3.19.15 \(\int \frac {a+b x}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=260 \[ -\frac {e^3 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^4}-\frac {4 b e^3 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {4 b e^3 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac {3 b e^2}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}+\frac {b e}{(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.19, antiderivative size = 260, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {770, 21, 44} \begin {gather*} -\frac {e^3 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^4}-\frac {3 b e^2}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {4 b e^3 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {4 b e^3 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {b e}{(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*e^2)/((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b/(3*(b*d - a*e)^2*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (b*e)/((b*d - a*e)^3*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^3*(a + b*x))/((b*d - a*e)^4*(d
+ e*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (4*b*e^3*(a + b*x)*Log[a + b*x])/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x +
b^2*x^2]) + (4*b*e^3*(a + b*x)*Log[d + e*x])/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.10, size = 144, normalized size = 0.55 \begin {gather*} \frac {\frac {3 e^3 (a+b x)^3 (a e-b d)}{d+e x}+12 b e^3 (a+b x)^3 \log (d+e x)-9 b e^2 (a+b x)^2 (b d-a e)+3 b e (a+b x) (b d-a e)^2-b (b d-a e)^3-12 b e^3 (a+b x)^3 \log (a+b x)}{3 \left ((a+b x)^2\right )^{3/2} (b d-a e)^5} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(b*(b*d - a*e)^3) + 3*b*e*(b*d - a*e)^2*(a + b*x) - 9*b*e^2*(b*d - a*e)*(a + b*x)^2 + (3*e^3*(-(b*d) + a*e)*
(a + b*x)^3)/(d + e*x) - 12*b*e^3*(a + b*x)^3*Log[a + b*x] + 12*b*e^3*(a + b*x)^3*Log[d + e*x])/(3*(b*d - a*e)
^5*((a + b*x)^2)^(3/2))

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 180.00, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

$Aborted

________________________________________________________________________________________

fricas [B]  time = 0.44, size = 751, normalized size = 2.89 \begin {gather*} -\frac {b^{4} d^{4} - 6 \, a b^{3} d^{3} e + 18 \, a^{2} b^{2} d^{2} e^{2} - 10 \, a^{3} b d e^{3} - 3 \, a^{4} e^{4} + 12 \, {\left (b^{4} d e^{3} - a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (b^{4} d^{2} e^{2} + 4 \, a b^{3} d e^{3} - 5 \, a^{2} b^{2} e^{4}\right )} x^{2} - 2 \, {\left (b^{4} d^{3} e - 9 \, a b^{3} d^{2} e^{2} - 3 \, a^{2} b^{2} d e^{3} + 11 \, a^{3} b e^{4}\right )} x + 12 \, {\left (b^{4} e^{4} x^{4} + a^{3} b d e^{3} + {\left (b^{4} d e^{3} + 3 \, a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + {\left (3 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \log \left (b x + a\right ) - 12 \, {\left (b^{4} e^{4} x^{4} + a^{3} b d e^{3} + {\left (b^{4} d e^{3} + 3 \, a b^{3} e^{4}\right )} x^{3} + 3 \, {\left (a b^{3} d e^{3} + a^{2} b^{2} e^{4}\right )} x^{2} + {\left (3 \, a^{2} b^{2} d e^{3} + a^{3} b e^{4}\right )} x\right )} \log \left (e x + d\right )}{3 \, {\left (a^{3} b^{5} d^{6} - 5 \, a^{4} b^{4} d^{5} e + 10 \, a^{5} b^{3} d^{4} e^{2} - 10 \, a^{6} b^{2} d^{3} e^{3} + 5 \, a^{7} b d^{2} e^{4} - a^{8} d e^{5} + {\left (b^{8} d^{5} e - 5 \, a b^{7} d^{4} e^{2} + 10 \, a^{2} b^{6} d^{3} e^{3} - 10 \, a^{3} b^{5} d^{2} e^{4} + 5 \, a^{4} b^{4} d e^{5} - a^{5} b^{3} e^{6}\right )} x^{4} + {\left (b^{8} d^{6} - 2 \, a b^{7} d^{5} e - 5 \, a^{2} b^{6} d^{4} e^{2} + 20 \, a^{3} b^{5} d^{3} e^{3} - 25 \, a^{4} b^{4} d^{2} e^{4} + 14 \, a^{5} b^{3} d e^{5} - 3 \, a^{6} b^{2} e^{6}\right )} x^{3} + 3 \, {\left (a b^{7} d^{6} - 4 \, a^{2} b^{6} d^{5} e + 5 \, a^{3} b^{5} d^{4} e^{2} - 5 \, a^{5} b^{3} d^{2} e^{4} + 4 \, a^{6} b^{2} d e^{5} - a^{7} b e^{6}\right )} x^{2} + {\left (3 \, a^{2} b^{6} d^{6} - 14 \, a^{3} b^{5} d^{5} e + 25 \, a^{4} b^{4} d^{4} e^{2} - 20 \, a^{5} b^{3} d^{3} e^{3} + 5 \, a^{6} b^{2} d^{2} e^{4} + 2 \, a^{7} b d e^{5} - a^{8} e^{6}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(b^4*d^4 - 6*a*b^3*d^3*e + 18*a^2*b^2*d^2*e^2 - 10*a^3*b*d*e^3 - 3*a^4*e^4 + 12*(b^4*d*e^3 - a*b^3*e^4)*x
^3 + 6*(b^4*d^2*e^2 + 4*a*b^3*d*e^3 - 5*a^2*b^2*e^4)*x^2 - 2*(b^4*d^3*e - 9*a*b^3*d^2*e^2 - 3*a^2*b^2*d*e^3 +
11*a^3*b*e^4)*x + 12*(b^4*e^4*x^4 + a^3*b*d*e^3 + (b^4*d*e^3 + 3*a*b^3*e^4)*x^3 + 3*(a*b^3*d*e^3 + a^2*b^2*e^4
)*x^2 + (3*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*log(b*x + a) - 12*(b^4*e^4*x^4 + a^3*b*d*e^3 + (b^4*d*e^3 + 3*a*b^3*e
^4)*x^3 + 3*(a*b^3*d*e^3 + a^2*b^2*e^4)*x^2 + (3*a^2*b^2*d*e^3 + a^3*b*e^4)*x)*log(e*x + d))/(a^3*b^5*d^6 - 5*
a^4*b^4*d^5*e + 10*a^5*b^3*d^4*e^2 - 10*a^6*b^2*d^3*e^3 + 5*a^7*b*d^2*e^4 - a^8*d*e^5 + (b^8*d^5*e - 5*a*b^7*d
^4*e^2 + 10*a^2*b^6*d^3*e^3 - 10*a^3*b^5*d^2*e^4 + 5*a^4*b^4*d*e^5 - a^5*b^3*e^6)*x^4 + (b^8*d^6 - 2*a*b^7*d^5
*e - 5*a^2*b^6*d^4*e^2 + 20*a^3*b^5*d^3*e^3 - 25*a^4*b^4*d^2*e^4 + 14*a^5*b^3*d*e^5 - 3*a^6*b^2*e^6)*x^3 + 3*(
a*b^7*d^6 - 4*a^2*b^6*d^5*e + 5*a^3*b^5*d^4*e^2 - 5*a^5*b^3*d^2*e^4 + 4*a^6*b^2*d*e^5 - a^7*b*e^6)*x^2 + (3*a^
2*b^6*d^6 - 14*a^3*b^5*d^5*e + 25*a^4*b^4*d^4*e^2 - 20*a^5*b^3*d^3*e^3 + 5*a^6*b^2*d^2*e^4 + 2*a^7*b*d*e^5 - a
^8*e^6)*x)

