3.12.19 \(\int \frac {(A+B x) (a+c x^2)}{(d+e x)^4} \, dx\)

Optimal. Leaf size=101 \[ -\frac {a B e^2-2 A c d e+3 B c d^2}{2 e^4 (d+e x)^2}+\frac {\left (a e^2+c d^2\right ) (B d-A e)}{3 e^4 (d+e x)^3}+\frac {c (3 B d-A e)}{e^4 (d+e x)}+\frac {B c \log (d+e x)}{e^4} \]

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Rubi [A]  time = 0.07, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 1, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {772} \begin {gather*} -\frac {a B e^2-2 A c d e+3 B c d^2}{2 e^4 (d+e x)^2}+\frac {\left (a e^2+c d^2\right ) (B d-A e)}{3 e^4 (d+e x)^3}+\frac {c (3 B d-A e)}{e^4 (d+e x)}+\frac {B c \log (d+e x)}{e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2))/(d + e*x)^4,x]

[Out]

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

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

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Mathematica [A]  time = 0.05, size = 98, normalized size = 0.97 \begin {gather*} \frac {-2 A e \left (a e^2+c \left (d^2+3 d e x+3 e^2 x^2\right )\right )+B \left (c d \left (11 d^2+27 d e x+18 e^2 x^2\right )-a e^2 (d+3 e x)\right )+6 B c (d+e x)^3 \log (d+e x)}{6 e^4 (d+e x)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2))/(d + e*x)^4,x]

[Out]

(-2*A*e*(a*e^2 + c*(d^2 + 3*d*e*x + 3*e^2*x^2)) + B*(-(a*e^2*(d + 3*e*x)) + c*d*(11*d^2 + 27*d*e*x + 18*e^2*x^
2)) + 6*B*c*(d + e*x)^3*Log[d + e*x])/(6*e^4*(d + e*x)^3)

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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 )}{(d+e x)^4} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2))/(d + e*x)^4,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(a + c*x^2))/(d + e*x)^4, x]

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fricas [A]  time = 0.41, size = 160, normalized size = 1.58 \begin {gather*} \frac {11 \, B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} - 2 \, A a e^{3} + 6 \, {\left (3 \, B c d e^{2} - A c e^{3}\right )} x^{2} + 3 \, {\left (9 \, B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x + 6 \, {\left (B c e^{3} x^{3} + 3 \, B c d e^{2} x^{2} + 3 \, B c d^{2} e x + B c d^{3}\right )} \log \left (e x + d\right )}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)/(e*x+d)^4,x, algorithm="fricas")

[Out]

1/6*(11*B*c*d^3 - 2*A*c*d^2*e - B*a*d*e^2 - 2*A*a*e^3 + 6*(3*B*c*d*e^2 - A*c*e^3)*x^2 + 3*(9*B*c*d^2*e - 2*A*c
*d*e^2 - B*a*e^3)*x + 6*(B*c*e^3*x^3 + 3*B*c*d*e^2*x^2 + 3*B*c*d^2*e*x + B*c*d^3)*log(e*x + d))/(e^7*x^3 + 3*d
*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4)

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giac [A]  time = 0.16, size = 103, normalized size = 1.02 \begin {gather*} B c e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {{\left (6 \, {\left (3 \, B c d e - A c e^{2}\right )} x^{2} + 3 \, {\left (9 \, B c d^{2} - 2 \, A c d e - B a e^{2}\right )} x + {\left (11 \, B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} - 2 \, A a e^{3}\right )} e^{\left (-1\right )}\right )} e^{\left (-3\right )}}{6 \, {\left (x e + d\right )}^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)/(e*x+d)^4,x, algorithm="giac")

[Out]

B*c*e^(-4)*log(abs(x*e + d)) + 1/6*(6*(3*B*c*d*e - A*c*e^2)*x^2 + 3*(9*B*c*d^2 - 2*A*c*d*e - B*a*e^2)*x + (11*
B*c*d^3 - 2*A*c*d^2*e - B*a*d*e^2 - 2*A*a*e^3)*e^(-1))*e^(-3)/(x*e + d)^3

