3.5.44 \(\int \frac {(d+e x)^m}{(b x+c x^2)^2} \, dx\) [444]

Optimal. Leaf size=180 \[ -\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{b^2 d (c d-b e) \left (b x+c x^2\right )}-\frac {c^2 (2 c d-b e (2-m)) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {c (d+e x)}{c d-b e}\right )}{b^3 (c d-b e)^2 (1+m)}+\frac {(2 c d-b e m) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;1+\frac {e x}{d}\right )}{b^3 d^2 (1+m)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {754, 844, 67, 70} \begin {gather*} -\frac {c^2 (d+e x)^{m+1} (2 c d-b e (2-m)) \, _2F_1\left (1,m+1;m+2;\frac {c (d+e x)}{c d-b e}\right )}{b^3 (m+1) (c d-b e)^2}+\frac {(d+e x)^{m+1} (2 c d-b e m) \, _2F_1\left (1,m+1;m+2;\frac {e x}{d}+1\right )}{b^3 d^2 (m+1)}-\frac {(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{b^2 d \left (b x+c x^2\right ) (c d-b e)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m/(b*x + c*x^2)^2,x]

[Out]

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

Rule 67

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

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 754

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(d + e*x)^(m + 1)*(b
*c*d - b^2*e + 2*a*c*e + c*(2*c*d - b*e)*x)*((a + b*x + c*x^2)^(p + 1)/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e +
a*e^2))), x] + Dist[1/((p + 1)*(b^2 - 4*a*c)*(c*d^2 - b*d*e + a*e^2)), Int[(d + e*x)^m*Simp[b*c*d*e*(2*p - m
+ 2) + b^2*e^2*(m + p + 2) - 2*c^2*d^2*(2*p + 3) - 2*a*c*e^2*(m + 2*p + 3) - c*e*(2*c*d - b*e)*(m + 2*p + 4)*x
, x]*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b
*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && LtQ[p, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 844

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Int[Exp
andIntegrand[(d + e*x)^m, (f + g*x)/(a + b*x + c*x^2), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b^2 -
4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] &&  !RationalQ[m]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.14, size = 174, normalized size = 0.97 \begin {gather*} -\frac {(d+e x)^{1+m} \left (b^2 d (c d-b e)^2 (1+m)+b c d (-2 c d+b e) (-c d+b e) (1+m) x+x (b+c x) \left (c^2 d^2 (2 c d+b e (-2+m)) \, _2F_1\left (1,1+m;2+m;\frac {c (d+e x)}{c d-b e}\right )-(c d-b e)^2 (2 c d-b e m) \, _2F_1\left (1,1+m;2+m;1+\frac {e x}{d}\right )\right )\right )}{b^3 d^2 (c d-b e)^2 (1+m) x (b+c x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^m/(b*x + c*x^2)^2,x]

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.15, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+b x \right )^{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^m/(c*x^2+b*x)^2,x)

[Out]

int((e*x+d)^m/(c*x^2+b*x)^2,x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{m}}{x^{2} \left (b + c x\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d + e*x)**m/(x**2*(b + c*x)**2), x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m/(b*x + c*x^2)^2,x)

[Out]

int((d + e*x)^m/(b*x + c*x^2)^2, x)

________________________________________________________________________________________