3.5.25 \(\int \frac {(d+e x)^5}{(a+c x^2)^2} \, dx\)

Optimal. Leaf size=190 \[ \frac {d \left (-15 a^2 e^4+10 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{5/2}}+\frac {e^3 \left (5 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{c^3}-\frac {3 d e^2 x \left (2 c d^2-5 a e^2\right )}{2 a c^2}-\frac {e^3 x^2 \left (2 c d^2-a e^2\right )}{a c^2}-\frac {d e^4 x^3}{2 a c}-\frac {(d+e x)^4 (a e-c d x)}{2 a c \left (a+c x^2\right )} \]

________________________________________________________________________________________

Rubi [A]  time = 0.18, antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {739, 801, 635, 205, 260} \begin {gather*} \frac {d \left (-15 a^2 e^4+10 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{5/2}}-\frac {e^3 x^2 \left (2 c d^2-a e^2\right )}{a c^2}+\frac {e^3 \left (5 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{c^3}-\frac {3 d e^2 x \left (2 c d^2-5 a e^2\right )}{2 a c^2}-\frac {d e^4 x^3}{2 a c}-\frac {(d+e x)^4 (a e-c d x)}{2 a c \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^5/(a + c*x^2)^2,x]

[Out]

(-3*d*e^2*(2*c*d^2 - 5*a*e^2)*x)/(2*a*c^2) - (e^3*(2*c*d^2 - a*e^2)*x^2)/(a*c^2) - (d*e^4*x^3)/(2*a*c) - ((a*e
- c*d*x)*(d + e*x)^4)/(2*a*c*(a + c*x^2)) + (d*(c^2*d^4 + 10*a*c*d^2*e^2 - 15*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqr
t[a]])/(2*a^(3/2)*c^(5/2)) + (e^3*(5*c*d^2 - a*e^2)*Log[a + c*x^2])/c^3

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[d, Int[1/(a + c*x^2), x], x] + Dist[e, Int[x/
(a + c*x^2), x], x] /; FreeQ[{a, c, d, e}, x] &&  !NiceSqrtQ[-(a*c)]

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

\begin {align*} \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx &=-\frac {(a e-c d x) (d+e x)^4}{2 a c \left (a+c x^2\right )}+\frac {\int \frac {(d+e x)^3 \left (c d^2+4 a e^2-3 c d e x\right )}{a+c x^2} \, dx}{2 a c}\\ &=-\frac {(a e-c d x) (d+e x)^4}{2 a c \left (a+c x^2\right )}+\frac {\int \left (-3 d e^2 \left (2 d^2-\frac {5 a e^2}{c}\right )-\frac {4 e^3 \left (2 c d^2-a e^2\right ) x}{c}-3 d e^4 x^2+\frac {c^2 d^5+10 a c d^3 e^2-15 a^2 d e^4+4 a e^3 \left (5 c d^2-a e^2\right ) x}{c \left (a+c x^2\right )}\right ) \, dx}{2 a c}\\ &=-\frac {3 d e^2 \left (2 c d^2-5 a e^2\right ) x}{2 a c^2}-\frac {e^3 \left (2 c d^2-a e^2\right ) x^2}{a c^2}-\frac {d e^4 x^3}{2 a c}-\frac {(a e-c d x) (d+e x)^4}{2 a c \left (a+c x^2\right )}+\frac {\int \frac {c^2 d^5+10 a c d^3 e^2-15 a^2 d e^4+4 a e^3 \left (5 c d^2-a e^2\right ) x}{a+c x^2} \, dx}{2 a c^2}\\ &=-\frac {3 d e^2 \left (2 c d^2-5 a e^2\right ) x}{2 a c^2}-\frac {e^3 \left (2 c d^2-a e^2\right ) x^2}{a c^2}-\frac {d e^4 x^3}{2 a c}-\frac {(a e-c d x) (d+e x)^4}{2 a c \left (a+c x^2\right )}+\frac {\left (2 e^3 \left (5 c d^2-a e^2\right )\right ) \int \frac {x}{a+c x^2} \, dx}{c^2}+\frac {\left (d \left (c^2 d^4+10 a c d^2 e^2-15 a^2 e^4\right )\right ) \int \frac {1}{a+c x^2} \, dx}{2 a c^2}\\ &=-\frac {3 d e^2 \left (2 c d^2-5 a e^2\right ) x}{2 a c^2}-\frac {e^3 \left (2 c d^2-a e^2\right ) x^2}{a c^2}-\frac {d e^4 x^3}{2 a c}-\frac {(a e-c d x) (d+e x)^4}{2 a c \left (a+c x^2\right )}+\frac {d \left (c^2 d^4+10 a c d^2 e^2-15 a^2 e^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} c^{5/2}}+\frac {e^3 \left (5 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{c^3}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.12, size = 164, normalized size = 0.86 \begin {gather*} \frac {\frac {\sqrt {c} d \left (-15 a^2 e^4+10 a c d^2 e^2+c^2 d^4\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{a^{3/2}}+\frac {-a^3 e^5+5 a^2 c d e^3 (2 d+e x)-5 a c^2 d^3 e (d+2 e x)+c^3 d^5 x}{a \left (a+c x^2\right )}+2 \left (5 c d^2 e^3-a e^5\right ) \log \left (a+c x^2\right )+10 c d e^4 x+c e^5 x^2}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(10*c*d*e^4*x + c*e^5*x^2 + (-(a^3*e^5) + c^3*d^5*x + 5*a^2*c*d*e^3*(2*d + e*x) - 5*a*c^2*d^3*e*(d + 2*e*x))/(
a*(a + c*x^2)) + (Sqrt[c]*d*(c^2*d^4 + 10*a*c*d^2*e^2 - 15*a^2*e^4)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(3/2) + 2*(
5*c*d^2*e^3 - a*e^5)*Log[a + c*x^2])/(2*c^3)

