3.12.31 \(\int \frac {(A+B x) (a+c x^2)^2}{(d+e x)^3} \, dx\)

Optimal. Leaf size=185 \[ -\frac {\left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^6 (d+e x)}+\frac {\left (a e^2+c d^2\right )^2 (B d-A e)}{2 e^6 (d+e x)^2}+\frac {c x \left (2 a B e^2-3 A c d e+6 B c d^2\right )}{e^5}-\frac {2 c \log (d+e x) \left (-a A e^3+3 a B d e^2-3 A c d^2 e+5 B c d^3\right )}{e^6}-\frac {c^2 x^2 (3 B d-A e)}{2 e^4}+\frac {B c^2 x^3}{3 e^3} \]

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Rubi [A]  time = 0.20, antiderivative size = 185, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {772} \begin {gather*} \frac {c x \left (2 a B e^2-3 A c d e+6 B c d^2\right )}{e^5}-\frac {\left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{e^6 (d+e x)}+\frac {\left (a e^2+c d^2\right )^2 (B d-A e)}{2 e^6 (d+e x)^2}-\frac {2 c \log (d+e x) \left (-a A e^3+3 a B d e^2-3 A c d^2 e+5 B c d^3\right )}{e^6}-\frac {c^2 x^2 (3 B d-A e)}{2 e^4}+\frac {B c^2 x^3}{3 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^2)/(d + e*x)^3,x]

[Out]

(c*(6*B*c*d^2 - 3*A*c*d*e + 2*a*B*e^2)*x)/e^5 - (c^2*(3*B*d - A*e)*x^2)/(2*e^4) + (B*c^2*x^3)/(3*e^3) + ((B*d
- A*e)*(c*d^2 + a*e^2)^2)/(2*e^6*(d + e*x)^2) - ((c*d^2 + a*e^2)*(5*B*c*d^2 - 4*A*c*d*e + a*B*e^2))/(e^6*(d +
e*x)) - (2*c*(5*B*c*d^3 - 3*A*c*d^2*e + 3*a*B*d*e^2 - a*A*e^3)*Log[d + e*x])/e^6

Rule 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx &=\int \left (-\frac {c \left (-6 B c d^2+3 A c d e-2 a B e^2\right )}{e^5}+\frac {c^2 (-3 B d+A e) x}{e^4}+\frac {B c^2 x^2}{e^3}+\frac {(-B d+A e) \left (c d^2+a e^2\right )^2}{e^5 (d+e x)^3}+\frac {\left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right )}{e^5 (d+e x)^2}+\frac {2 c \left (-5 B c d^3+3 A c d^2 e-3 a B d e^2+a A e^3\right )}{e^5 (d+e x)}\right ) \, dx\\ &=\frac {c \left (6 B c d^2-3 A c d e+2 a B e^2\right ) x}{e^5}-\frac {c^2 (3 B d-A e) x^2}{2 e^4}+\frac {B c^2 x^3}{3 e^3}+\frac {(B d-A e) \left (c d^2+a e^2\right )^2}{2 e^6 (d+e x)^2}-\frac {\left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right )}{e^6 (d+e x)}-\frac {2 c \left (5 B c d^3-3 A c d^2 e+3 a B d e^2-a A e^3\right ) \log (d+e x)}{e^6}\\ \end {align*}

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Mathematica [A]  time = 0.16, size = 174, normalized size = 0.94 \begin {gather*} \frac {6 c e x \left (2 a B e^2-3 A c d e+6 B c d^2\right )-\frac {6 \left (a e^2+c d^2\right ) \left (a B e^2-4 A c d e+5 B c d^2\right )}{d+e x}+\frac {3 \left (a e^2+c d^2\right )^2 (B d-A e)}{(d+e x)^2}+12 c \log (d+e x) \left (a A e^3-3 a B d e^2+3 A c d^2 e-5 B c d^3\right )+3 c^2 e^2 x^2 (A e-3 B d)+2 B c^2 e^3 x^3}{6 e^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^2)/(d + e*x)^3,x]

[Out]

