3.6.94 \(\int \frac {(a+\frac {b}{x})^m}{(c+d x)^4} \, dx\) [594]

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

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 185, normalized size of antiderivative = 1.00, 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 = {445, 457, 91, 79, 70} \begin {gather*} -\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} \end {gather*}

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 70

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

Rule 79

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] || L
tQ[p, n]))))

Rule 91

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 445

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 457

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]]

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\\ &=-\text {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 {\text {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 ) \text {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*}

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Mathematica [A]
time = 0.22, size = 155, normalized size = 0.84 \begin {gather*} \frac {\left (a+\frac {b}{x}\right )^{1+m} \left (\frac {2 d^2 (a c-b d) x^3}{(c+d x)^3}+\frac {d (-6 a c+b d (4+m)) x^2}{(c+d x)^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 ) \, _2F_1\left (2,1+m;2+m;\frac {b c+a c x}{a c x-b d x}\right )}{(a c-b d)^2 (1+m)}\right )}{6 c^2 (a c-b d)^2} \end {gather*}

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 = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (a +\frac {b}{x}\right )^{m}}{\left (d x +c \right )^{4}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + \frac {b}{x}\right )^{m}}{\left (c + d x\right )^{4}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+\frac {b}{x}\right )}^m}{{\left (c+d\,x\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^m/(c + d*x)^4,x)

[Out]

int((a + b/x)^m/(c + d*x)^4, x)

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