3.977 \(\int \frac {(a+b x)^n (c+d x)^{-n}}{x^4} \, dx\)

Optimal. Leaf size=194 \[ \frac {(a+b x)^{n+1} (c+d x)^{1-n} (a d (n+2)+b c (2-n))}{6 a^2 c^2 x^2}+\frac {(b c-a d) (a+b x)^{n+1} (c+d x)^{-n-1} \left (a^2 d^2 \left (n^2+3 n+2\right )+2 a b c d \left (1-n^2\right )+b^2 c^2 \left (n^2-3 n+2\right )\right ) \, _2F_1\left (2,n+1;n+2;\frac {c (a+b x)}{a (c+d x)}\right )}{6 a^4 c^2 (n+1)}-\frac {(a+b x)^{n+1} (c+d x)^{1-n}}{3 a c x^3} \]

[Out]

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

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Rubi [A]  time = 0.14, antiderivative size = 194, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {129, 151, 12, 131} \[ \frac {(b c-a d) (a+b x)^{n+1} (c+d x)^{-n-1} \left (a^2 d^2 \left (n^2+3 n+2\right )+2 a b c d \left (1-n^2\right )+b^2 c^2 \left (n^2-3 n+2\right )\right ) \, _2F_1\left (2,n+1;n+2;\frac {c (a+b x)}{a (c+d x)}\right )}{6 a^4 c^2 (n+1)}+\frac {(a+b x)^{n+1} (c+d x)^{1-n} (a d (n+2)+b c (2-n))}{6 a^2 c^2 x^2}-\frac {(a+b x)^{n+1} (c+d x)^{1-n}}{3 a c x^3} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^n/(x^4*(c + d*x)^n),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 129

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, m, n, p}, x] && ILtQ[m + n
+ p + 2, 0] && NeQ[m, -1] && (SumSimplerQ[m, 1] || ( !(NeQ[n, -1] && SumSimplerQ[n, 1]) &&  !(NeQ[p, -1] && S
umSimplerQ[p, 1])))

Rule 131

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*c -
a*d)^n*(a + b*x)^(m + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

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]

Rubi steps

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

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Mathematica [A]  time = 0.18, size = 156, normalized size = 0.80 \[ \frac {(a+b x)^{n+1} (c+d x)^{-n-1} \left (-\frac {2 a^3 c (c+d x)^2}{x^3}+\frac {(b c-a d) \left (a^2 d^2 \left (n^2+3 n+2\right )-2 a b c d \left (n^2-1\right )+b^2 c^2 \left (n^2-3 n+2\right )\right ) \, _2F_1\left (2,n+1;n+2;\frac {c (a+b x)}{a (c+d x)}\right )}{n+1}+\frac {a^2 (c+d x)^2 (a d (n+2)-b c (n-2))}{x^2}\right )}{6 a^4 c^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^n/(x^4*(c + d*x)^n),x]

[Out]

((a + b*x)^(1 + n)*(c + d*x)^(-1 - n)*((-2*a^3*c*(c + d*x)^2)/x^3 + (a^2*(-(b*c*(-2 + n)) + a*d*(2 + n))*(c +
d*x)^2)/x^2 + ((b*c - a*d)*(-2*a*b*c*d*(-1 + n^2) + b^2*c^2*(2 - 3*n + n^2) + a^2*d^2*(2 + 3*n + n^2))*Hyperge
ometric2F1[2, 1 + n, 2 + n, (c*(a + b*x))/(a*(c + d*x))])/(1 + n)))/(6*a^4*c^2)

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fricas [F]  time = 0.99, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{n} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x + a)^n/((d*x + c)^n*x^4), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{n} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.14, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{n} \left (d x +c \right )^{-n}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{n}}{{\left (d x + c\right )}^{n} x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+b\,x\right )}^n}{x^4\,{\left (c+d\,x\right )}^n} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^n/(x^4*(c + d*x)^n),x)

[Out]

int((a + b*x)^n/(x^4*(c + d*x)^n), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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