Optimal. Leaf size=224 \[ \frac {d^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d) (d e-c f) (d g-c h)}-\frac {f^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f) (d e-c f) (f g-e h)}+\frac {h^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {h (a+b x)}{b g-a h}\right )}{(m+1) (b g-a h) (d g-c h) (f g-e h)} \]
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Rubi [A] time = 0.19, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {180, 68} \[ \frac {d^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d) (d e-c f) (d g-c h)}-\frac {f^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f) (d e-c f) (f g-e h)}+\frac {h^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac {h (a+b x)}{b g-a h}\right )}{(m+1) (b g-a h) (d g-c h) (f g-e h)} \]
Antiderivative was successfully verified.
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Rule 68
Rule 180
Rubi steps
\begin {align*} \int \frac {(a+b x)^m}{(c+d x) (e+f x) (g+h x)} \, dx &=\int \left (\frac {d^2 (a+b x)^m}{(d e-c f) (d g-c h) (c+d x)}+\frac {f^2 (a+b x)^m}{(d e-c f) (-f g+e h) (e+f x)}+\frac {h^2 (a+b x)^m}{(d g-c h) (f g-e h) (g+h x)}\right ) \, dx\\ &=\frac {d^2 \int \frac {(a+b x)^m}{c+d x} \, dx}{(d e-c f) (d g-c h)}-\frac {f^2 \int \frac {(a+b x)^m}{e+f x} \, dx}{(d e-c f) (f g-e h)}+\frac {h^2 \int \frac {(a+b x)^m}{g+h x} \, dx}{(d g-c h) (f g-e h)}\\ &=\frac {d^2 (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{(b c-a d) (d e-c f) (d g-c h) (1+m)}-\frac {f^2 (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {f (a+b x)}{b e-a f}\right )}{(b e-a f) (d e-c f) (f g-e h) (1+m)}+\frac {h^2 (a+b x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {h (a+b x)}{b g-a h}\right )}{(b g-a h) (d g-c h) (f g-e h) (1+m)}\\ \end {align*}
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Mathematica [A] time = 0.37, size = 193, normalized size = 0.86 \[ \frac {(a+b x)^{m+1} \left (\frac {d^2 \, _2F_1\left (1,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )}{(b c-a d) (c f-d e) (c h-d g)}+\frac {f^2 \, _2F_1\left (1,m+1;m+2;\frac {f (a+b x)}{a f-b e}\right )}{(b e-a f) (d e-c f) (e h-f g)}+\frac {h^2 \, _2F_1\left (1,m+1;m+2;\frac {h (a+b x)}{a h-b g}\right )}{(b g-a h) (d g-c h) (f g-e h)}\right )}{m+1} \]
Antiderivative was successfully verified.
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fricas [F] time = 0.85, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m}}{d f h x^{3} + c e g + {\left (d f g + {\left (d e + c f\right )} h\right )} x^{2} + {\left (c e h + {\left (d e + c f\right )} g\right )} x}, x\right ) \]
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 \[ \int \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )} {\left (f x + e\right )} {\left (h x + g\right )}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.26, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m}}{\left (d x +c \right ) \left (f x +e \right ) \left (h x +g \right )}\, dx \]
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 \[ \int \frac {{\left (b x + a\right )}^{m}}{{\left (d x + c\right )} {\left (f x + e\right )} {\left (h x + g\right )}}\,{d x} \]
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 \[ \int \frac {{\left (a+b\,x\right )}^m}{\left (e+f\,x\right )\,\left (g+h\,x\right )\,\left (c+d\,x\right )} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]
Verification of antiderivative is not currently implemented for this CAS.
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