3.14.7 \(\int \frac {(A+B x) (a+c x^2)^2}{(d+e x)^5} \, dx\) [1307]

Optimal. Leaf size=189 \[ \frac {B c^2 x}{e^5}+\frac {(B d-A e) \left (c d^2+a e^2\right )^2}{4 e^6 (d+e x)^4}-\frac {\left (c d^2+a e^2\right ) \left (5 B c d^2-4 A c d e+a B e^2\right )}{3 e^6 (d+e x)^3}+\frac {c \left (5 B c d^3-3 A c d^2 e+3 a B d e^2-a A e^3\right )}{e^6 (d+e x)^2}-\frac {2 c \left (5 B c d^2-2 A c d e+a B e^2\right )}{e^6 (d+e x)}-\frac {c^2 (5 B d-A e) \log (d+e x)}{e^6} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 189, 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 = {786} \begin {gather*} -\frac {2 c \left (a B e^2-2 A c d e+5 B c d^2\right )}{e^6 (d+e x)}-\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 )}{3 e^6 (d+e x)^3}+\frac {\left (a e^2+c d^2\right )^2 (B d-A e)}{4 e^6 (d+e x)^4}+\frac {c \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 (d+e x)^2}-\frac {c^2 (5 B d-A e) \log (d+e x)}{e^6}+\frac {B c^2 x}{e^5} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(a + c*x^2)^2)/(d + e*x)^5,x]

[Out]

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

Rule 786

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

________________________________________________________________________________________

Mathematica [A]
time = 0.08, size = 221, normalized size = 1.17 \begin {gather*} \frac {A e \left (-3 a^2 e^4-2 a c e^2 \left (d^2+4 d e x+6 e^2 x^2\right )+c^2 d \left (25 d^3+88 d^2 e x+108 d e^2 x^2+48 e^3 x^3\right )\right )-B \left (a^2 e^4 (d+4 e x)+6 a c e^2 \left (d^3+4 d^2 e x+6 d e^2 x^2+4 e^3 x^3\right )+c^2 \left (77 d^5+248 d^4 e x+252 d^3 e^2 x^2+48 d^2 e^3 x^3-48 d e^4 x^4-12 e^5 x^5\right )\right )-12 c^2 (5 B d-A e) (d+e x)^4 \log (d+e x)}{12 e^6 (d+e x)^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(a + c*x^2)^2)/(d + e*x)^5,x]

[Out]

(A*e*(-3*a^2*e^4 - 2*a*c*e^2*(d^2 + 4*d*e*x + 6*e^2*x^2) + c^2*d*(25*d^3 + 88*d^2*e*x + 108*d*e^2*x^2 + 48*e^3
*x^3)) - B*(a^2*e^4*(d + 4*e*x) + 6*a*c*e^2*(d^3 + 4*d^2*e*x + 6*d*e^2*x^2 + 4*e^3*x^3) + c^2*(77*d^5 + 248*d^
4*e*x + 252*d^3*e^2*x^2 + 48*d^2*e^3*x^3 - 48*d*e^4*x^4 - 12*e^5*x^5)) - 12*c^2*(5*B*d - A*e)*(d + e*x)^4*Log[
d + e*x])/(12*e^6*(d + e*x)^4)