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maple [A]  time = 0.06, size = 151, normalized size = 1.50 \begin {gather*} -\frac {A a}{3 \left (e x +d \right )^{3} e}-\frac {A c \,d^{2}}{3 \left (e x +d \right )^{3} e^{3}}+\frac {B a d}{3 \left (e x +d \right )^{3} e^{2}}+\frac {B c \,d^{3}}{3 \left (e x +d \right )^{3} e^{4}}+\frac {A c d}{\left (e x +d \right )^{2} e^{3}}-\frac {B a}{2 \left (e x +d \right )^{2} e^{2}}-\frac {3 B c \,d^{2}}{2 \left (e x +d \right )^{2} e^{4}}-\frac {A c}{\left (e x +d \right ) e^{3}}+\frac {3 B c d}{\left (e x +d \right ) e^{4}}+\frac {B c \ln \left (e x +d \right )}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)/(e*x+d)^4,x)

[Out]

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

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maxima [A]  time = 0.46, size = 129, normalized size = 1.28 \begin {gather*} \frac {11 \, B c d^{3} - 2 \, A c d^{2} e - B a d e^{2} - 2 \, A a e^{3} + 6 \, {\left (3 \, B c d e^{2} - A c e^{3}\right )} x^{2} + 3 \, {\left (9 \, B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x}{6 \, {\left (e^{7} x^{3} + 3 \, d e^{6} x^{2} + 3 \, d^{2} e^{5} x + d^{3} e^{4}\right )}} + \frac {B c \log \left (e x + d\right )}{e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x^2+a)/(e*x+d)^4,x, algorithm="maxima")

[Out]

1/6*(11*B*c*d^3 - 2*A*c*d^2*e - B*a*d*e^2 - 2*A*a*e^3 + 6*(3*B*c*d*e^2 - A*c*e^3)*x^2 + 3*(9*B*c*d^2*e - 2*A*c
*d*e^2 - B*a*e^3)*x)/(e^7*x^3 + 3*d*e^6*x^2 + 3*d^2*e^5*x + d^3*e^4) + B*c*log(e*x + d)/e^4

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mupad [B]  time = 0.10, size = 122, normalized size = 1.21 \begin {gather*} \frac {B\,c\,\ln \left (d+e\,x\right )}{e^4}-\frac {\frac {-11\,B\,c\,d^3+2\,A\,c\,d^2\,e+B\,a\,d\,e^2+2\,A\,a\,e^3}{6\,e^4}+\frac {x\,\left (-9\,B\,c\,d^2+2\,A\,c\,d\,e+B\,a\,e^2\right )}{2\,e^3}+\frac {c\,x^2\,\left (A\,e-3\,B\,d\right )}{e^2}}{d^3+3\,d^2\,e\,x+3\,d\,e^2\,x^2+e^3\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + c*x^2)*(A + B*x))/(d + e*x)^4,x)

[Out]

(B*c*log(d + e*x))/e^4 - ((2*A*a*e^3 - 11*B*c*d^3 + B*a*d*e^2 + 2*A*c*d^2*e)/(6*e^4) + (x*(B*a*e^2 - 9*B*c*d^2
+ 2*A*c*d*e))/(2*e^3) + (c*x^2*(A*e - 3*B*d))/e^2)/(d^3 + e^3*x^3 + 3*d*e^2*x^2 + 3*d^2*e*x)

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sympy [A]  time = 2.60, size = 138, normalized size = 1.37 \begin {gather*} \frac {B c \log {\left (d + e x \right )}}{e^{4}} + \frac {- 2 A a e^{3} - 2 A c d^{2} e - B a d e^{2} + 11 B c d^{3} + x^{2} \left (- 6 A c e^{3} + 18 B c d e^{2}\right ) + x \left (- 6 A c d e^{2} - 3 B a e^{3} + 27 B c d^{2} e\right )}{6 d^{3} e^{4} + 18 d^{2} e^{5} x + 18 d e^{6} x^{2} + 6 e^{7} x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(c*x**2+a)/(e*x+d)**4,x)

[Out]

B*c*log(d + e*x)/e**4 + (-2*A*a*e**3 - 2*A*c*d**2*e - B*a*d*e**2 + 11*B*c*d**3 + x**2*(-6*A*c*e**3 + 18*B*c*d*
e**2) + x*(-6*A*c*d*e**2 - 3*B*a*e**3 + 27*B*c*d**2*e))/(6*d**3*e**4 + 18*d**2*e**5*x + 18*d*e**6*x**2 + 6*e**
7*x**3)

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