivative was successfully verified.
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Rule 181
Rule 159
Rule 140
Rule 139
Rule 138
Rubi steps
\begin{align*} \int (a+b x)^m (c+d x)^n (e+f x)^p (g+h x)^2 \, dx &=\frac{h \int (a+b x)^{1+m} (c+d x)^n (e+f x)^p (g+h x) \, dx}{b}+\frac{(b g-a h) \int (a+b x)^m (c+d x)^n (e+f x)^p (g+h x) \, dx}{b}\\ &=\frac{h^2 \int (a+b x)^{2+m} (c+d x)^n (e+f x)^p \, dx}{b^2}+2 \frac{(h (b g-a h)) \int (a+b x)^{1+m} (c+d x)^n (e+f x)^p \, dx}{b^2}+\frac{(b g-a h)^2 \int (a+b x)^m (c+d x)^n (e+f x)^p \, dx}{b^2}\\ &=\frac{\left (h^2 (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n}\right ) \int (a+b x)^{2+m} \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^n (e+f x)^p \, dx}{b^2}+2 \frac{\left (h (b g-a h) (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n}\right ) \int (a+b x)^{1+m} \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^n (e+f x)^p \, dx}{b^2}+\frac{\left ((b g-a h)^2 (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n}\right ) \int (a+b x)^m \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^n (e+f x)^p \, dx}{b^2}\\ &=\frac{\left (h^2 (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac{b (e+f x)}{b e-a f}\right )^{-p}\right ) \int (a+b x)^{2+m} \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^n \left (\frac{b e}{b e-a f}+\frac{b f x}{b e-a f}\right )^p \, dx}{b^2}+2 \frac{\left (h (b g-a h) (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac{b (e+f x)}{b e-a f}\right )^{-p}\right ) \int (a+b x)^{1+m} \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^n \left (\frac{b e}{b e-a f}+\frac{b f x}{b e-a f}\right )^p \, dx}{b^2}+\frac{\left ((b g-a h)^2 (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac{b (e+f x)}{b e-a f}\right )^{-p}\right ) \int (a+b x)^m \left (\frac{b c}{b c-a d}+\frac{b d x}{b c-a d}\right )^n \left (\frac{b e}{b e-a f}+\frac{b f x}{b e-a f}\right )^p \, dx}{b^2}\\ &=\frac{(b g-a h)^2 (a+b x)^{1+m} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (1+m;-n,-p;2+m;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^3 (1+m)}+\frac{2 h (b g-a h) (a+b x)^{2+m} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (2+m;-n,-p;3+m;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^3 (2+m)}+\frac{h^2 (a+b x)^{3+m} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} (e+f x)^p \left (\frac{b (e+f x)}{b e-a f}\right )^{-p} F_1\left (3+m;-n,-p;4+m;-\frac{d (a+b x)}{b c-a d},-\frac{f (a+b x)}{b e-a f}\right )}{b^3 (3+m)}\\ \end{align*}
Mathematica [F] time = 1.486, size = 0, normalized size = 0. \[ \int (a+b x)^m (c+d x)^n (e+f x)^p (g+h x)^2 \, dx \]
Verification is Not applicable to the result.
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Maple [F] time = 0.173, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{n} \left ( fx+e \right ) ^{p} \left ( hx+g \right ) ^{2}\, 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 (h x + g\right )}^{2}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{n}{\left (f x + e\right )}^{p}\,{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 (h^{2} x^{2} + 2 \, g h x + g^{2}\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{n}{\left (f x + e\right )}^{p}, 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(-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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