3.121 \(\int \frac{(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx\)

Optimal. Leaf size=140 \[ \frac{(a+b x)^{m+1} (d g-c h) \, _2F_1\left (1,m+1;m+2;-\frac{h (a+b x)}{b g-a h}\right )}{(m+1) (b g-a h) (f g-e h)}-\frac{(a+b x)^{m+1} (d e-c f) \, _2F_1\left (1,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f) (f g-e h)} \]

[Out]

-(((d*e - c*f)*(a + b*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((f*(a + b*x))/(b*e - a*f))])/((b*e - a*f
)*(f*g - e*h)*(1 + m))) + ((d*g - c*h)*(a + b*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((h*(a + b*x))/(b
*g - a*h))])/((b*g - a*h)*(f*g - e*h)*(1 + m))

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Rubi [A]  time = 0.0591633, antiderivative size = 140, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {156, 68} \[ \frac{(a+b x)^{m+1} (d g-c h) \, _2F_1\left (1,m+1;m+2;-\frac{h (a+b x)}{b g-a h}\right )}{(m+1) (b g-a h) (f g-e h)}-\frac{(a+b x)^{m+1} (d e-c f) \, _2F_1\left (1,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f) (f g-e h)} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^m*(c + d*x))/((e + f*x)*(g + h*x)),x]

[Out]

-(((d*e - c*f)*(a + b*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((f*(a + b*x))/(b*e - a*f))])/((b*e - a*f
)*(f*g - e*h)*(1 + m))) + ((d*g - c*h)*(a + b*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((h*(a + b*x))/(b
*g - a*h))])/((b*g - a*h)*(f*g - e*h)*(1 + m))

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(a+b x)^m (c+d x)}{(e+f x) (g+h x)} \, dx &=-\frac{(d e-c f) \int \frac{(a+b x)^m}{e+f x} \, dx}{f g-e h}+\frac{(d g-c h) \int \frac{(a+b x)^m}{g+h x} \, dx}{f g-e h}\\ &=-\frac{(d e-c f) (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) (f g-e h) (1+m)}+\frac{(d g-c h) (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) (f g-e h) (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0939513, size = 115, normalized size = 0.82 \[ \frac{(a+b x)^{m+1} \left (\frac{(d g-c h) \, _2F_1\left (1,m+1;m+2;\frac{h (a+b x)}{a h-b g}\right )}{b g-a h}-\frac{(d e-c f) \, _2F_1\left (1,m+1;m+2;\frac{f (a+b x)}{a f-b e}\right )}{b e-a f}\right )}{(m+1) (f g-e h)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^m*(c + d*x))/((e + f*x)*(g + h*x)),x]

[Out]

((a + b*x)^(1 + m)*(-(((d*e - c*f)*Hypergeometric2F1[1, 1 + m, 2 + m, (f*(a + b*x))/(-(b*e) + a*f)])/(b*e - a*
f)) + ((d*g - c*h)*Hypergeometric2F1[1, 1 + m, 2 + m, (h*(a + b*x))/(-(b*g) + a*h)])/(b*g - a*h)))/((f*g - e*h
)*(1 + m))

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Maple [F]  time = 0.066, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) \left ( bx+a \right ) ^{m}}{ \left ( fx+e \right ) \left ( hx+g \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x)

[Out]

int((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}{\left (h x + g\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x, algorithm="maxima")

[Out]

integrate((d*x + c)*(b*x + a)^m/((f*x + e)*(h*x + g)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d x + c\right )}{\left (b x + a\right )}^{m}}{f h x^{2} + e g +{\left (f g + e h\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x, algorithm="fricas")

[Out]

integral((d*x + c)*(b*x + a)^m/(f*h*x^2 + e*g + (f*g + e*h)*x), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{m} \left (c + d x\right )}{\left (e + f x\right ) \left (g + h x\right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(d*x+c)/(f*x+e)/(h*x+g),x)

[Out]

Integral((a + b*x)**m*(c + d*x)/((e + f*x)*(g + h*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}{\left (b x + a\right )}^{m}}{{\left (f x + e\right )}{\left (h x + g\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)/(f*x+e)/(h*x+g),x, algorithm="giac")

[Out]

integrate((d*x + c)*(b*x + a)^m/((f*x + e)*(h*x + g)), x)