3.384 \(\int \frac{x^m}{(a+b x) (c+d x)^3} \, dx\)

Optimal. Leaf size=206 \[ \frac{d x^{m+1} \left (a^2 d^2 (1-m) m-2 a b c d (2-m) m-b^2 c^2 \left (m^2-3 m+2\right )\right ) \, _2F_1\left (1,m+1;m+2;-\frac{d x}{c}\right )}{2 c^3 (m+1) (b c-a d)^3}+\frac{b^3 x^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a (m+1) (b c-a d)^3}+\frac{d x^{m+1} (a d (1-m)-b c (3-m))}{2 c^2 (c+d x) (b c-a d)^2}-\frac{d x^{m+1}}{2 c (c+d x)^2 (b c-a d)} \]

[Out]

-(d*x^(1 + m))/(2*c*(b*c - a*d)*(c + d*x)^2) + (d*(a*d*(1 - m) - b*c*(3 - m))*x^(1 + m))/(2*c^2*(b*c - a*d)^2*
(c + d*x)) + (b^3*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(a*(b*c - a*d)^3*(1 + m)) + (d*(a^
2*d^2*(1 - m)*m - 2*a*b*c*d*(2 - m)*m - b^2*c^2*(2 - 3*m + m^2))*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m,
-((d*x)/c)])/(2*c^3*(b*c - a*d)^3*(1 + m))

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Rubi [A]  time = 0.239848, antiderivative size = 206, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {103, 151, 156, 64} \[ \frac{d x^{m+1} \left (a^2 d^2 (1-m) m-2 a b c d (2-m) m-b^2 c^2 \left (m^2-3 m+2\right )\right ) \, _2F_1\left (1,m+1;m+2;-\frac{d x}{c}\right )}{2 c^3 (m+1) (b c-a d)^3}+\frac{b^3 x^{m+1} \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a (m+1) (b c-a d)^3}+\frac{d x^{m+1} (a d (1-m)-b c (3-m))}{2 c^2 (c+d x) (b c-a d)^2}-\frac{d x^{m+1}}{2 c (c+d x)^2 (b c-a d)} \]

Antiderivative was successfully verified.

[In]

Int[x^m/((a + b*x)*(c + d*x)^3),x]

[Out]

-(d*x^(1 + m))/(2*c*(b*c - a*d)*(c + d*x)^2) + (d*(a*d*(1 - m) - b*c*(3 - m))*x^(1 + m))/(2*c^2*(b*c - a*d)^2*
(c + d*x)) + (b^3*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(a*(b*c - a*d)^3*(1 + m)) + (d*(a^
2*d^2*(1 - m)*m - 2*a*b*c*d*(2 - m)*m - b^2*c^2*(2 - 3*m + m^2))*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m,
-((d*x)/c)])/(2*c^3*(b*c - a*d)^3*(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 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(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*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[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 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)^3} \, dx &=-\frac{d x^{1+m}}{2 c (b c-a d) (c+d x)^2}-\frac{\int \frac{x^m (-2 b c+a d (1-m)+b d (1-m) x)}{(a+b x) (c+d x)^2} \, dx}{2 c (b c-a d)}\\ &=-\frac{d x^{1+m}}{2 c (b c-a d) (c+d x)^2}+\frac{d (a d (1-m)-b c (3-m)) x^{1+m}}{2 c^2 (b c-a d)^2 (c+d x)}+\frac{\int \frac{x^m \left (2 b^2 c^2-a^2 d^2 (1-m) m+a b c d (3-m) m-b d (a d (1-m)-b c (3-m)) m x\right )}{(a+b x) (c+d x)} \, dx}{2 c^2 (b c-a d)^2}\\ &=-\frac{d x^{1+m}}{2 c (b c-a d) (c+d x)^2}+\frac{d (a d (1-m)-b c (3-m)) x^{1+m}}{2 c^2 (b c-a d)^2 (c+d x)}+\frac{b^3 \int \frac{x^m}{a+b x} \, dx}{(b c-a d)^3}+\frac{\left (d \left (a^2 d^2 (1-m) m-2 a b c d (2-m) m-b^2 c^2 \left (2-3 m+m^2\right )\right )\right ) \int \frac{x^m}{c+d x} \, dx}{2 c^2 (b c-a d)^3}\\ &=-\frac{d x^{1+m}}{2 c (b c-a d) (c+d x)^2}+\frac{d (a d (1-m)-b c (3-m)) x^{1+m}}{2 c^2 (b c-a d)^2 (c+d x)}+\frac{b^3 x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{b x}{a}\right )}{a (b c-a d)^3 (1+m)}+\frac{d \left (a^2 d^2 (1-m) m-2 a b c d (2-m) m-b^2 c^2 \left (2-3 m+m^2\right )\right ) x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{d x}{c}\right )}{2 c^3 (b c-a d)^3 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.20357, size = 169, normalized size = 0.82 \[ \frac{x^{m+1} \left (\frac{a d \left (a^2 d^2 (m-1) m-2 a b c d (m-2) m+b^2 c^2 \left (m^2-3 m+2\right )\right ) \, _2F_1\left (1,m+1;m+2;-\frac{d x}{c}\right )-2 b^3 c^3 \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a c^2 (m+1) (b c-a d)^2}+\frac{d (a d (m-1)-b c (m-3))}{c (c+d x) (b c-a d)}+\frac{d}{(c+d x)^2}\right )}{2 c (a d-b c)} \]

Antiderivative was successfully verified.

[In]

Integrate[x^m/((a + b*x)*(c + d*x)^3),x]

[Out]

(x^(1 + m)*(d/(c + d*x)^2 + (d*(-(b*c*(-3 + m)) + a*d*(-1 + m)))/(c*(b*c - a*d)*(c + d*x)) + (-2*b^3*c^3*Hyper
geometric2F1[1, 1 + m, 2 + m, -((b*x)/a)] + a*d*(-2*a*b*c*d*(-2 + m)*m + a^2*d^2*(-1 + m)*m + b^2*c^2*(2 - 3*m
+ m^2))*Hypergeometric2F1[1, 1 + m, 2 + m, -((d*x)/c)])/(a*c^2*(b*c - a*d)^2*(1 + m))))/(2*c*(-(b*c) + a*d))

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Maple [F]  time = 0.075, size = 0, normalized size = 0. \begin{align*} \int{\frac{{x}^{m}}{ \left ( bx+a \right ) \left ( dx+c \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int(x^m/(b*x+a)/(d*x+c)^3,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 )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x+a)/(d*x+c)^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{x^{m}}{b d^{3} x^{4} + a c^{3} +{\left (3 \, b c d^{2} + a d^{3}\right )} x^{3} + 3 \,{\left (b c^{2} d + a c d^{2}\right )} x^{2} +{\left (b c^{3} + 3 \, a c^{2} 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)^3,x, algorithm="fricas")

[Out]

integral(x^m/(b*d^3*x^4 + a*c^3 + (3*b*c*d^2 + a*d^3)*x^3 + 3*(b*c^2*d + a*c*d^2)*x^2 + (b*c^3 + 3*a*c^2*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)**3,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 )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^m/(b*x+a)/(d*x+c)^3,x, algorithm="giac")

[Out]

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