3.1291 \(\int \frac {(A+B x) (a+c x^2)}{(d+e x)^2} \, dx\)

Optimal. Leaf size=89 \[ \frac {\left (a e^2+c d^2\right ) (B d-A e)}{e^4 (d+e x)}+\frac {\log (d+e x) \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^4}-\frac {c x (2 B d-A e)}{e^3}+\frac {B c x^2}{2 e^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 89, 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} \[ \frac {\left (a e^2+c d^2\right ) (B d-A e)}{e^4 (d+e x)}+\frac {\log (d+e x) \left (a B e^2-2 A c d e+3 B c d^2\right )}{e^4}-\frac {c x (2 B d-A e)}{e^3}+\frac {B c x^2}{2 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.06, size = 86, normalized size = 0.97 \[ \frac {\frac {2 \left (a e^2+c d^2\right ) (B d-A e)}{d+e x}+2 \log (d+e x) \left (a B e^2-2 A c d e+3 B c d^2\right )+2 c e x (A e-2 B d)+B c e^2 x^2}{2 e^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.74, size = 152, normalized size = 1.71 \[ \frac {B c e^{3} x^{3} + 2 \, B c d^{3} - 2 \, A c d^{2} e + 2 \, B a d e^{2} - 2 \, A a e^{3} - {\left (3 \, B c d e^{2} - 2 \, A c e^{3}\right )} x^{2} - 2 \, {\left (2 \, B c d^{2} e - A c d e^{2}\right )} x + 2 \, {\left (3 \, B c d^{3} - 2 \, A c d^{2} e + B a d e^{2} + {\left (3 \, B c d^{2} e - 2 \, A c d e^{2} + B a e^{3}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x + d e^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 151, normalized size = 1.70 \[ \frac {1}{2} \, {\left (B c - \frac {2 \, {\left (3 \, B c d e - A c e^{2}\right )} e^{\left (-1\right )}}{x e + d}\right )} {\left (x e + d\right )}^{2} e^{\left (-4\right )} - {\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} e^{\left (-4\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) + {\left (\frac {B c d^{3} e^{2}}{x e + d} - \frac {A c d^{2} e^{3}}{x e + d} + \frac {B a d e^{4}}{x e + d} - \frac {A a e^{5}}{x e + d}\right )} e^{\left (-6\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 131, normalized size = 1.47 \[ \frac {B c \,x^{2}}{2 e^{2}}-\frac {A a}{\left (e x +d \right ) e}-\frac {A c \,d^{2}}{\left (e x +d \right ) e^{3}}-\frac {2 A c d \ln \left (e x +d \right )}{e^{3}}+\frac {A c x}{e^{2}}+\frac {B a d}{\left (e x +d \right ) e^{2}}+\frac {B a \ln \left (e x +d \right )}{e^{2}}+\frac {B c \,d^{3}}{\left (e x +d \right ) e^{4}}+\frac {3 B c \,d^{2} \ln \left (e x +d \right )}{e^{4}}-\frac {2 B c d x}{e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.54, size = 101, normalized size = 1.13 \[ \frac {B c d^{3} - A c d^{2} e + B a d e^{2} - A a e^{3}}{e^{5} x + d e^{4}} + \frac {B c e x^{2} - 2 \, {\left (2 \, B c d - A c e\right )} x}{2 \, e^{3}} + \frac {{\left (3 \, B c d^{2} - 2 \, A c d e + B a e^{2}\right )} \log \left (e x + d\right )}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.08, size = 105, normalized size = 1.18 \[ x\,\left (\frac {A\,c}{e^2}-\frac {2\,B\,c\,d}{e^3}\right )-\frac {-B\,c\,d^3+A\,c\,d^2\,e-B\,a\,d\,e^2+A\,a\,e^3}{e\,\left (x\,e^4+d\,e^3\right )}+\frac {\ln \left (d+e\,x\right )\,\left (3\,B\,c\,d^2-2\,A\,c\,d\,e+B\,a\,e^2\right )}{e^4}+\frac {B\,c\,x^2}{2\,e^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.59, size = 104, normalized size = 1.17 \[ \frac {B c x^{2}}{2 e^{2}} + x \left (\frac {A c}{e^{2}} - \frac {2 B c d}{e^{3}}\right ) + \frac {- A a e^{3} - A c d^{2} e + B a d e^{2} + B c d^{3}}{d e^{4} + e^{5} x} + \frac {\left (- 2 A c d e + B a e^{2} + 3 B c d^{2}\right ) \log {\left (d + e x \right )}}{e^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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