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(d+e x)^5}{\left (a+c x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 561, normalized size = 2.95 \begin {gather*} \left [\frac {2 \, a^{2} c^{2} e^{5} x^{4} + 20 \, a^{2} c^{2} d e^{4} x^{3} + 2 \, a^{3} c e^{5} x^{2} - 10 \, a^{2} c^{2} d^{4} e + 20 \, a^{3} c d^{2} e^{3} - 2 \, a^{4} e^{5} + {\left (a c^{2} d^{5} + 10 \, a^{2} c d^{3} e^{2} - 15 \, a^{3} d e^{4} + {\left (c^{3} d^{5} + 10 \, a c^{2} d^{3} e^{2} - 15 \, a^{2} c d e^{4}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 2 \, {\left (a c^{3} d^{5} - 10 \, a^{2} c^{2} d^{3} e^{2} + 15 \, a^{3} c d e^{4}\right )} x + 4 \, {\left (5 \, a^{3} c d^{2} e^{3} - a^{4} e^{5} + {\left (5 \, a^{2} c^{2} d^{2} e^{3} - a^{3} c e^{5}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}, \frac {a^{2} c^{2} e^{5} x^{4} + 10 \, a^{2} c^{2} d e^{4} x^{3} + a^{3} c e^{5} x^{2} - 5 \, a^{2} c^{2} d^{4} e + 10 \, a^{3} c d^{2} e^{3} - a^{4} e^{5} + {\left (a c^{2} d^{5} + 10 \, a^{2} c d^{3} e^{2} - 15 \, a^{3} d e^{4} + {\left (c^{3} d^{5} + 10 \, a c^{2} d^{3} e^{2} - 15 \, a^{2} c d e^{4}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + {\left (a c^{3} d^{5} - 10 \, a^{2} c^{2} d^{3} e^{2} + 15 \, a^{3} c d e^{4}\right )} x + 2 \, {\left (5 \, a^{3} c d^{2} e^{3} - a^{4} e^{5} + {\left (5 \, a^{2} c^{2} d^{2} e^{3} - a^{3} c e^{5}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(c*x^2+a)^2,x, algorithm="fricas")

[Out]