________________________________________________________________________________________

Maple [A]
time = 0.63, size = 240, normalized size = 1.27

method result size
default \(\frac {B \,c^{2} x}{e^{5}}+\frac {2 c \left (2 A c d e -B \,e^{2} a -5 B c \,d^{2}\right )}{e^{6} \left (e x +d \right )}-\frac {c \left (A a \,e^{3}+3 A c \,d^{2} e -3 a B d \,e^{2}-5 B c \,d^{3}\right )}{e^{6} \left (e x +d \right )^{2}}+\frac {c^{2} \left (A e -5 B d \right ) \ln \left (e x +d \right )}{e^{6}}-\frac {A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}+A \,c^{2} d^{4} e -B \,a^{2} d \,e^{4}-2 B a c \,d^{3} e^{2}-B \,c^{2} d^{5}}{4 e^{6} \left (e x +d \right )^{4}}-\frac {-4 A a c d \,e^{3}-4 A \,c^{2} d^{3} e +B \,e^{4} a^{2}+6 B a c \,d^{2} e^{2}+5 B \,c^{2} d^{4}}{3 e^{6} \left (e x +d \right )^{3}}\) \(240\)
norman \(\frac {\frac {B \,c^{2} x^{5}}{e}-\frac {3 A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}-25 A \,c^{2} d^{4} e +B \,a^{2} d \,e^{4}+6 B a c \,d^{3} e^{2}+125 B \,c^{2} d^{5}}{12 e^{6}}+\frac {2 \left (2 A \,c^{2} d e -B \,e^{2} a c -10 B \,c^{2} d^{2}\right ) x^{3}}{e^{3}}-\frac {\left (A a c \,e^{3}-9 A \,c^{2} d^{2} e +3 B a c d \,e^{2}+45 B \,c^{2} d^{3}\right ) x^{2}}{e^{4}}-\frac {\left (2 A a c d \,e^{3}-22 A \,c^{2} d^{3} e +B \,e^{4} a^{2}+6 B a c \,d^{2} e^{2}+110 B \,c^{2} d^{4}\right ) x}{3 e^{5}}}{\left (e x +d \right )^{4}}+\frac {c^{2} \left (A e -5 B d \right ) \ln \left (e x +d \right )}{e^{6}}\) \(240\)
risch \(\frac {B \,c^{2} x}{e^{5}}+\frac {\left (4 A \,c^{2} d \,e^{3}-2 B \,e^{4} a c -10 B \,c^{2} d^{2} e^{2}\right ) x^{3}-e c \left (A a \,e^{3}-9 A c \,d^{2} e +3 a B d \,e^{2}+25 B c \,d^{3}\right ) x^{2}+\left (-\frac {2}{3} A a c d \,e^{3}+\frac {22}{3} A \,c^{2} d^{3} e -\frac {1}{3} B \,e^{4} a^{2}-2 B a c \,d^{2} e^{2}-\frac {65}{3} B \,c^{2} d^{4}\right ) x -\frac {3 A \,a^{2} e^{5}+2 A a c \,d^{2} e^{3}-25 A \,c^{2} d^{4} e +B \,a^{2} d \,e^{4}+6 B a c \,d^{3} e^{2}+77 B \,c^{2} d^{5}}{12 e}}{e^{5} \left (e x +d \right )^{4}}+\frac {c^{2} \ln \left (e x +d \right ) A}{e^{5}}-\frac {5 c^{2} \ln \left (e x +d \right ) B d}{e^{6}}\) \(241\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(c*x^2+a)^2/(e*x+d)^5,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 261, normalized size = 1.38 \begin {gather*} B c^{2} x e^{\left (-5\right )} - {\left (5 \, B c^{2} d - A c^{2} e\right )} e^{\left (-6\right )} \log \left (x e + d\right ) - \frac {77 \, B c^{2} d^{5} - 25 \, A c^{2} d^{4} e + 6 \, B a c d^{3} e^{2} + 2 \, A a c d^{2} e^{3} + B a^{2} d e^{4} + 24 \, {\left (5 \, B c^{2} d^{2} e^{3} - 2 \, A c^{2} d e^{4} + B a c e^{5}\right )} x^{3} + 3 \, A a^{2} e^{5} + 12 \, {\left (25 \, B c^{2} d^{3} e^{2} - 9 \, A c^{2} d^{2} e^{3} + 3 \, B a c d e^{4} + A a c e^{5}\right )} x^{2} + 4 \, {\left (65 \, B c^{2} d^{4} e - 22 \, A c^{2} d^{3} e^{2} + 6 \, B a c d^{2} e^{3} + 2 \, A a c d e^{4} + B a^{2} e^{5}\right )} x}{12 \, {\left (x^{4} e^{10} + 4 \, d x^{3} e^{9} + 6 \, d^{2} x^{2} e^{8} + 4 \, d^{3} x e^{7} + d^{4} 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)^5,x, algorithm="maxima")

[Out]