(6*c*e*(6*B*c*d^2 - 3*A*c*d*e + 2*a*B*e^2)*x + 3*c^2*e^2*(-3*B*d + A*e)*x^2 + 2*B*c^2*e^3*x^3 + (3*(B*d - A*e)
*(c*d^2 + a*e^2)^2)/(d + e*x)^2 - (6*(c*d^2 + a*e^2)*(5*B*c*d^2 - 4*A*c*d*e + a*B*e^2))/(d + e*x) + 12*c*(-5*B
*c*d^3 + 3*A*c*d^2*e - 3*a*B*d*e^2 + a*A*e^3)*Log[d + e*x])/(6*e^6)

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(A+B x) \left (a+c x^2\right )^2}{(d+e x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2)^2)/(d + e*x)^3,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2)^2)/(d + e*x)^3, x]

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fricas [B]  time = 0.40, size = 394, normalized size = 2.13 \begin {gather*} \frac {2 \, B c^{2} e^{5} x^{5} - 27 \, B c^{2} d^{5} + 21 \, A c^{2} d^{4} e - 30 \, B a c d^{3} e^{2} + 18 \, A a c d^{2} e^{3} - 3 \, B a^{2} d e^{4} - 3 \, A a^{2} e^{5} - {\left (5 \, B c^{2} d e^{4} - 3 \, A c^{2} e^{5}\right )} x^{4} + 4 \, {\left (5 \, B c^{2} d^{2} e^{3} - 3 \, A c^{2} d e^{4} + 3 \, B a c e^{5}\right )} x^{3} + 3 \, {\left (21 \, B c^{2} d^{3} e^{2} - 11 \, A c^{2} d^{2} e^{3} + 8 \, B a c d e^{4}\right )} x^{2} + 6 \, {\left (B c^{2} d^{4} e + A c^{2} d^{3} e^{2} - 4 \, B a c d^{2} e^{3} + 4 \, A a c d e^{4} - B a^{2} e^{5}\right )} x - 12 \, {\left (5 \, B c^{2} d^{5} - 3 \, A c^{2} d^{4} e + 3 \, B a c d^{3} e^{2} - A a c d^{2} e^{3} + {\left (5 \, B c^{2} d^{3} e^{2} - 3 \, A c^{2} d^{2} e^{3} + 3 \, B a c d e^{4} - A a c e^{5}\right )} x^{2} + 2 \, {\left (5 \, B c^{2} d^{4} e - 3 \, A c^{2} d^{3} e^{2} + 3 \, B a c d^{2} e^{3} - A a c d e^{4}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2/(e*x+d)^3,x, algorithm="fricas")

[Out]

1/6*(2*B*c^2*e^5*x^5 - 27*B*c^2*d^5 + 21*A*c^2*d^4*e - 30*B*a*c*d^3*e^2 + 18*A*a*c*d^2*e^3 - 3*B*a^2*d*e^4 - 3
*A*a^2*e^5 - (5*B*c^2*d*e^4 - 3*A*c^2*e^5)*x^4 + 4*(5*B*c^2*d^2*e^3 - 3*A*c^2*d*e^4 + 3*B*a*c*e^5)*x^3 + 3*(21
*B*c^2*d^3*e^2 - 11*A*c^2*d^2*e^3 + 8*B*a*c*d*e^4)*x^2 + 6*(B*c^2*d^4*e + A*c^2*d^3*e^2 - 4*B*a*c*d^2*e^3 + 4*
A*a*c*d*e^4 - B*a^2*e^5)*x - 12*(5*B*c^2*d^5 - 3*A*c^2*d^4*e + 3*B*a*c*d^3*e^2 - A*a*c*d^2*e^3 + (5*B*c^2*d^3*
e^2 - 3*A*c^2*d^2*e^3 + 3*B*a*c*d*e^4 - A*a*c*e^5)*x^2 + 2*(5*B*c^2*d^4*e - 3*A*c^2*d^3*e^2 + 3*B*a*c*d^2*e^3
- A*a*c*d*e^4)*x)*log(e*x + d))/(e^8*x^2 + 2*d*e^7*x + d^2*e^6)

