3.1303 \(\int \frac {(A+B x) (a+c x^2)^2}{d+e x} \, dx\)

Optimal. Leaf size=169 \[ -\frac {\left (a e^2+c d^2\right )^2 (B d-A e) \log (d+e x)}{e^6}+\frac {x \left (B \left (a e^2+c d^2\right )^2-A c d e \left (2 a e^2+c d^2\right )\right )}{e^5}-\frac {c x^2 \left (2 a e^2+c d^2\right ) (B d-A e)}{2 e^4}+\frac {c x^3 \left (2 a B e^2-A c d e+B c d^2\right )}{3 e^3}-\frac {c^2 x^4 (B d-A e)}{4 e^2}+\frac {B c^2 x^5}{5 e} \]

[Out]

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

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Rubi [A]  time = 0.19, antiderivative size = 169, 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} \[ \frac {c x^3 \left (2 a B e^2-A c d e+B c d^2\right )}{3 e^3}-\frac {c x^2 \left (2 a e^2+c d^2\right ) (B d-A e)}{2 e^4}+\frac {x \left (B \left (a e^2+c d^2\right )^2-A c d e \left (2 a e^2+c d^2\right )\right )}{e^5}-\frac {\left (a e^2+c d^2\right )^2 (B d-A e) \log (d+e x)}{e^6}-\frac {c^2 x^4 (B d-A e)}{4 e^2}+\frac {B c^2 x^5}{5 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.10, size = 174, normalized size = 1.03 \[ \frac {e x \left (B \left (60 a^2 e^4+20 a c e^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )+c^2 \left (60 d^4-30 d^3 e x+20 d^2 e^2 x^2-15 d e^3 x^3+12 e^4 x^4\right )\right )+5 A c e \left (12 a e^2 (e x-2 d)+c \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )\right )\right )-60 \left (a e^2+c d^2\right )^2 (B d-A e) \log (d+e x)}{60 e^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(e*x*(5*A*c*e*(12*a*e^2*(-2*d + e*x) + c*(-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2 + 3*e^3*x^3)) + B*(60*a^2*e^4 + 20
*a*c*e^2*(6*d^2 - 3*d*e*x + 2*e^2*x^2) + c^2*(60*d^4 - 30*d^3*e*x + 20*d^2*e^2*x^2 - 15*d*e^3*x^3 + 12*e^4*x^4
))) - 60*(B*d - A*e)*(c*d^2 + a*e^2)^2*Log[d + e*x])/(60*e^6)

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fricas [A]  time = 0.56, size = 243, normalized size = 1.44 \[ \frac {12 \, B c^{2} e^{5} x^{5} - 15 \, {\left (B c^{2} d e^{4} - A c^{2} e^{5}\right )} x^{4} + 20 \, {\left (B c^{2} d^{2} e^{3} - A c^{2} d e^{4} + 2 \, B a c e^{5}\right )} x^{3} - 30 \, {\left (B c^{2} d^{3} e^{2} - A c^{2} d^{2} e^{3} + 2 \, B a c d e^{4} - 2 \, A a c e^{5}\right )} x^{2} + 60 \, {\left (B c^{2} d^{4} e - A c^{2} d^{3} e^{2} + 2 \, B a c d^{2} e^{3} - 2 \, A a c d e^{4} + B a^{2} e^{5}\right )} x - 60 \, {\left (B c^{2} d^{5} - A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} - A a^{2} e^{5}\right )} \log \left (e x + d\right )}{60 \, e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*c^2*e^5*x^5 - 15*(B*c^2*d*e^4 - A*c^2*e^5)*x^4 + 20*(B*c^2*d^2*e^3 - A*c^2*d*e^4 + 2*B*a*c*e^5)*x^3
- 30*(B*c^2*d^3*e^2 - A*c^2*d^2*e^3 + 2*B*a*c*d*e^4 - 2*A*a*c*e^5)*x^2 + 60*(B*c^2*d^4*e - A*c^2*d^3*e^2 + 2*
B*a*c*d^2*e^3 - 2*A*a*c*d*e^4 + B*a^2*e^5)*x - 60*(B*c^2*d^5 - A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 2*A*a*c*d^2*e^3
+ B*a^2*d*e^4 - A*a^2*e^5)*log(e*x + d))/e^6

