3.21.40 \(\int \frac {a+b x}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\) [2040]

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

[Out]

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

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Rubi [A]
time = 0.13, 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 = {784, 21, 46} \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 {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} \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 46

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] && Lt
Q[m + n + 2, 0])

Rule 784

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*}

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Mathematica [A]
time = 0.07, size = 144, normalized size = 0.55 \begin {gather*} \frac {-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+\frac {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 \left ((a+b x)^2\right )^{3/2}} \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))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(482\) vs. \(2(192)=384\).
time = 0.13, size = 483, normalized size = 1.86

method	result	size
default	\(\frac {\left (-36 \ln \left (e x +d \right ) a^{2} b^{2} d \,e^{3} x -36 \ln \left (e x +d \right ) a \,b^{3} d \,e^{3} x^{2}+36 \ln \left (b x +a \right ) a^{2} b^{2} d \,e^{3} x +36 \ln \left (b x +a \right ) a \,b^{3} d \,e^{3} x^{2}+12 \ln \left (b x +a \right ) b^{4} d \,e^{3} x^{3}+36 \ln \left (b x +a \right ) a^{2} b^{2} e^{4} x^{2}+36 \ln \left (b x +a \right ) a \,b^{3} e^{4} x^{3}+12 \ln \left (b x +a \right ) a^{3} b d \,e^{3}+12 \ln \left (b x +a \right ) a^{3} b \,e^{4} x -36 \ln \left (e x +d \right ) a \,b^{3} e^{4} x^{3}-12 \ln \left (e x +d \right ) b^{4} d \,e^{3} x^{3}-36 \ln \left (e x +d \right ) a^{2} b^{2} e^{4} x^{2}-12 \ln \left (e x +d \right ) a^{3} b \,e^{4} x +18 a \,b^{3} d^{2} e^{2} x -12 a \,b^{3} e^{4} x^{3}+12 b^{4} d \,e^{3} x^{3}-30 a^{2} b^{2} e^{4} x^{2}+6 b^{4} d^{2} e^{2} x^{2}-22 a^{3} b \,e^{4} x -2 b^{4} d^{3} e x +12 \ln \left (b x +a \right ) b^{4} e^{4} x^{4}+b^{4} d^{4}-3 a^{4} e^{4}-12 \ln \left (e x +d \right ) b^{4} e^{4} x^{4}-6 a \,b^{3} d^{3} e +18 a^{2} b^{2} d^{2} e^{2}-10 a^{3} b d \,e^{3}+24 a \,b^{3} d \,e^{3} x^{2}+6 a^{2} b^{2} d \,e^{3} x -12 \ln \left (e x +d \right ) a^{3} b d \,e^{3}\right ) \left (b x +a \right )^{2}}{3 \left (e x +d \right ) \left (a e -b d \right )^{5} \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}}}\)	\(483\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}\, \left (-\frac {4 e^{3} b^{3} x^{3}}{a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}-\frac {2 b^{2} \left (5 a e +b d \right ) e^{2} x^{2}}{a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}}-\frac {2 \left (11 a^{2} e^{2}+8 a b d e -b^{2} d^{2}\right ) b e x}{3 \left (a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}-\frac {3 a^{3} e^{3}+13 a^{2} b d \,e^{2}-5 a \,b^{2} d^{2} e +b^{3} d^{3}}{3 \left (a^{4} e^{4}-4 a^{3} b d \,e^{3}+6 a^{2} b^{2} d^{2} e^{2}-4 a \,b^{3} d^{3} e +b^{4} d^{4}\right )}\right )}{\left (b x +a \right )^{4} \left (e x +d \right )}+\frac {4 \sqrt {\left (b x +a \right )^{2}}\, e^{3} b \ln \left (-b x -a \right )}{\left (b x +a \right ) \left (a^{5} e^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}-\frac {4 \sqrt {\left (b x +a \right )^{2}}\, e^{3} b \ln \left (e x +d \right )}{\left (b x +a \right ) \left (a^{5} e^{5}-5 a^{4} b d \,e^{4}+10 a^{3} b^{2} d^{2} e^{3}-10 a^{2} b^{3} d^{3} e^{2}+5 a \,b^{4} d^{4} e -b^{5} d^{5}\right )}\)	\(518\)