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giac [A]  time = 0.19, size = 237, normalized size = 1.28 \begin {gather*} -2 \, {\left (5 \, B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 3 \, B a c d e^{2} - A a c e^{3}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{6} \, {\left (2 \, B c^{2} x^{3} e^{6} - 9 \, B c^{2} d x^{2} e^{5} + 36 \, B c^{2} d^{2} x e^{4} + 3 \, A c^{2} x^{2} e^{6} - 18 \, A c^{2} d x e^{5} + 12 \, B a c x e^{6}\right )} e^{\left (-9\right )} - \frac {{\left (9 \, B c^{2} d^{5} - 7 \, A c^{2} d^{4} e + 10 \, B a c d^{3} e^{2} - 6 \, A a c d^{2} e^{3} + B a^{2} d e^{4} + A a^{2} e^{5} + 2 \, {\left (5 \, B c^{2} d^{4} e - 4 \, A c^{2} d^{3} e^{2} + 6 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} + B a^{2} e^{5}\right )} x\right )} e^{\left (-6\right )}}{2 \, {\left (x e + d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2/(e*x+d)^3,x, algorithm="giac")

[Out]

-2*(5*B*c^2*d^3 - 3*A*c^2*d^2*e + 3*B*a*c*d*e^2 - A*a*c*e^3)*e^(-6)*log(abs(x*e + d)) + 1/6*(2*B*c^2*x^3*e^6 -
9*B*c^2*d*x^2*e^5 + 36*B*c^2*d^2*x*e^4 + 3*A*c^2*x^2*e^6 - 18*A*c^2*d*x*e^5 + 12*B*a*c*x*e^6)*e^(-9) - 1/2*(9
*B*c^2*d^5 - 7*A*c^2*d^4*e + 10*B*a*c*d^3*e^2 - 6*A*a*c*d^2*e^3 + B*a^2*d*e^4 + A*a^2*e^5 + 2*(5*B*c^2*d^4*e -
4*A*c^2*d^3*e^2 + 6*B*a*c*d^2*e^3 - 4*A*a*c*d*e^4 + B*a^2*e^5)*x)*e^(-6)/(x*e + d)^2

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maple [A]  time = 0.06, size = 331, normalized size = 1.79 \begin {gather*} \frac {B \,c^{2} x^{3}}{3 e^{3}}-\frac {A \,a^{2}}{2 \left (e x +d \right )^{2} e}-\frac {A a c \,d^{2}}{\left (e x +d \right )^{2} e^{3}}-\frac {A \,c^{2} d^{4}}{2 \left (e x +d \right )^{2} e^{5}}+\frac {A \,c^{2} x^{2}}{2 e^{3}}+\frac {B \,a^{2} d}{2 \left (e x +d \right )^{2} e^{2}}+\frac {B a c \,d^{3}}{\left (e x +d \right )^{2} e^{4}}+\frac {B \,c^{2} d^{5}}{2 \left (e x +d \right )^{2} e^{6}}-\frac {3 B \,c^{2} d \,x^{2}}{2 e^{4}}+\frac {4 A a c d}{\left (e x +d \right ) e^{3}}+\frac {2 A a c \ln \left (e x +d \right )}{e^{3}}+\frac {4 A \,c^{2} d^{3}}{\left (e x +d \right ) e^{5}}+\frac {6 A \,c^{2} d^{2} \ln \left (e x +d \right )}{e^{5}}-\frac {3 A \,c^{2} d x}{e^{4}}-\frac {B \,a^{2}}{\left (e x +d \right ) e^{2}}-\frac {6 B a c \,d^{2}}{\left (e x +d \right ) e^{4}}-\frac {6 B a c d \ln \left (e x +d \right )}{e^{4}}+\frac {2 B a c x}{e^{3}}-\frac {5 B \,c^{2} d^{4}}{\left (e x +d \right ) e^{6}}-\frac {10 B \,c^{2} d^{3} \ln \left (e x +d \right )}{e^{6}}+\frac {6 B \,c^{2} d^{2} x}{e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^2/(e*x+d)^3,x)

[Out]