[1/4*(2*a^2*c^2*e^5*x^4 + 20*a^2*c^2*d*e^4*x^3 + 2*a^3*c*e^5*x^2 - 10*a^2*c^2*d^4*e + 20*a^3*c*d^2*e^3 - 2*a^4
*e^5 + (a*c^2*d^5 + 10*a^2*c*d^3*e^2 - 15*a^3*d*e^4 + (c^3*d^5 + 10*a*c^2*d^3*e^2 - 15*a^2*c*d*e^4)*x^2)*sqrt(
-a*c)*log((c*x^2 + 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(a*c^3*d^5 - 10*a^2*c^2*d^3*e^2 + 15*a^3*c*d*e^4)*x +
4*(5*a^3*c*d^2*e^3 - a^4*e^5 + (5*a^2*c^2*d^2*e^3 - a^3*c*e^5)*x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3), 1
/2*(a^2*c^2*e^5*x^4 + 10*a^2*c^2*d*e^4*x^3 + a^3*c*e^5*x^2 - 5*a^2*c^2*d^4*e + 10*a^3*c*d^2*e^3 - a^4*e^5 + (a
*c^2*d^5 + 10*a^2*c*d^3*e^2 - 15*a^3*d*e^4 + (c^3*d^5 + 10*a*c^2*d^3*e^2 - 15*a^2*c*d*e^4)*x^2)*sqrt(a*c)*arct
an(sqrt(a*c)*x/a) + (a*c^3*d^5 - 10*a^2*c^2*d^3*e^2 + 15*a^3*c*d*e^4)*x + 2*(5*a^3*c*d^2*e^3 - a^4*e^5 + (5*a^
2*c^2*d^2*e^3 - a^3*c*e^5)*x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3)]

________________________________________________________________________________________

giac [A]  time = 0.17, size = 175, normalized size = 0.92 \begin {gather*} \frac {{\left (5 \, c d^{2} e^{3} - a e^{5}\right )} \log \left (c x^{2} + a\right )}{c^{3}} + \frac {{\left (c^{2} d^{5} + 10 \, a c d^{3} e^{2} - 15 \, a^{2} d e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} + \frac {c^{2} x^{2} e^{5} + 10 \, c^{2} d x e^{4}}{2 \, c^{4}} - \frac {5 \, a c^{2} d^{4} e - 10 \, a^{2} c d^{2} e^{3} + a^{3} e^{5} - {\left (c^{3} d^{5} - 10 \, a c^{2} d^{3} e^{2} + 5 \, a^{2} c d e^{4}\right )} x}{2 \, {\left (c x^{2} + a\right )} a c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(c*x^2+a)^2,x, algorithm="giac")

[Out]

(5*c*d^2*e^3 - a*e^5)*log(c*x^2 + a)/c^3 + 1/2*(c^2*d^5 + 10*a*c*d^3*e^2 - 15*a^2*d*e^4)*arctan(c*x/sqrt(a*c))
/(sqrt(a*c)*a*c^2) + 1/2*(c^2*x^2*e^5 + 10*c^2*d*x*e^4)/c^4 - 1/2*(5*a*c^2*d^4*e - 10*a^2*c*d^2*e^3 + a^3*e^5
- (c^3*d^5 - 10*a*c^2*d^3*e^2 + 5*a^2*c*d*e^4)*x)/((c*x^2 + a)*a*c^3)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 248, normalized size = 1.31 \begin {gather*} \frac {5 a d \,e^{4} x}{2 \left (c \,x^{2}+a \right ) c^{2}}-\frac {15 a d \,e^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c^{2}}+\frac {d^{5} x}{2 \left (c \,x^{2}+a \right ) a}+\frac {d^{5} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, a}-\frac {5 d^{3} e^{2} x}{\left (c \,x^{2}+a \right ) c}+\frac {5 d^{3} e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {e^{5} x^{2}}{2 c^{2}}-\frac {a^{2} e^{5}}{2 \left (c \,x^{2}+a \right ) c^{3}}+\frac {5 a \,d^{2} e^{3}}{\left (c \,x^{2}+a \right ) c^{2}}-\frac {a \,e^{5} \ln \left (c \,x^{2}+a \right )}{c^{3}}-\frac {5 d^{4} e}{2 \left (c \,x^{2}+a \right ) c}+\frac {5 d^{2} e^{3} \ln \left (c \,x^{2}+a \right )}{c^{2}}+\frac {5 d \,e^{4} x}{c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^5/(c*x^2+a)^2,x)