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,method=_RETURNVERBOSE)

[Out]

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

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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(b*d-%e*a>0)', see `assume?` fo
r more detai

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 708 vs. \(2 (197) = 394\).
time = 2.38, size = 708, normalized size = 2.72 \begin {gather*} -\frac {b^{4} d^{4} - {\left (12 \, a b^{3} x^{3} + 30 \, a^{2} b^{2} x^{2} + 22 \, a^{3} b x + 3 \, a^{4}\right )} e^{4} + 2 \, {\left (6 \, b^{4} d x^{3} + 12 \, a b^{3} d x^{2} + 3 \, a^{2} b^{2} d x - 5 \, a^{3} b d\right )} e^{3} + 6 \, {\left (b^{4} d^{2} x^{2} + 3 \, a b^{3} d^{2} x + 3 \, a^{2} b^{2} d^{2}\right )} e^{2} - 2 \, {\left (b^{4} d^{3} x + 3 \, a b^{3} d^{3}\right )} e + 12 \, {\left ({\left (b^{4} x^{4} + 3 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + a^{3} b x\right )} e^{4} + {\left (b^{4} d x^{3} + 3 \, a b^{3} d x^{2} + 3 \, a^{2} b^{2} d x + a^{3} b d\right )} e^{3}\right )} \log \left (b x + a\right ) - 12 \, {\left ({\left (b^{4} x^{4} + 3 \, a b^{3} x^{3} + 3 \, a^{2} b^{2} x^{2} + a^{3} b x\right )} e^{4} + {\left (b^{4} d x^{3} + 3 \, a b^{3} d x^{2} + 3 \, a^{2} b^{2} d x + a^{3} b d\right )} e^{3}\right )} \log \left (x e + d\right )}{3 \, {\left (b^{8} d^{6} x^{3} + 3 \, a b^{7} d^{6} x^{2} + 3 \, a^{2} b^{6} d^{6} x + a^{3} b^{5} d^{6} - {\left (a^{5} b^{3} x^{4} + 3 \, a^{6} b^{2} x^{3} + 3 \, a^{7} b x^{2} + a^{8} x\right )} e^{6} + {\left (5 \, a^{4} b^{4} d x^{4} + 14 \, a^{5} b^{3} d x^{3} + 12 \, a^{6} b^{2} d x^{2} + 2 \, a^{7} b d x - a^{8} d\right )} e^{5} - 5 \, {\left (2 \, a^{3} b^{5} d^{2} x^{4} + 5 \, a^{4} b^{4} d^{2} x^{3} + 3 \, a^{5} b^{3} d^{2} x^{2} - a^{6} b^{2} d^{2} x - a^{7} b d^{2}\right )} e^{4} + 10 \, {\left (a^{2} b^{6} d^{3} x^{4} + 2 \, a^{3} b^{5} d^{3} x^{3} - 2 \, a^{5} b^{3} d^{3} x - a^{6} b^{2} d^{3}\right )} e^{3} - 5 \, {\left (a b^{7} d^{4} x^{4} + a^{2} b^{6} d^{4} x^{3} - 3 \, a^{3} b^{5} d^{4} x^{2} - 5 \, a^{4} b^{4} d^{4} x - 2 \, a^{5} b^{3} d^{4}\right )} e^{2} + {\left (b^{8} d^{5} x^{4} - 2 \, a b^{7} d^{5} x^{3} - 12 \, a^{2} b^{6} d^{5} x^{2} - 14 \, a^{3} b^{5} d^{5} x - 5 \, a^{4} b^{4} d^{5}\right )} e\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 - (12*a*b^3*x^3 + 30*a^2*b^2*x^2 + 22*a^3*b*x + 3*a^4)*e^4 + 2*(6*b^4*d*x^3 + 12*a*b^3*d*x^2 + 3
*a^2*b^2*d*x - 5*a^3*b*d)*e^3 + 6*(b^4*d^2*x^2 + 3*a*b^3*d^2*x + 3*a^2*b^2*d^2)*e^2 - 2*(b^4*d^3*x + 3*a*b^3*d
^3)*e + 12*((b^4*x^4 + 3*a*b^3*x^3 + 3*a^2*b^2*x^2 + a^3*b*x)*e^4 + (b^4*d*x^3 + 3*a*b^3*d*x^2 + 3*a^2*b^2*d*x
+ a^3*b*d)*e^3)*log(b*x + a) - 12*((b^4*x^4 + 3*a*b^3*x^3 + 3*a^2*b^2*x^2 + a^3*b*x)*e^4 + (b^4*d*x^3 + 3*a*b
^3*d*x^2 + 3*a^2*b^2*d*x + a^3*b*d)*e^3)*log(x*e + d))/(b^8*d^6*x^3 + 3*a*b^7*d^6*x^2 + 3*a^2*b^6*d^6*x + a^3*
b^5*d^6 - (a^5*b^3*x^4 + 3*a^6*b^2*x^3 + 3*a^7*b*x^2 + a^8*x)*e^6 + (5*a^4*b^4*d*x^4 + 14*a^5*b^3*d*x^3 + 12*a
^6*b^2*d*x^2 + 2*a^7*b*d*x - a^8*d)*e^5 - 5*(2*a^3*b^5*d^2*x^4 + 5*a^4*b^4*d^2*x^3 + 3*a^5*b^3*d^2*x^2 - a^6*b
^2*d^2*x - a^7*b*d^2)*e^4 + 10*(a^2*b^6*d^3*x^4 + 2*a^3*b^5*d^3*x^3 - 2*a^5*b^3*d^3*x - a^6*b^2*d^3)*e^3 - 5*(
a*b^7*d^4*x^4 + a^2*b^6*d^4*x^3 - 3*a^3*b^5*d^4*x^2 - 5*a^4*b^4*d^4*x - 2*a^5*b^3*d^4)*e^2 + (b^8*d^5*x^4 - 2*
a*b^7*d^5*x^3 - 12*a^2*b^6*d^5*x^2 - 14*a^3*b^5*d^5*x - 5*a^4*b^4*d^5)*e)

