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

Optimal. Leaf size=94 \[ \frac {\left (a e^2+c d^2\right ) (B d-A e)}{2 e^4 (d+e x)^2}-\frac {a B e^2-2 A c d e+3 B c d^2}{e^4 (d+e x)}-\frac {c (3 B d-A e) \log (d+e x)}{e^4}+\frac {B c x}{e^3} \]

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Rubi [A]  time = 0.07, antiderivative size = 94, 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 {\left (a e^2+c d^2\right ) (B d-A e)}{2 e^4 (d+e x)^2}-\frac {a B e^2-2 A c d e+3 B c d^2}{e^4 (d+e x)}-\frac {c (3 B d-A e) \log (d+e x)}{e^4}+\frac {B c x}{e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.08, size = 88, normalized size = 0.94 \begin {gather*} \frac {\frac {\left (a e^2+c d^2\right ) (B d-A e)}{(d+e x)^2}-\frac {2 \left (a B e^2-2 A c d e+3 B c d^2\right )}{d+e x}+2 \log (d+e x) (A c e-3 B c d)+2 B c e x}{2 e^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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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)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 168, normalized size = 1.79 \begin {gather*} \frac {2 \, B c e^{3} x^{3} + 4 \, B c d e^{2} x^{2} - 5 \, B c d^{3} + 3 \, A c d^{2} e - B a d e^{2} - A a e^{3} - 2 \, {\left (2 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x - 2 \, {\left (3 \, B c d^{3} - A c d^{2} e + {\left (3 \, B c d e^{2} - A c e^{3}\right )} x^{2} + 2 \, {\left (3 \, B c d^{2} e - A c d e^{2}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} 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)^3,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 96, normalized size = 1.02 \begin {gather*} B c x e^{\left (-3\right )} - {\left (3 \, B c d - A c e\right )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) - \frac {{\left (5 \, B c d^{3} - 3 \, A c d^{2} e + B a d e^{2} + A a e^{3} + 2 \, {\left (3 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x\right )} e^{\left (-4\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)/(e*x+d)^3,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 144, normalized size = 1.53 \begin {gather*} -\frac {A a}{2 \left (e x +d \right )^{2} e}-\frac {A c \,d^{2}}{2 \left (e x +d \right )^{2} e^{3}}+\frac {B a d}{2 \left (e x +d \right )^{2} e^{2}}+\frac {B c \,d^{3}}{2 \left (e x +d \right )^{2} e^{4}}+\frac {2 A c d}{\left (e x +d \right ) e^{3}}+\frac {A c \ln \left (e x +d \right )}{e^{3}}-\frac {B a}{\left (e x +d \right ) e^{2}}-\frac {3 B c \,d^{2}}{\left (e x +d \right ) e^{4}}-\frac {3 B c d \ln \left (e x +d \right )}{e^{4}}+\frac {B c x}{e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.63, size = 111, normalized size = 1.18 \begin {gather*} -\frac {5 \, B c d^{3} - 3 \, A c d^{2} e + B a d e^{2} + A a e^{3} + 2 \, {\left (3 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x}{2 \, {\left (e^{6} x^{2} + 2 \, d e^{5} x + d^{2} e^{4}\right )}} + \frac {B c x}{e^{3}} - \frac {{\left (3 \, B c d - A c e\right )} \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)^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.76, size = 111, normalized size = 1.18 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (A\,c\,e-3\,B\,c\,d\right )}{e^4}-\frac {\frac {5\,B\,c\,d^3-3\,A\,c\,d^2\,e+B\,a\,d\,e^2+A\,a\,e^3}{2\,e}+x\,\left (3\,B\,c\,d^2-2\,A\,c\,d\,e+B\,a\,e^2\right )}{d^2\,e^3+2\,d\,e^4\,x+e^5\,x^2}+\frac {B\,c\,x}{e^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)^3,x)

[Out]

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

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sympy [A]  time = 1.27, size = 117, normalized size = 1.24 \begin {gather*} \frac {B c x}{e^{3}} - \frac {c \left (- A e + 3 B d\right ) \log {\left (d + e x \right )}}{e^{4}} + \frac {- A a e^{3} + 3 A c d^{2} e - B a d e^{2} - 5 B c d^{3} + x \left (4 A c d e^{2} - 2 B a e^{3} - 6 B c d^{2} e\right )}{2 d^{2} e^{4} + 4 d e^{5} x + 2 e^{6} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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