3.4.83 \(\int \frac {x^m}{(a+b x) (c+d x)^2} \, dx\) [383]

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

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 125, normalized size of antiderivative = 1.00, 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 = {105, 162, 66} \begin {gather*} \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)} \end {gather*}

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 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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])))

Rule 105

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] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 162

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]

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*}

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Mathematica [A]
time = 0.11, size = 113, normalized size = 0.90 \begin {gather*} \frac {x^{1+m} \left (b^2 c^2 (c+d x) \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )+a d \left (-c (b c-a d) (1+m)+(b c (-1+m)-a d m) (c+d x) \, _2F_1\left (1,1+m;2+m;-\frac {d x}{c}\right )\right )\right )}{a c^2 (b c-a d)^2 (1+m) (c+d x)} \end {gather*}

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 = 0.04, size = 0, normalized size = 0.00 \[\int \frac {x^{m}}{\left (b x +a \right ) \left (d x +c \right )^{2}}\, dx\]

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 = 0.00, 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(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 = 0.00, 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(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(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: HeuristicGCDFailed} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: HeuristicGCDFailed >> no luck

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^m}{\left (a+b\,x\right )\,{\left (c+d\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m/((a + b*x)*(c + d*x)^2),x)

[Out]

int(x^m/((a + b*x)*(c + d*x)^2), x)

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