3.445 \(\int \frac {(d+e x)^m}{(b x+c x^2)^3} \, dx\)

Optimal. Leaf size=350 \[ -\frac {(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{2 b^2 d \left (b x+c x^2\right )^2 (c d-b e)}-\frac {(d+e x)^{m+1} \left (-b^2 e^2 (1-m) m-6 b c d e m+12 c^2 d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac {e x}{d}+1\right )}{2 b^5 d^3 (m+1)}+\frac {c^3 (d+e x)^{m+1} \left (b^2 e^2 \left (m^2-7 m+12\right )-6 b c d e (4-m)+12 c^2 d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac {c (d+e x)}{c d-b e}\right )}{2 b^5 (m+1) (c d-b e)^3}+\frac {(d+e x)^{m+1} \left (c x (2 c d-b e) \left (-b^2 e^2 (1-m)-6 b c d e+6 c^2 d^2\right )+b (c d-b e) \left (-b^2 e^2 (1-m)-b c d e (m+4)+6 c^2 d^2\right )\right )}{2 b^4 d^2 \left (b x+c x^2\right ) (c d-b e)^2} \]

[Out]

-1/2*(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)^2+1/2*(e*x+d)^(1+m)*(b*(-b*e+c
*d)*(6*c^2*d^2-b^2*e^2*(1-m)-b*c*d*e*(4+m))+c*(-b*e+2*c*d)*(6*c^2*d^2-6*b*c*d*e-b^2*e^2*(1-m))*x)/b^4/d^2/(-b*
e+c*d)^2/(c*x^2+b*x)+1/2*c^3*(12*c^2*d^2-6*b*c*d*e*(4-m)+b^2*e^2*(m^2-7*m+12))*(e*x+d)^(1+m)*hypergeom([1, 1+m
],[2+m],c*(e*x+d)/(-b*e+c*d))/b^5/(-b*e+c*d)^3/(1+m)-1/2*(12*c^2*d^2-6*b*c*d*e*m-b^2*e^2*(1-m)*m)*(e*x+d)^(1+m
)*hypergeom([1, 1+m],[2+m],1+e*x/d)/b^5/d^3/(1+m)

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Rubi [A]  time = 0.43, antiderivative size = 350, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {740, 822, 830, 65, 68} \[ \frac {c^3 (d+e x)^{m+1} \left (b^2 e^2 \left (m^2-7 m+12\right )-6 b c d e (4-m)+12 c^2 d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac {c (d+e x)}{c d-b e}\right )}{2 b^5 (m+1) (c d-b e)^3}-\frac {(d+e x)^{m+1} \left (-b^2 e^2 (1-m) m-6 b c d e m+12 c^2 d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac {e x}{d}+1\right )}{2 b^5 d^3 (m+1)}+\frac {(d+e x)^{m+1} \left (c x (2 c d-b e) \left (-b^2 e^2 (1-m)-6 b c d e+6 c^2 d^2\right )+b (c d-b e) \left (-b^2 e^2 (1-m)-b c d e (m+4)+6 c^2 d^2\right )\right )}{2 b^4 d^2 \left (b x+c x^2\right ) (c d-b e)^2}-\frac {(d+e x)^{m+1} (c x (2 c d-b e)+b (c d-b e))}{2 b^2 d \left (b x+c x^2\right )^2 (c d-b e)} \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^m/(b*x + c*x^2)^3,x]

[Out]

-((d + e*x)^(1 + m)*(b*(c*d - b*e) + c*(2*c*d - b*e)*x))/(2*b^2*d*(c*d - b*e)*(b*x + c*x^2)^2) + ((d + e*x)^(1
+ m)*(b*(c*d - b*e)*(6*c^2*d^2 - b^2*e^2*(1 - m) - b*c*d*e*(4 + m)) + c*(2*c*d - b*e)*(6*c^2*d^2 - 6*b*c*d*e
- b^2*e^2*(1 - m))*x))/(2*b^4*d^2*(c*d - b*e)^2*(b*x + c*x^2)) + (c^3*(12*c^2*d^2 - 6*b*c*d*e*(4 - m) + b^2*e^
2*(12 - 7*m + m^2))*(d + e*x)^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(d + e*x))/(c*d - b*e)])/(2*b^5*(c
*d - b*e)^3*(1 + m)) - ((12*c^2*d^2 - 6*b*c*d*e*m - b^2*e^2*(1 - m)*m)*(d + e*x)^(1 + m)*Hypergeometric2F1[1,
1 + m, 2 + m, 1 + (e*x)/d])/(2*b^5*d^3*(1 + m))

Rule 65

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

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]