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Sympy [F(-1)] Timed out
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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 413 vs. \(2 (197) = 394\).
time = 1.60, size = 413, normalized size = 1.59 \begin {gather*} -\frac {4 \, b^{2} e^{3} \log \left ({\left | b x + a \right |}\right )}{b^{6} d^{5} \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{5} d^{4} e \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{4} d^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{3} d^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b^{2} d e^{4} \mathrm {sgn}\left (b x + a\right ) - a^{5} b e^{5} \mathrm {sgn}\left (b x + a\right )} + \frac {4 \, b e^{4} \log \left ({\left | x e + d \right |}\right )}{b^{5} d^{5} e \mathrm {sgn}\left (b x + a\right ) - 5 \, a b^{4} d^{4} e^{2} \mathrm {sgn}\left (b x + a\right ) + 10 \, a^{2} b^{3} d^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 10 \, a^{3} b^{2} d^{2} e^{4} \mathrm {sgn}\left (b x + a\right ) + 5 \, a^{4} b d e^{5} \mathrm {sgn}\left (b x + a\right ) - a^{5} e^{6} \mathrm {sgn}\left (b x + a\right )} - \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}{3 \, {\left (b d - a e\right )}^{5} {\left (b x + a\right )}^{3} {\left (x e + d\right )} \mathrm {sgn}\left (b x + a\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^2*e^3*log(abs(b*x + a))/(b^6*d^5*sgn(b*x + a) - 5*a*b^5*d^4*e*sgn(b*x + a) + 10*a^2*b^4*d^3*e^2*sgn(b*x +
a) - 10*a^3*b^3*d^2*e^3*sgn(b*x + a) + 5*a^4*b^2*d*e^4*sgn(b*x + a) - a^5*b*e^5*sgn(b*x + a)) + 4*b*e^4*log(a
bs(x*e + d))/(b^5*d^5*e*sgn(b*x + a) - 5*a*b^4*d^4*e^2*sgn(b*x + a) + 10*a^2*b^3*d^3*e^3*sgn(b*x + a) - 10*a^3
*b^2*d^2*e^4*sgn(b*x + a) + 5*a^4*b*d*e^5*sgn(b*x + a) - a^5*e^6*sgn(b*x + a)) - 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)/((b*d - a*e)^
5*(b*x + a)^3*(x*e + d)*sgn(b*x + a))

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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)

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