________________________________________________________________________________________

giac [B]  time = 0.51, size = 774, normalized size = 2.98 \begin {gather*} \frac {4 \, b e^{4} \log \left ({\left | -b + \frac {b d}{x e + d} - \frac {a e}{x e + d} \right |}\right )}{b^{5} d^{5} e \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - 5 \, a b^{4} d^{4} e^{2} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) + 10 \, a^{2} b^{3} d^{3} e^{3} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - 10 \, a^{3} b^{2} d^{2} e^{4} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) + 5 \, a^{4} b d e^{5} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - a^{5} e^{6} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )} + \frac {e^{7}}{{\left (b^{4} d^{4} e^{4} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - 4 \, a b^{3} d^{3} e^{5} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) + 6 \, a^{2} b^{2} d^{2} e^{6} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) - 4 \, a^{3} b d e^{7} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right ) + a^{4} e^{8} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )\right )} {\left (x e + d\right )}} + \frac {13 \, b^{4} e^{3} - \frac {30 \, {\left (b^{4} d e^{4} - a b^{3} e^{5}\right )} e^{\left (-1\right )}}{x e + d} + \frac {18 \, {\left (b^{4} d^{2} e^{5} - 2 \, a b^{3} d e^{6} + a^{2} b^{2} e^{7}\right )} e^{\left (-2\right )}}{{\left (x e + d\right )}^{2}}}{3 \, {\left (b d - a e\right )}^{5} {\left (b - \frac {b d}{x e + d} + \frac {a e}{x e + d}\right )}^{3} \mathrm {sgn}\left (-\frac {b e}{x e + d} + \frac {b d e}{{\left (x e + d\right )}^{2}} - \frac {a e^{2}}{{\left (x e + d\right )}^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