Rule 740

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 822

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

Rule 830

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 )^3} \, dx &=-\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}-\frac {\int \frac {(d+e x)^m \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)+c e (2 c d-b e) (2-m) x\right )}{\left (b x+c x^2\right )^2} \, dx}{2 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)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (b (c d-b e) \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right )+c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) x\right )}{2 b^4 d^2 (c d-b e)^2 \left (b x+c x^2\right )}+\frac {\int \frac {(d+e x)^m \left ((c d-b e)^2 \left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right )-c e (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) m x\right )}{b x+c x^2} \, dx}{2 b^4 d^2 (c d-b e)^2}\\ &=-\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (b (c d-b e) \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right )+c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) x\right )}{2 b^4 d^2 (c d-b e)^2 \left (b x+c x^2\right )}+\frac {\int \left (\frac {(c d-b e)^2 \left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right ) (d+e x)^m}{b x}+\frac {c^3 d^2 \left (-12 c^2 d^2+6 b c d e (4-m)-b^2 e^2 \left (12-7 m+m^2\right )\right ) (d+e x)^m}{b (b+c x)}\right ) \, dx}{2 b^4 d^2 (c d-b e)^2}\\ &=-\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (b (c d-b e) \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right )+c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) x\right )}{2 b^4 d^2 (c d-b e)^2 \left (b x+c x^2\right )}+\frac {\left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right ) \int \frac {(d+e x)^m}{x} \, dx}{2 b^5 d^2}-\frac {\left (c^3 \left (12 c^2 d^2-6 b c d e (4-m)+b^2 e^2 \left (12-7 m+m^2\right )\right )\right ) \int \frac {(d+e x)^m}{b+c x} \, dx}{2 b^5 (c d-b e)^2}\\ &=-\frac {(d+e x)^{1+m} (b (c d-b e)+c (2 c d-b e) x)}{2 b^2 d (c d-b e) \left (b x+c x^2\right )^2}+\frac {(d+e x)^{1+m} \left (b (c d-b e) \left (6 c^2 d^2-b^2 e^2 (1-m)-b c d e (4+m)\right )+c (2 c d-b e) \left (6 c^2 d^2-6 b c d e-b^2 e^2 (1-m)\right ) x\right )}{2 b^4 d^2 (c d-b e)^2 \left (b x+c x^2\right )}+\frac {c^3 \left (12 c^2 d^2-6 b c d e (4-m)+b^2 e^2 \left (12-7 m+m^2\right )\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;\frac {c (d+e x)}{c d-b e}\right )}{2 b^5 (c d-b e)^3 (1+m)}-\frac {\left (12 c^2 d^2-6 b c d e m-b^2 e^2 (1-m) m\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;1+\frac {e x}{d}\right )}{2 b^5 d^3 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.48, size = 337, normalized size = 0.96 \[ -\frac {(d+e x)^{m+1} \left (2 b^4 d^2 (m+1) (c d-b e)^3-2 b^3 d (m+1) x (c d-b e)^3 (4 c d-b e (m-1))+x^2 \left (-2 b^2 c d (m+1) (c d-b e)^2 \left (b^2 e^2 (m-1)-b c d e (m+4)+6 c^2 d^2\right )-(b+c x) \left ((b+c x) \left (2 c^3 d^3 \left (b^2 e^2 \left (m^2-7 m+12\right )+6 b c d e (m-4)+12 c^2 d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac {c (d+e x)}{c d-b e}\right )-2 (c d-b e)^3 \left (b^2 e^2 (m-1) m-6 b c d e m+12 c^2 d^2\right ) \, _2F_1\left (1,m+1;m+2;\frac {e x}{d}+1\right )\right )-2 b c d (m+1) (2 c d-b e) (b e-c d) \left (b^2 e^2 (m-1)-6 b c d e+6 c^2 d^2\right )\right )\right )\right )}{4 b^5 d^3 (m+1) x^2 (b+c x)^2 (c d-b e)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((d + e*x)^(1 + m)*(2*b^4*d^2*(c*d - b*e)^3*(1 + m) - 2*b^3*d*(c*d - b*e)^3*(4*c*d - b*e*(-1 + m))*(1 + m
)*x + x^2*(-2*b^2*c*d*(c*d - b*e)^2*(1 + m)*(6*c^2*d^2 + b^2*e^2*(-1 + m) - b*c*d*e*(4 + m)) - (b + c*x)*(-2*b
*c*d*(2*c*d - b*e)*(-(c*d) + b*e)*(6*c^2*d^2 - 6*b*c*d*e + b^2*e^2*(-1 + m))*(1 + m) + (b + c*x)*(2*c^3*d^3*(1
2*c^2*d^2 + 6*b*c*d*e*(-4 + m) + b^2*e^2*(12 - 7*m + m^2))*Hypergeometric2F1[1, 1 + m, 2 + m, (c*(d + e*x))/(c
*d - b*e)] - 2*(c*d - b*e)^3*(12*c^2*d^2 - 6*b*c*d*e*m + b^2*e^2*(-1 + m)*m)*Hypergeometric2F1[1, 1 + m, 2 + m
, 1 + (e*x)/d])))))/(b^5*d^3*(c*d - b*e)^3*(1 + m)*x^2*(b + c*x)^2)

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fricas [F]  time = 1.28, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (e x + d\right )}^{m}}{c^{3} x^{6} + 3 \, b c^{2} x^{5} + 3 \, b^{2} c x^{4} + b^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(c*x^2+b*x)^3,x, algorithm="fricas")

[Out]

integral((e*x + d)^m/(c^3*x^6 + 3*b*c^2*x^5 + 3*b^2*c*x^4 + b^3*x^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(c*x^2+b*x)^3,x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.73, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x +d \right )^{m}}{\left (c \,x^{2}+b x \right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (e x + d\right )}^{m}}{{\left (c x^{2} + b x\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^m/(c*x^2+b*x)^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (d+e\,x\right )}^m}{{\left (c\,x^2+b\,x\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^m/(b*x + c*x^2)^3,x)

[Out]

int((d + e*x)^m/(b*x + c*x^2)^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (d + e x\right )^{m}}{x^{3} \left (b + c x\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m/(c*x**2+b*x)**3,x)

[Out]

Integral((d + e*x)**m/(x**3*(b + c*x)**3), x)

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