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

Optimal. Leaf size=224 \[ \frac{d^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{(m+1) (b c-a d) (d e-c f) (d g-c h)}-\frac{f^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{f (a+b x)}{b e-a f}\right )}{(m+1) (b e-a f) (d e-c f) (f g-e h)}+\frac{h^2 (a+b x)^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{h (a+b x)}{b g-a h}\right )}{(m+1) (b g-a h) (d g-c h) (f g-e h)} \]

[Out]

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

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Rubi [A]  time = 0.194952, antiderivative size = 224, normalized size of antiderivative = 1., 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.

[In]

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

[Out]

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

Rule 180

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_))^(q_), x
_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p*(g + h*x)^q, x], x] /; FreeQ[{a, b, c, d,
e, f, g, h, m, n}, x] && IntegersQ[p, q]

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 &=\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*}

Mathematica [A]  time = 0.395389, 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.

[In]

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

[Out]

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

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Maple [F]  time = 0.101, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( bx+a \right ) ^{m}}{ \left ( hx+g \right ) \left ( fx+e \right ) \left ( dx+c \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 (b x + a\right )}^{m}}{{\left (d x + c\right )}{\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((b*x + a)^m/((d*x + c)*(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 (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 ) \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((b*x + a)^m/(d*f*h*x^3 + c*e*g + (d*f*g + (d*e + c*f)*h)*x^2 + (c*e*h + (d*e + c*f)*g)*x), x)

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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.

[In]

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

[Out]

Timed out

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