B*c^2*x*e^(-5) - (5*B*c^2*d - A*c^2*e)*e^(-6)*log(x*e + d) - 1/12*(77*B*c^2*d^5 - 25*A*c^2*d^4*e + 6*B*a*c*d^3
*e^2 + 2*A*a*c*d^2*e^3 + B*a^2*d*e^4 + 24*(5*B*c^2*d^2*e^3 - 2*A*c^2*d*e^4 + B*a*c*e^5)*x^3 + 3*A*a^2*e^5 + 12
*(25*B*c^2*d^3*e^2 - 9*A*c^2*d^2*e^3 + 3*B*a*c*d*e^4 + A*a*c*e^5)*x^2 + 4*(65*B*c^2*d^4*e - 22*A*c^2*d^3*e^2 +
 6*B*a*c*d^2*e^3 + 2*A*a*c*d*e^4 + B*a^2*e^5)*x)/(x^4*e^10 + 4*d*x^3*e^9 + 6*d^2*x^2*e^8 + 4*d^3*x*e^7 + d^4*e
^6)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 380 vs. \(2 (183) = 366\).
time = 4.18, size = 380, normalized size = 2.01 \begin {gather*} -\frac {77 \, B c^{2} d^{5} - {\left (12 \, B c^{2} x^{5} - 24 \, B a c x^{3} - 12 \, A a c x^{2} - 4 \, B a^{2} x - 3 \, A a^{2}\right )} e^{5} - {\left (48 \, B c^{2} d x^{4} + 48 \, A c^{2} d x^{3} - 36 \, B a c d x^{2} - 8 \, A a c d x - B a^{2} d\right )} e^{4} + 2 \, {\left (24 \, B c^{2} d^{2} x^{3} - 54 \, A c^{2} d^{2} x^{2} + 12 \, B a c d^{2} x + A a c d^{2}\right )} e^{3} + 2 \, {\left (126 \, B c^{2} d^{3} x^{2} - 44 \, A c^{2} d^{3} x + 3 \, B a c d^{3}\right )} e^{2} + {\left (248 \, B c^{2} d^{4} x - 25 \, A c^{2} d^{4}\right )} e + 12 \, {\left (5 \, B c^{2} d^{5} - A c^{2} x^{4} e^{5} + {\left (5 \, B c^{2} d x^{4} - 4 \, A c^{2} d x^{3}\right )} e^{4} + 2 \, {\left (10 \, B c^{2} d^{2} x^{3} - 3 \, A c^{2} d^{2} x^{2}\right )} e^{3} + 2 \, {\left (15 \, B c^{2} d^{3} x^{2} - 2 \, A c^{2} d^{3} x\right )} e^{2} + {\left (20 \, B c^{2} d^{4} x - A c^{2} d^{4}\right )} e\right )} \log \left (x e + d\right )}{12 \, {\left (x^{4} e^{10} + 4 \, d x^{3} e^{9} + 6 \, d^{2} x^{2} e^{8} + 4 \, d^{3} x e^{7} + d^{4} 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)^5,x, algorithm="fricas")

[Out]

-1/12*(77*B*c^2*d^5 - (12*B*c^2*x^5 - 24*B*a*c*x^3 - 12*A*a*c*x^2 - 4*B*a^2*x - 3*A*a^2)*e^5 - (48*B*c^2*d*x^4
 + 48*A*c^2*d*x^3 - 36*B*a*c*d*x^2 - 8*A*a*c*d*x - B*a^2*d)*e^4 + 2*(24*B*c^2*d^2*x^3 - 54*A*c^2*d^2*x^2 + 12*
B*a*c*d^2*x + A*a*c*d^2)*e^3 + 2*(126*B*c^2*d^3*x^2 - 44*A*c^2*d^3*x + 3*B*a*c*d^3)*e^2 + (248*B*c^2*d^4*x - 2
5*A*c^2*d^4)*e + 12*(5*B*c^2*d^5 - A*c^2*x^4*e^5 + (5*B*c^2*d*x^4 - 4*A*c^2*d*x^3)*e^4 + 2*(10*B*c^2*d^2*x^3 -
 3*A*c^2*d^2*x^2)*e^3 + 2*(15*B*c^2*d^3*x^2 - 2*A*c^2*d^3*x)*e^2 + (20*B*c^2*d^4*x - A*c^2*d^4)*e)*log(x*e + d
))/(x^4*e^10 + 4*d*x^3*e^9 + 6*d^2*x^2*e^8 + 4*d^3*x*e^7 + d^4*e^6)

