3.594 \(\int \frac{(a+\frac{b}{x})^m}{(c+d x)^4} \, dx\)

Optimal. Leaf size=185 \[ -\frac{b \left (a+\frac{b}{x}\right )^{m+1} \left (6 a^2 c^2-6 a b c d (m+1)+b^2 d^2 \left (m^2+3 m+2\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac{c \left (a+\frac{b}{x}\right )}{a c-b d}\right )}{6 c^2 (m+1) (a c-b d)^4}+\frac{d^2 \left (a+\frac{b}{x}\right )^{m+1}}{3 c^2 \left (\frac{c}{x}+d\right )^3 (a c-b d)}-\frac{d \left (a+\frac{b}{x}\right )^{m+1} (6 a c-b d (m+4))}{6 c^2 \left (\frac{c}{x}+d\right )^2 (a c-b d)^2} \]

[Out]

(d^2*(a + b/x)^(1 + m))/(3*c^2*(a*c - b*d)*(d + c/x)^3) - (d*(6*a*c - b*d*(4 + m))*(a + b/x)^(1 + m))/(6*c^2*(
a*c - b*d)^2*(d + c/x)^2) - (b*(6*a^2*c^2 - 6*a*b*c*d*(1 + m) + b^2*d^2*(2 + 3*m + m^2))*(a + b/x)^(1 + m)*Hyp
ergeometric2F1[2, 1 + m, 2 + m, (c*(a + b/x))/(a*c - b*d)])/(6*c^2*(a*c - b*d)^4*(1 + m))

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Rubi [A]  time = 0.18285, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {434, 446, 89, 78, 68} \[ -\frac{b \left (a+\frac{b}{x}\right )^{m+1} \left (6 a^2 c^2-6 a b c d (m+1)+b^2 d^2 \left (m^2+3 m+2\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac{c \left (a+\frac{b}{x}\right )}{a c-b d}\right )}{6 c^2 (m+1) (a c-b d)^4}+\frac{d^2 \left (a+\frac{b}{x}\right )^{m+1}}{3 c^2 \left (\frac{c}{x}+d\right )^3 (a c-b d)}-\frac{d \left (a+\frac{b}{x}\right )^{m+1} (6 a c-b d (m+4))}{6 c^2 \left (\frac{c}{x}+d\right )^2 (a c-b d)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^m/(c + d*x)^4,x]

[Out]

(d^2*(a + b/x)^(1 + m))/(3*c^2*(a*c - b*d)*(d + c/x)^3) - (d*(6*a*c - b*d*(4 + m))*(a + b/x)^(1 + m))/(6*c^2*(
a*c - b*d)^2*(d + c/x)^2) - (b*(6*a^2*c^2 - 6*a*b*c*d*(1 + m) + b^2*d^2*(2 + 3*m + m^2))*(a + b/x)^(1 + m)*Hyp
ergeometric2F1[2, 1 + m, 2 + m, (c*(a + b/x))/(a*c - b*d)])/(6*c^2*(a*c - b*d)^4*(1 + m))