1/3*B*c^2/e^3*x^3+1/2*c^2/e^3*A*x^2-3/2*c^2/e^4*B*x^2*d-3*c^2/e^4*A*x*d+2*c/e^3*B*x*a+6*c^2/e^5*B*x*d^2+4/e^3/
(e*x+d)*A*a*c*d+4/e^5/(e*x+d)*A*c^2*d^3-1/e^2/(e*x+d)*B*a^2-6/e^4/(e*x+d)*B*a*c*d^2-5/e^6/(e*x+d)*B*c^2*d^4-1/
2/e/(e*x+d)^2*A*a^2-1/e^3/(e*x+d)^2*A*d^2*a*c-1/2/e^5/(e*x+d)^2*A*c^2*d^4+1/2/e^2/(e*x+d)^2*B*d*a^2+1/e^4/(e*x
+d)^2*B*a*c*d^3+1/2/e^6/(e*x+d)^2*B*c^2*d^5+2*c/e^3*ln(e*x+d)*a*A+6*c^2/e^5*ln(e*x+d)*A*d^2-6*c/e^4*ln(e*x+d)*
a*B*d-10*c^2/e^6*ln(e*x+d)*B*d^3

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maxima [A]  time = 0.56, size = 258, normalized size = 1.39 \begin {gather*} -\frac {9 \, B c^{2} d^{5} - 7 \, A c^{2} d^{4} e + 10 \, B a c d^{3} e^{2} - 6 \, A a c d^{2} e^{3} + B a^{2} d e^{4} + A a^{2} e^{5} + 2 \, {\left (5 \, B c^{2} d^{4} e - 4 \, A c^{2} d^{3} e^{2} + 6 \, B a c d^{2} e^{3} - 4 \, A a c d e^{4} + B a^{2} e^{5}\right )} x}{2 \, {\left (e^{8} x^{2} + 2 \, d e^{7} x + d^{2} e^{6}\right )}} + \frac {2 \, B c^{2} e^{2} x^{3} - 3 \, {\left (3 \, B c^{2} d e - A c^{2} e^{2}\right )} x^{2} + 6 \, {\left (6 \, B c^{2} d^{2} - 3 \, A c^{2} d e + 2 \, B a c e^{2}\right )} x}{6 \, e^{5}} - \frac {2 \, {\left (5 \, B c^{2} d^{3} - 3 \, A c^{2} d^{2} e + 3 \, B a c d e^{2} - A a c e^{3}\right )} \log \left (e x + d\right )}{e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)^2/(e*x+d)^3,x, algorithm="maxima")

[Out]

-1/2*(9*B*c^2*d^5 - 7*A*c^2*d^4*e + 10*B*a*c*d^3*e^2 - 6*A*a*c*d^2*e^3 + B*a^2*d*e^4 + A*a^2*e^5 + 2*(5*B*c^2*
d^4*e - 4*A*c^2*d^3*e^2 + 6*B*a*c*d^2*e^3 - 4*A*a*c*d*e^4 + B*a^2*e^5)*x)/(e^8*x^2 + 2*d*e^7*x + d^2*e^6) + 1/
6*(2*B*c^2*e^2*x^3 - 3*(3*B*c^2*d*e - A*c^2*e^2)*x^2 + 6*(6*B*c^2*d^2 - 3*A*c^2*d*e + 2*B*a*c*e^2)*x)/e^5 - 2*
(5*B*c^2*d^3 - 3*A*c^2*d^2*e + 3*B*a*c*d*e^2 - A*a*c*e^3)*log(e*x + d)/e^6