________________________________________________________________________________________

Sympy [A]
time = 79.02, size = 304, normalized size = 1.61 \begin {gather*} \frac {B c^{2} x}{e^{5}} - \frac {c^{2} \left (- A e + 5 B d\right ) \log {\left (d + e x \right )}}{e^{6}} + \frac {- 3 A a^{2} e^{5} - 2 A a c d^{2} e^{3} + 25 A c^{2} d^{4} e - B a^{2} d e^{4} - 6 B a c d^{3} e^{2} - 77 B c^{2} d^{5} + x^{3} \cdot \left (48 A c^{2} d e^{4} - 24 B a c e^{5} - 120 B c^{2} d^{2} e^{3}\right ) + x^{2} \left (- 12 A a c e^{5} + 108 A c^{2} d^{2} e^{3} - 36 B a c d e^{4} - 300 B c^{2} d^{3} e^{2}\right ) + x \left (- 8 A a c d e^{4} + 88 A c^{2} d^{3} e^{2} - 4 B a^{2} e^{5} - 24 B a c d^{2} e^{3} - 260 B c^{2} d^{4} e\right )}{12 d^{4} e^{6} + 48 d^{3} e^{7} x + 72 d^{2} e^{8} x^{2} + 48 d e^{9} x^{3} + 12 e^{10} x^{4}} \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)**5,x)

[Out]

B*c**2*x/e**5 - c**2*(-A*e + 5*B*d)*log(d + e*x)/e**6 + (-3*A*a**2*e**5 - 2*A*a*c*d**2*e**3 + 25*A*c**2*d**4*e
 - B*a**2*d*e**4 - 6*B*a*c*d**3*e**2 - 77*B*c**2*d**5 + x**3*(48*A*c**2*d*e**4 - 24*B*a*c*e**5 - 120*B*c**2*d*
*2*e**3) + x**2*(-12*A*a*c*e**5 + 108*A*c**2*d**2*e**3 - 36*B*a*c*d*e**4 - 300*B*c**2*d**3*e**2) + x*(-8*A*a*c
*d*e**4 + 88*A*c**2*d**3*e**2 - 4*B*a**2*e**5 - 24*B*a*c*d**2*e**3 - 260*B*c**2*d**4*e))/(12*d**4*e**6 + 48*d*
*3*e**7*x + 72*d**2*e**8*x**2 + 48*d*e**9*x**3 + 12*e**10*x**4)

________________________________________________________________________________________

Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 372 vs. \(2 (183) = 366\).
time = 1.17, size = 372, normalized size = 1.97 \begin {gather*} {\left (x e + d\right )} B c^{2} e^{\left (-6\right )} + {\left (5 \, B c^{2} d - A c^{2} e\right )} e^{\left (-6\right )} \log \left (\frac {{\left | x e + d \right |} e^{\left (-1\right )}}{{\left (x e + d\right )}^{2}}\right ) - \frac {1}{12} \, {\left (\frac {120 \, B c^{2} d^{2} e^{22}}{x e + d} - \frac {60 \, B c^{2} d^{3} e^{22}}{{\left (x e + d\right )}^{2}} + \frac {20 \, B c^{2} d^{4} e^{22}}{{\left (x e + d\right )}^{3}} - \frac {3 \, B c^{2} d^{5} e^{22}}{{\left (x e + d\right )}^{4}} - \frac {48 \, A c^{2} d e^{23}}{x e + d} + \frac {36 \, A c^{2} d^{2} e^{23}}{{\left (x e + d\right )}^{2}} - \frac {16 \, A c^{2} d^{3} e^{23}}{{\left (x e + d\right )}^{3}} + \frac {3 \, A c^{2} d^{4} e^{23}}{{\left (x e + d\right )}^{4}} + \frac {24 \, B a c e^{24}}{x e + d} - \frac {36 \, B a c d e^{24}}{{\left (x e + d\right )}^{2}} + \frac {24 \, B a c d^{2} e^{24}}{{\left (x e + d\right )}^{3}} - \frac {6 \, B a c d^{3} e^{24}}{{\left (x e + d\right )}^{4}} + \frac {12 \, A a c e^{25}}{{\left (x e + d\right )}^{2}} - \frac {16 \, A a c d e^{25}}{{\left (x e + d\right )}^{3}} + \frac {6 \, A a c d^{2} e^{25}}{{\left (x e + d\right )}^{4}} + \frac {4 \, B a^{2} e^{26}}{{\left (x e + d\right )}^{3}} - \frac {3 \, B a^{2} d e^{26}}{{\left (x e + d\right )}^{4}} + \frac {3 \, A a^{2} e^{27}}{{\left (x e + d\right )}^{4}}\right )} e^{\left (-28\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)^5,x, algorithm="giac")