4*b*e^4*log(abs(-b + b*d/(x*e + d) - a*e/(x*e + d)))/(b^5*d^5*e*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2
/(x*e + d)^2) - 5*a*b^4*d^4*e^2*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) + 10*a^2*b^3*d^3*e
^3*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) - 10*a^3*b^2*d^2*e^4*sgn(-b*e/(x*e + d) + b*d*e
/(x*e + d)^2 - a*e^2/(x*e + d)^2) + 5*a^4*b*d*e^5*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2)
- a^5*e^6*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2)) + e^7/((b^4*d^4*e^4*sgn(-b*e/(x*e + d)
+ b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) - 4*a*b^3*d^3*e^5*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e
+ d)^2) + 6*a^2*b^2*d^2*e^6*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) - 4*a^3*b*d*e^7*sgn(-
b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2) + a^4*e^8*sgn(-b*e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^
2/(x*e + d)^2))*(x*e + d)) + 1/3*(13*b^4*e^3 - 30*(b^4*d*e^4 - a*b^3*e^5)*e^(-1)/(x*e + d) + 18*(b^4*d^2*e^5 -
2*a*b^3*d*e^6 + a^2*b^2*e^7)*e^(-2)/(x*e + d)^2)/((b*d - a*e)^5*(b - b*d/(x*e + d) + a*e/(x*e + d))^3*sgn(-b*
e/(x*e + d) + b*d*e/(x*e + d)^2 - a*e^2/(x*e + d)^2))

________________________________________________________________________________________

maple [B]  time = 0.07, size = 483, normalized size = 1.86 \begin {gather*} \frac {\left (12 b^{4} e^{4} x^{4} \ln \left (b x +a \right )-12 b^{4} e^{4} x^{4} \ln \left (e x +d \right )+36 a \,b^{3} e^{4} x^{3} \ln \left (b x +a \right )-36 a \,b^{3} e^{4} x^{3} \ln \left (e x +d \right )+12 b^{4} d \,e^{3} x^{3} \ln \left (b x +a \right )-12 b^{4} d \,e^{3} x^{3} \ln \left (e x +d \right )+36 a^{2} b^{2} e^{4} x^{2} \ln \left (b x +a \right )-36 a^{2} b^{2} e^{4} x^{2} \ln \left (e x +d \right )+36 a \,b^{3} d \,e^{3} x^{2} \ln \left (b x +a \right )-36 a \,b^{3} d \,e^{3} x^{2} \ln \left (e x +d \right )-12 a \,b^{3} e^{4} x^{3}+12 b^{4} d \,e^{3} x^{3}+12 a^{3} b \,e^{4} x \ln \left (b x +a \right )-12 a^{3} b \,e^{4} x \ln \left (e x +d \right )+36 a^{2} b^{2} d \,e^{3} x \ln \left (b x +a \right )-36 a^{2} b^{2} d \,e^{3} x \ln \left (e x +d \right )-30 a^{2} b^{2} e^{4} x^{2}+24 a \,b^{3} d \,e^{3} x^{2}+6 b^{4} d^{2} e^{2} x^{2}+12 a^{3} b d \,e^{3} \ln \left (b x +a \right )-12 a^{3} b d \,e^{3} \ln \left (e x +d \right )-22 a^{3} b \,e^{4} x +6 a^{2} b^{2} d \,e^{3} x +18 a \,b^{3} d^{2} e^{2} x -2 b^{4} d^{3} e x -3 a^{4} e^{4}-10 a^{3} b d \,e^{3}+18 a^{2} b^{2} d^{2} e^{2}-6 a \,b^{3} d^{3} e +b^{4} d^{4}\right ) \left (b x +a \right )^{2}}{3 \left (e x +d \right ) \left (a e -b d \right )^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(36*a^2*b^2*d*e^3*x*ln(b*x+a)+36*a*b^3*d*e^3*x^2*ln(b*x+a)-6*a*b^3*d^3*e+18*a^2*b^2*d^2*e^2-36*a*b^3*d*e^3
*x^2*ln(e*x+d)-36*a^2*b^2*d*e^3*x*ln(e*x+d)-12*b^4*e^4*x^4*ln(e*x+d)+b^4*d^4-3*a^4*e^4+6*b^4*d^2*e^2*x^2-22*a^
3*b*e^4*x-2*b^4*d^3*e*x-30*a^2*b^2*e^4*x^2-12*a*b^3*e^4*x^3+12*b^4*d*e^3*x^3-10*a^3*b*d*e^3+6*a^2*b^2*d*e^3*x+
18*a*b^3*d^2*e^2*x-12*a^3*b*d*e^3*ln(e*x+d)+36*a*b^3*e^4*x^3*ln(b*x+a)+24*a*b^3*d*e^3*x^2+12*ln(b*x+a)*x^4*b^4
*e^4+12*a^3*b*e^4*x*ln(b*x+a)-12*a^3*b*e^4*x*ln(e*x+d)+12*a^3*b*d*e^3*ln(b*x+a)-12*b^4*d*e^3*x^3*ln(e*x+d)+12*
b^4*d*e^3*x^3*ln(b*x+a)+36*a^2*b^2*e^4*x^2*ln(b*x+a)-36*a^2*b^2*e^4*x^2*ln(e*x+d)-36*a*b^3*e^4*x^3*ln(e*x+d))*
(b*x+a)^2/(e*x+d)/(a*e-b*d)^5/((b*x+a)^2)^(5/2)

________________________________________________________________________________________

maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+b\,x}{{\left (d+e\,x\right )}^2\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________