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giac [A]  time = 0.16, size = 244, normalized size = 1.44 \[ -{\left (B c^{2} d^{5} - A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} - A a^{2} e^{5}\right )} e^{\left (-6\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{60} \, {\left (12 \, B c^{2} x^{5} e^{4} - 15 \, B c^{2} d x^{4} e^{3} + 20 \, B c^{2} d^{2} x^{3} e^{2} - 30 \, B c^{2} d^{3} x^{2} e + 60 \, B c^{2} d^{4} x + 15 \, A c^{2} x^{4} e^{4} - 20 \, A c^{2} d x^{3} e^{3} + 30 \, A c^{2} d^{2} x^{2} e^{2} - 60 \, A c^{2} d^{3} x e + 40 \, B a c x^{3} e^{4} - 60 \, B a c d x^{2} e^{3} + 120 \, B a c d^{2} x e^{2} + 60 \, A a c x^{2} e^{4} - 120 \, A a c d x e^{3} + 60 \, B a^{2} x e^{4}\right )} e^{\left (-5\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(B*c^2*d^5 - A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 2*A*a*c*d^2*e^3 + B*a^2*d*e^4 - A*a^2*e^5)*e^(-6)*log(abs(x*e +
d)) + 1/60*(12*B*c^2*x^5*e^4 - 15*B*c^2*d*x^4*e^3 + 20*B*c^2*d^2*x^3*e^2 - 30*B*c^2*d^3*x^2*e + 60*B*c^2*d^4*x
+ 15*A*c^2*x^4*e^4 - 20*A*c^2*d*x^3*e^3 + 30*A*c^2*d^2*x^2*e^2 - 60*A*c^2*d^3*x*e + 40*B*a*c*x^3*e^4 - 60*B*a
*c*d*x^2*e^3 + 120*B*a*c*d^2*x*e^2 + 60*A*a*c*x^2*e^4 - 120*A*a*c*d*x*e^3 + 60*B*a^2*x*e^4)*e^(-5)

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maple [A]  time = 0.05, size = 285, normalized size = 1.69 \[ \frac {B \,c^{2} x^{5}}{5 e}+\frac {A \,c^{2} x^{4}}{4 e}-\frac {B \,c^{2} d \,x^{4}}{4 e^{2}}-\frac {A \,c^{2} d \,x^{3}}{3 e^{2}}+\frac {2 B a c \,x^{3}}{3 e}+\frac {B \,c^{2} d^{2} x^{3}}{3 e^{3}}+\frac {A a c \,x^{2}}{e}+\frac {A \,c^{2} d^{2} x^{2}}{2 e^{3}}-\frac {B a c d \,x^{2}}{e^{2}}-\frac {B \,c^{2} d^{3} x^{2}}{2 e^{4}}+\frac {A \,a^{2} \ln \left (e x +d \right )}{e}+\frac {2 A a c \,d^{2} \ln \left (e x +d \right )}{e^{3}}-\frac {2 A a c d x}{e^{2}}+\frac {A \,c^{2} d^{4} \ln \left (e x +d \right )}{e^{5}}-\frac {A \,c^{2} d^{3} x}{e^{4}}-\frac {B \,a^{2} d \ln \left (e x +d \right )}{e^{2}}+\frac {B \,a^{2} x}{e}-\frac {2 B a c \,d^{3} \ln \left (e x +d \right )}{e^{4}}+\frac {2 B a c \,d^{2} x}{e^{3}}-\frac {B \,c^{2} d^{5} \ln \left (e x +d \right )}{e^{6}}+\frac {B \,c^{2} d^{4} x}{e^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.56, size = 242, normalized size = 1.43 \[ \frac {12 \, B c^{2} e^{4} x^{5} - 15 \, {\left (B c^{2} d e^{3} - A c^{2} e^{4}\right )} x^{4} + 20 \, {\left (B c^{2} d^{2} e^{2} - A c^{2} d e^{3} + 2 \, B a c e^{4}\right )} x^{3} - 30 \, {\left (B c^{2} d^{3} e - A c^{2} d^{2} e^{2} + 2 \, B a c d e^{3} - 2 \, A a c e^{4}\right )} x^{2} + 60 \, {\left (B c^{2} d^{4} - A c^{2} d^{3} e + 2 \, B a c d^{2} e^{2} - 2 \, A a c d e^{3} + B a^{2} e^{4}\right )} x}{60 \, e^{5}} - \frac {{\left (B c^{2} d^{5} - A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} - A a^{2} e^{5}\right )} \log \left (e x + d\right )}{e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/60*(12*B*c^2*e^4*x^5 - 15*(B*c^2*d*e^3 - A*c^2*e^4)*x^4 + 20*(B*c^2*d^2*e^2 - A*c^2*d*e^3 + 2*B*a*c*e^4)*x^3
- 30*(B*c^2*d^3*e - A*c^2*d^2*e^2 + 2*B*a*c*d*e^3 - 2*A*a*c*e^4)*x^2 + 60*(B*c^2*d^4 - A*c^2*d^3*e + 2*B*a*c*
d^2*e^2 - 2*A*a*c*d*e^3 + B*a^2*e^4)*x)/e^5 - (B*c^2*d^5 - A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 2*A*a*c*d^2*e^3 + B
*a^2*d*e^4 - A*a^2*e^5)*log(e*x + d)/e^6