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mupad [B]  time = 1.75, size = 275, normalized size = 1.49 \begin {gather*} x^2\,\left (\frac {A\,c^2}{2\,e^3}-\frac {3\,B\,c^2\,d}{2\,e^4}\right )-\frac {x\,\left (B\,a^2\,e^4+6\,B\,a\,c\,d^2\,e^2-4\,A\,a\,c\,d\,e^3+5\,B\,c^2\,d^4-4\,A\,c^2\,d^3\,e\right )+\frac {B\,a^2\,d\,e^4+A\,a^2\,e^5+10\,B\,a\,c\,d^3\,e^2-6\,A\,a\,c\,d^2\,e^3+9\,B\,c^2\,d^5-7\,A\,c^2\,d^4\,e}{2\,e}}{d^2\,e^5+2\,d\,e^6\,x+e^7\,x^2}-x\,\left (\frac {3\,d\,\left (\frac {A\,c^2}{e^3}-\frac {3\,B\,c^2\,d}{e^4}\right )}{e}-\frac {2\,B\,a\,c}{e^3}+\frac {3\,B\,c^2\,d^2}{e^5}\right )-\frac {\ln \left (d+e\,x\right )\,\left (10\,B\,c^2\,d^3-6\,A\,c^2\,d^2\,e+6\,B\,a\,c\,d\,e^2-2\,A\,a\,c\,e^3\right )}{e^6}+\frac {B\,c^2\,x^3}{3\,e^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)^2*(A + B*x))/(d + e*x)^3,x)

[Out]

x^2*((A*c^2)/(2*e^3) - (3*B*c^2*d)/(2*e^4)) - (x*(B*a^2*e^4 + 5*B*c^2*d^4 - 4*A*c^2*d^3*e - 4*A*a*c*d*e^3 + 6*
B*a*c*d^2*e^2) + (A*a^2*e^5 + 9*B*c^2*d^5 + B*a^2*d*e^4 - 7*A*c^2*d^4*e - 6*A*a*c*d^2*e^3 + 10*B*a*c*d^3*e^2)/
(2*e))/(d^2*e^5 + e^7*x^2 + 2*d*e^6*x) - x*((3*d*((A*c^2)/e^3 - (3*B*c^2*d)/e^4))/e - (2*B*a*c)/e^3 + (3*B*c^2
*d^2)/e^5) - (log(d + e*x)*(10*B*c^2*d^3 - 2*A*a*c*e^3 - 6*A*c^2*d^2*e + 6*B*a*c*d*e^2))/e^6 + (B*c^2*x^3)/(3*
e^3)

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sympy [A]  time = 3.33, size = 282, normalized size = 1.52 \begin {gather*} \frac {B c^{2} x^{3}}{3 e^{3}} - \frac {2 c \left (- A a e^{3} - 3 A c d^{2} e + 3 B a d e^{2} + 5 B c d^{3}\right ) \log {\left (d + e x \right )}}{e^{6}} + x^{2} \left (\frac {A c^{2}}{2 e^{3}} - \frac {3 B c^{2} d}{2 e^{4}}\right ) + x \left (- \frac {3 A c^{2} d}{e^{4}} + \frac {2 B a c}{e^{3}} + \frac {6 B c^{2} d^{2}}{e^{5}}\right ) + \frac {- A a^{2} e^{5} + 6 A a c d^{2} e^{3} + 7 A c^{2} d^{4} e - B a^{2} d e^{4} - 10 B a c d^{3} e^{2} - 9 B c^{2} d^{5} + x \left (8 A a c d e^{4} + 8 A c^{2} d^{3} e^{2} - 2 B a^{2} e^{5} - 12 B a c d^{2} e^{3} - 10 B c^{2} d^{4} e\right )}{2 d^{2} e^{6} + 4 d e^{7} x + 2 e^{8} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)**2/(e*x+d)**3,x)

[Out]

B*c**2*x**3/(3*e**3) - 2*c*(-A*a*e**3 - 3*A*c*d**2*e + 3*B*a*d*e**2 + 5*B*c*d**3)*log(d + e*x)/e**6 + x**2*(A*
c**2/(2*e**3) - 3*B*c**2*d/(2*e**4)) + x*(-3*A*c**2*d/e**4 + 2*B*a*c/e**3 + 6*B*c**2*d**2/e**5) + (-A*a**2*e**
5 + 6*A*a*c*d**2*e**3 + 7*A*c**2*d**4*e - B*a**2*d*e**4 - 10*B*a*c*d**3*e**2 - 9*B*c**2*d**5 + x*(8*A*a*c*d*e*
*4 + 8*A*c**2*d**3*e**2 - 2*B*a**2*e**5 - 12*B*a*c*d**2*e**3 - 10*B*c**2*d**4*e))/(2*d**2*e**6 + 4*d*e**7*x +
2*e**8*x**2)

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