Rule 434

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[((a + b*x^n)^p*(d + c*x
^n)^q)/x^(n*q), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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{\left (a+\frac{b}{x}\right )^m}{(c+d x)^4} \, dx &=\int \frac{\left (a+\frac{b}{x}\right )^m}{\left (d+\frac{c}{x}\right )^4 x^4} \, dx\\ &=-\operatorname{Subst}\left (\int \frac{x^2 (a+b x)^m}{(d+c x)^4} \, dx,x,\frac{1}{x}\right )\\ &=\frac{d^2 \left (a+\frac{b}{x}\right )^{1+m}}{3 c^2 (a c-b d) \left (d+\frac{c}{x}\right )^3}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^m (-d (3 a c-b d (1+m))+3 c (a c-b d) x)}{(d+c x)^3} \, dx,x,\frac{1}{x}\right )}{3 c^2 (a c-b d)}\\ &=\frac{d^2 \left (a+\frac{b}{x}\right )^{1+m}}{3 c^2 (a c-b d) \left (d+\frac{c}{x}\right )^3}-\frac{d (6 a c-b d (4+m)) \left (a+\frac{b}{x}\right )^{1+m}}{6 c^2 (a c-b d)^2 \left (d+\frac{c}{x}\right )^2}-\frac{\left (6 a^2 c^2-6 a b c d (1+m)+b^2 d^2 \left (2+3 m+m^2\right )\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^m}{(d+c x)^2} \, dx,x,\frac{1}{x}\right )}{6 c^2 (a c-b d)^2}\\ &=\frac{d^2 \left (a+\frac{b}{x}\right )^{1+m}}{3 c^2 (a c-b d) \left (d+\frac{c}{x}\right )^3}-\frac{d (6 a c-b d (4+m)) \left (a+\frac{b}{x}\right )^{1+m}}{6 c^2 (a c-b d)^2 \left (d+\frac{c}{x}\right )^2}-\frac{b \left (6 a^2 c^2-6 a b c d (1+m)+b^2 d^2 \left (2+3 m+m^2\right )\right ) \left (a+\frac{b}{x}\right )^{1+m} \, _2F_1\left (2,1+m;2+m;\frac{c \left (a+\frac{b}{x}\right )}{a c-b d}\right )}{6 c^2 (a c-b d)^4 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.144135, size = 155, normalized size = 0.84 \[ \frac{\left (a+\frac{b}{x}\right )^{m+1} \left (-\frac{b \left (6 a^2 c^2-6 a b c d (m+1)+b^2 d^2 \left (m^2+3 m+2\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac{b c+a x c}{a c x-b d x}\right )}{(m+1) (a c-b d)^2}+\frac{2 d^2 x^3 (a c-b d)}{(c+d x)^3}+\frac{d x^2 (b d (m+4)-6 a c)}{(c+d x)^2}\right )}{6 c^2 (a c-b d)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^m/(c + d*x)^4,x]

[Out]

((a + b/x)^(1 + m)*((2*d^2*(a*c - b*d)*x^3)/(c + d*x)^3 + (d*(-6*a*c + b*d*(4 + m))*x^2)/(c + d*x)^2 - (b*(6*a
^2*c^2 - 6*a*b*c*d*(1 + m) + b^2*d^2*(2 + 3*m + m^2))*Hypergeometric2F1[2, 1 + m, 2 + m, (b*c + a*c*x)/(a*c*x
- b*d*x)])/((a*c - b*d)^2*(1 + m))))/(6*c^2*(a*c - b*d)^2)

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Maple [F]  time = 0.088, size = 0, normalized size = 0. \begin{align*} \int{\frac{1}{ \left ( dx+c \right ) ^{4}} \left ( a+{\frac{b}{x}} \right ) ^{m}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b/x)^m/(d*x+c)^4,x)

[Out]

int((a+b/x)^m/(d*x+c)^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a + \frac{b}{x}\right )}^{m}}{{\left (d x + c\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^m/(d*x+c)^4,x, algorithm="maxima")

[Out]

integrate((a + b/x)^m/(d*x + c)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (\frac{a x + b}{x}\right )^{m}}{d^{4} x^{4} + 4 \, c d^{3} x^{3} + 6 \, c^{2} d^{2} x^{2} + 4 \, c^{3} d x + c^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^m/(d*x+c)^4,x, algorithm="fricas")

[Out]

integral(((a*x + b)/x)^m/(d^4*x^4 + 4*c*d^3*x^3 + 6*c^2*d^2*x^2 + 4*c^3*d*x + c^4), 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((a+b/x)**m/(d*x+c)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (a + \frac{b}{x}\right )}^{m}}{{\left (d x + c\right )}^{4}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b/x)^m/(d*x+c)^4,x, algorithm="giac")

[Out]

integrate((a + b/x)^m/(d*x + c)^4, x)