3.383 \(\int \frac{x^m}{(a+b x) (c+d x)^2} \, dx\)

Optimal. Leaf size=125 \[ \frac{b^2 x^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a (m+1) (b c-a d)^2}-\frac{d x^{m+1} (a d m+b c (1-m)) \, _2F_1\left (1,m+1;m+2;-\frac{d x}{c}\right )}{c^2 (m+1) (b c-a d)^2}-\frac{d x^{m+1}}{c (c+d x) (b c-a d)} \]

[Out]

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

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Rubi [A]  time = 0.0982408, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {103, 156, 64} \[ \frac{b^2 x^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a (m+1) (b c-a d)^2}-\frac{d x^{m+1} (a d m+b c (1-m)) \, _2F_1\left (1,m+1;m+2;-\frac{d x}{c}\right )}{c^2 (m+1) (b c-a d)^2}-\frac{d x^{m+1}}{c (c+d x) (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x^m/((a + b*x)*(c + d*x)^2),x]

[Out]

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

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

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 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int \frac{x^m}{(a+b x) (c+d x)^2} \, dx &=-\frac{d x^{1+m}}{c (b c-a d) (c+d x)}-\frac{\int \frac{x^m (-b c-a d m-b d m x)}{(a+b x) (c+d x)} \, dx}{c (b c-a d)}\\ &=-\frac{d x^{1+m}}{c (b c-a d) (c+d x)}+\frac{b^2 \int \frac{x^m}{a+b x} \, dx}{(b c-a d)^2}-\frac{(d (a d m+b (c-c m))) \int \frac{x^m}{c+d x} \, dx}{c (b c-a d)^2}\\ &=-\frac{d x^{1+m}}{c (b c-a d) (c+d x)}+\frac{b^2 x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{b x}{a}\right )}{a (b c-a d)^2 (1+m)}-\frac{d (a d m+b (c-c m)) x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{d x}{c}\right )}{c^2 (b c-a d)^2 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.0723619, size = 113, normalized size = 0.9 \[ \frac{x^{m+1} \left (b^2 c^2 (c+d x) \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )+a d \left ((c+d x) (b c (m-1)-a d m) \, _2F_1\left (1,m+1;m+2;-\frac{d x}{c}\right )-c (m+1) (b c-a d)\right )\right )}{a c^2 (m+1) (c+d x) (b c-a d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/((a + b*x)*(c + d*x)^2),x]

[Out]

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

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Maple [F]  time = 0.056, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( bx+a \right ) \left ( dx+c \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/(b*x+a)/(d*x+c)^2,x)

[Out]

int(x^m/(b*x+a)/(d*x+c)^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x + a\right )}{\left (d x + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x+a)/(d*x+c)^2,x, algorithm="maxima")

[Out]

integrate(x^m/((b*x + a)*(d*x + c)^2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{m}}{b d^{2} x^{3} + a c^{2} +{\left (2 \, b c d + a d^{2}\right )} x^{2} +{\left (b c^{2} + 2 \, a c d\right )} x}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x+a)/(d*x+c)^2,x, algorithm="fricas")

[Out]

integral(x^m/(b*d^2*x^3 + a*c^2 + (2*b*c*d + a*d^2)*x^2 + (b*c^2 + 2*a*c*d)*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(x**m/(b*x+a)/(d*x+c)**2,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{m}}{{\left (b x + a\right )}{\left (d x + c\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x+a)/(d*x+c)^2,x, algorithm="giac")

[Out]

integrate(x^m/((b*x + a)*(d*x + c)^2), x)