[Out]

(x*e + d)*B*c^2*e^(-6) + (5*B*c^2*d - A*c^2*e)*e^(-6)*log(abs(x*e + d)*e^(-1)/(x*e + d)^2) - 1/12*(120*B*c^2*d
^2*e^22/(x*e + d) - 60*B*c^2*d^3*e^22/(x*e + d)^2 + 20*B*c^2*d^4*e^22/(x*e + d)^3 - 3*B*c^2*d^5*e^22/(x*e + d)
^4 - 48*A*c^2*d*e^23/(x*e + d) + 36*A*c^2*d^2*e^23/(x*e + d)^2 - 16*A*c^2*d^3*e^23/(x*e + d)^3 + 3*A*c^2*d^4*e
^23/(x*e + d)^4 + 24*B*a*c*e^24/(x*e + d) - 36*B*a*c*d*e^24/(x*e + d)^2 + 24*B*a*c*d^2*e^24/(x*e + d)^3 - 6*B*
a*c*d^3*e^24/(x*e + d)^4 + 12*A*a*c*e^25/(x*e + d)^2 - 16*A*a*c*d*e^25/(x*e + d)^3 + 6*A*a*c*d^2*e^25/(x*e + d
)^4 + 4*B*a^2*e^26/(x*e + d)^3 - 3*B*a^2*d*e^26/(x*e + d)^4 + 3*A*a^2*e^27/(x*e + d)^4)*e^(-28)

________________________________________________________________________________________

Mupad [B]
time = 1.84, size = 277, normalized size = 1.47 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (A\,c^2\,e-5\,B\,c^2\,d\right )}{e^6}-\frac {x^3\,\left (10\,B\,c^2\,d^2\,e^2-4\,A\,c^2\,d\,e^3+2\,B\,a\,c\,e^4\right )+x\,\left (\frac {B\,a^2\,e^4}{3}+2\,B\,a\,c\,d^2\,e^2+\frac {2\,A\,a\,c\,d\,e^3}{3}+\frac {65\,B\,c^2\,d^4}{3}-\frac {22\,A\,c^2\,d^3\,e}{3}\right )+\frac {B\,a^2\,d\,e^4+3\,A\,a^2\,e^5+6\,B\,a\,c\,d^3\,e^2+2\,A\,a\,c\,d^2\,e^3+77\,B\,c^2\,d^5-25\,A\,c^2\,d^4\,e}{12\,e}+x^2\,\left (25\,B\,c^2\,d^3\,e-9\,A\,c^2\,d^2\,e^2+3\,B\,a\,c\,d\,e^3+A\,a\,c\,e^4\right )}{d^4\,e^5+4\,d^3\,e^6\,x+6\,d^2\,e^7\,x^2+4\,d\,e^8\,x^3+e^9\,x^4}+\frac {B\,c^2\,x}{e^5} \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)^5,x)

[Out]

(log(d + e*x)*(A*c^2*e - 5*B*c^2*d))/e^6 - (x^3*(2*B*a*c*e^4 - 4*A*c^2*d*e^3 + 10*B*c^2*d^2*e^2) + x*((B*a^2*e
^4)/3 + (65*B*c^2*d^4)/3 - (22*A*c^2*d^3*e)/3 + (2*A*a*c*d*e^3)/3 + 2*B*a*c*d^2*e^2) + (3*A*a^2*e^5 + 77*B*c^2
*d^5 + B*a^2*d*e^4 - 25*A*c^2*d^4*e + 2*A*a*c*d^2*e^3 + 6*B*a*c*d^3*e^2)/(12*e) + x^2*(A*a*c*e^4 + 25*B*c^2*d^
3*e - 9*A*c^2*d^2*e^2 + 3*B*a*c*d*e^3))/(d^4*e^5 + e^9*x^4 + 4*d^3*e^6*x + 4*d*e^8*x^3 + 6*d^2*e^7*x^2) + (B*c
^2*x)/e^5

________________________________________________________________________________________