[Out]

1/2/c^2*e^5*x^2+5/c^2*d*e^4*x+5/2/c^2/(c*x^2+a)*d*a*x*e^4-5/c/(c*x^2+a)*d^3*x*e^2+1/2/(c*x^2+a)*d^5/a*x-1/2/c^
3/(c*x^2+a)*e^5*a^2+5/c^2/(c*x^2+a)*a*d^2*e^3-5/2/c/(c*x^2+a)*e*d^4-1/c^3*a*ln(c*x^2+a)*e^5+5/c^2*ln(c*x^2+a)*
d^2*e^3-15/2/c^2*a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*d*e^4+5/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*d^3*e
^2+1/2/a/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*d^5

________________________________________________________________________________________

maxima [A]  time = 2.97, size = 181, normalized size = 0.95 \begin {gather*} -\frac {5 \, a c^{2} d^{4} e - 10 \, a^{2} c d^{2} e^{3} + a^{3} e^{5} - {\left (c^{3} d^{5} - 10 \, a c^{2} d^{3} e^{2} + 5 \, a^{2} c d e^{4}\right )} x}{2 \, {\left (a c^{4} x^{2} + a^{2} c^{3}\right )}} + \frac {e^{5} x^{2} + 10 \, d e^{4} x}{2 \, c^{2}} + \frac {{\left (5 \, c d^{2} e^{3} - a e^{5}\right )} \log \left (c x^{2} + a\right )}{c^{3}} + \frac {{\left (c^{2} d^{5} + 10 \, a c d^{3} e^{2} - 15 \, a^{2} d e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^5/(c*x^2+a)^2,x, algorithm="maxima")

[Out]

-1/2*(5*a*c^2*d^4*e - 10*a^2*c*d^2*e^3 + a^3*e^5 - (c^3*d^5 - 10*a*c^2*d^3*e^2 + 5*a^2*c*d*e^4)*x)/(a*c^4*x^2
+ a^2*c^3) + 1/2*(e^5*x^2 + 10*d*e^4*x)/c^2 + (5*c*d^2*e^3 - a*e^5)*log(c*x^2 + a)/c^3 + 1/2*(c^2*d^5 + 10*a*c
*d^3*e^2 - 15*a^2*d*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a*c^2)

________________________________________________________________________________________

mupad [B]  time = 0.16, size = 191, normalized size = 1.01 \begin {gather*} \frac {e^5\,x^2}{2\,c^2}-\frac {\frac {a^2\,e^5-10\,a\,c\,d^2\,e^3+5\,c^2\,d^4\,e}{2\,c}-\frac {x\,\left (5\,a^2\,d\,e^4-10\,a\,c\,d^3\,e^2+c^2\,d^5\right )}{2\,a}}{c^3\,x^2+a\,c^2}-\frac {\ln \left (c\,x^2+a\right )\,\left (32\,a^4\,c^3\,e^5-160\,a^3\,c^4\,d^2\,e^3\right )}{32\,a^3\,c^6}+\frac {5\,d\,e^4\,x}{c^2}+\frac {d\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (-15\,a^2\,e^4+10\,a\,c\,d^2\,e^2+c^2\,d^4\right )}{2\,a^{3/2}\,c^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^5/(a + c*x^2)^2,x)

[Out]

(e^5*x^2)/(2*c^2) - ((a^2*e^5 + 5*c^2*d^4*e - 10*a*c*d^2*e^3)/(2*c) - (x*(c^2*d^5 + 5*a^2*d*e^4 - 10*a*c*d^3*e
^2))/(2*a))/(a*c^2 + c^3*x^2) - (log(a + c*x^2)*(32*a^4*c^3*e^5 - 160*a^3*c^4*d^2*e^3))/(32*a^3*c^6) + (5*d*e^
4*x)/c^2 + (d*atan((c^(1/2)*x)/a^(1/2))*(c^2*d^4 - 15*a^2*e^4 + 10*a*c*d^2*e^2))/(2*a^(3/2)*c^(5/2))