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mupad [B]  time = 0.07, size = 260, normalized size = 1.54 \[ x\,\left (\frac {B\,a^2}{e}-\frac {d\,\left (\frac {d\,\left (\frac {d\,\left (\frac {A\,c^2}{e}-\frac {B\,c^2\,d}{e^2}\right )}{e}-\frac {2\,B\,a\,c}{e}\right )}{e}+\frac {2\,A\,a\,c}{e}\right )}{e}\right )+x^4\,\left (\frac {A\,c^2}{4\,e}-\frac {B\,c^2\,d}{4\,e^2}\right )-x^3\,\left (\frac {d\,\left (\frac {A\,c^2}{e}-\frac {B\,c^2\,d}{e^2}\right )}{3\,e}-\frac {2\,B\,a\,c}{3\,e}\right )+x^2\,\left (\frac {d\,\left (\frac {d\,\left (\frac {A\,c^2}{e}-\frac {B\,c^2\,d}{e^2}\right )}{e}-\frac {2\,B\,a\,c}{e}\right )}{2\,e}+\frac {A\,a\,c}{e}\right )+\frac {\ln \left (d+e\,x\right )\,\left (-B\,a^2\,d\,e^4+A\,a^2\,e^5-2\,B\,a\,c\,d^3\,e^2+2\,A\,a\,c\,d^2\,e^3-B\,c^2\,d^5+A\,c^2\,d^4\,e\right )}{e^6}+\frac {B\,c^2\,x^5}{5\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.58, size = 207, normalized size = 1.22 \[ \frac {B c^{2} x^{5}}{5 e} + x^{4} \left (\frac {A c^{2}}{4 e} - \frac {B c^{2} d}{4 e^{2}}\right ) + x^{3} \left (- \frac {A c^{2} d}{3 e^{2}} + \frac {2 B a c}{3 e} + \frac {B c^{2} d^{2}}{3 e^{3}}\right ) + x^{2} \left (\frac {A a c}{e} + \frac {A c^{2} d^{2}}{2 e^{3}} - \frac {B a c d}{e^{2}} - \frac {B c^{2} d^{3}}{2 e^{4}}\right ) + x \left (- \frac {2 A a c d}{e^{2}} - \frac {A c^{2} d^{3}}{e^{4}} + \frac {B a^{2}}{e} + \frac {2 B a c d^{2}}{e^{3}} + \frac {B c^{2} d^{4}}{e^{5}}\right ) - \frac {\left (- A e + B d\right ) \left (a e^{2} + c d^{2}\right )^{2} \log {\left (d + e x \right )}}{e^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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