________________________________________________________________________________________

sympy [B]  time = 2.16, size = 515, normalized size = 2.71 \begin {gather*} \left (- \frac {e^{3} \left (a e^{2} - 5 c d^{2}\right )}{c^{3}} - \frac {d \sqrt {- a^{3} c^{7}} \left (15 a^{2} e^{4} - 10 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{6}}\right ) \log {\left (x + \frac {- 4 a^{3} e^{5} - 4 a^{2} c^{3} \left (- \frac {e^{3} \left (a e^{2} - 5 c d^{2}\right )}{c^{3}} - \frac {d \sqrt {- a^{3} c^{7}} \left (15 a^{2} e^{4} - 10 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{6}}\right ) + 20 a^{2} c d^{2} e^{3}}{15 a^{2} c d e^{4} - 10 a c^{2} d^{3} e^{2} - c^{3} d^{5}} \right )} + \left (- \frac {e^{3} \left (a e^{2} - 5 c d^{2}\right )}{c^{3}} + \frac {d \sqrt {- a^{3} c^{7}} \left (15 a^{2} e^{4} - 10 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{6}}\right ) \log {\left (x + \frac {- 4 a^{3} e^{5} - 4 a^{2} c^{3} \left (- \frac {e^{3} \left (a e^{2} - 5 c d^{2}\right )}{c^{3}} + \frac {d \sqrt {- a^{3} c^{7}} \left (15 a^{2} e^{4} - 10 a c d^{2} e^{2} - c^{2} d^{4}\right )}{4 a^{3} c^{6}}\right ) + 20 a^{2} c d^{2} e^{3}}{15 a^{2} c d e^{4} - 10 a c^{2} d^{3} e^{2} - c^{3} d^{5}} \right )} + \frac {- a^{3} e^{5} + 10 a^{2} c d^{2} e^{3} - 5 a c^{2} d^{4} e + x \left (5 a^{2} c d e^{4} - 10 a c^{2} d^{3} e^{2} + c^{3} d^{5}\right )}{2 a^{2} c^{3} + 2 a c^{4} x^{2}} + \frac {5 d e^{4} x}{c^{2}} + \frac {e^{5} x^{2}}{2 c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**5/(c*x**2+a)**2,x)

[Out]

(-e**3*(a*e**2 - 5*c*d**2)/c**3 - d*sqrt(-a**3*c**7)*(15*a**2*e**4 - 10*a*c*d**2*e**2 - c**2*d**4)/(4*a**3*c**
6))*log(x + (-4*a**3*e**5 - 4*a**2*c**3*(-e**3*(a*e**2 - 5*c*d**2)/c**3 - d*sqrt(-a**3*c**7)*(15*a**2*e**4 - 1
0*a*c*d**2*e**2 - c**2*d**4)/(4*a**3*c**6)) + 20*a**2*c*d**2*e**3)/(15*a**2*c*d*e**4 - 10*a*c**2*d**3*e**2 - c
**3*d**5)) + (-e**3*(a*e**2 - 5*c*d**2)/c**3 + d*sqrt(-a**3*c**7)*(15*a**2*e**4 - 10*a*c*d**2*e**2 - c**2*d**4
)/(4*a**3*c**6))*log(x + (-4*a**3*e**5 - 4*a**2*c**3*(-e**3*(a*e**2 - 5*c*d**2)/c**3 + d*sqrt(-a**3*c**7)*(15*
a**2*e**4 - 10*a*c*d**2*e**2 - c**2*d**4)/(4*a**3*c**6)) + 20*a**2*c*d**2*e**3)/(15*a**2*c*d*e**4 - 10*a*c**2*
d**3*e**2 - c**3*d**5)) + (-a**3*e**5 + 10*a**2*c*d**2*e**3 - 5*a*c**2*d**4*e + x*(5*a**2*c*d*e**4 - 10*a*c**2
*d**3*e**2 + c**3*d**5))/(2*a**2*c**3 + 2*a*c**4*x**2) + 5*d*e**4*x/c**2 + e**5*x**2/(2*c**2)

________________________________________________________________________________________