3.19.16 \(\int \frac {a+b x}{(d+e x)^3 (a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

Optimal. Leaf size=323 \[ -\frac {4 b e^3 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^5}-\frac {e^3 (a+b x)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^4}-\frac {10 b^2 e^3 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^6}+\frac {10 b^2 e^3 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^6}-\frac {6 b^2 e^2}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}+\frac {3 b^2 e}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {b^2}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

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Rubi [A]  time = 0.24, antiderivative size = 323, 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 {4 b e^3 (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^5}-\frac {e^3 (a+b x)}{2 \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)^4}-\frac {6 b^2 e^2}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^5}-\frac {10 b^2 e^3 (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^6}+\frac {10 b^2 e^3 (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^6}+\frac {3 b^2 e}{2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4}-\frac {b^2}{3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(-6*b^2*e^2)/((b*d - a*e)^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - b^2/(3*(b*d - a*e)^3*(a + b*x)^2*Sqrt[a^2 + 2*a*b
*x + b^2*x^2]) + (3*b^2*e)/(2*(b*d - a*e)^4*(a + b*x)*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (e^3*(a + b*x))/(2*(b*d
- a*e)^4*(d + e*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (4*b*e^3*(a + b*x))/((b*d - a*e)^5*(d + e*x)*Sqrt[a^2 +
2*a*b*x + b^2*x^2]) - (10*b^2*e^3*(a + b*x)*Log[a + b*x])/((b*d - a*e)^6*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (10
*b^2*e^3*(a + b*x)*Log[d + e*x])/((b*d - a*e)^6*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)^3 \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)^3} \, 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)^3} \, 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^3}{(b d-a e)^3 (a+b x)^4}-\frac {3 b^3 e}{(b d-a e)^4 (a+b x)^3}+\frac {6 b^3 e^2}{(b d-a e)^5 (a+b x)^2}-\frac {10 b^3 e^3}{(b d-a e)^6 (a+b x)}+\frac {e^4}{(b d-a e)^4 (d+e x)^3}+\frac {4 b e^4}{(b d-a e)^5 (d+e x)^2}+\frac {10 b^2 e^4}{(b d-a e)^6 (d+e x)}\right ) \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {6 b^2 e^2}{(b d-a e)^5 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b^2}{3 (b d-a e)^3 (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 b^2 e}{2 (b d-a e)^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e^3 (a+b x)}{2 (b d-a e)^4 (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {4 b e^3 (a+b x)}{(b d-a e)^5 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {10 b^2 e^3 (a+b x) \log (a+b x)}{(b d-a e)^6 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {10 b^2 e^3 (a+b x) \log (d+e x)}{(b d-a e)^6 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 184, normalized size = 0.57 \begin {gather*} \frac {60 b^2 e^3 (a+b x)^3 \log (d+e x)-36 b^2 e^2 (a+b x)^2 (b d-a e)+9 b^2 e (a+b x) (b d-a e)^2-2 b^2 (b d-a e)^3-60 b^2 e^3 (a+b x)^3 \log (a+b x)-\frac {3 e^3 (a+b x)^3 (b d-a e)^2}{(d+e x)^2}-\frac {24 b e^3 (a+b x)^3 (b d-a e)}{d+e x}}{6 \left ((a+b x)^2\right )^{3/2} (b d-a e)^6} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

(-2*b^2*(b*d - a*e)^3 + 9*b^2*e*(b*d - a*e)^2*(a + b*x) - 36*b^2*e^2*(b*d - a*e)*(a + b*x)^2 - (3*e^3*(b*d - a
*e)^2*(a + b*x)^3)/(d + e*x)^2 - (24*b*e^3*(b*d - a*e)*(a + b*x)^3)/(d + e*x) - 60*b^2*e^3*(a + b*x)^3*Log[a +
b*x] + 60*b^2*e^3*(a + b*x)^3*Log[d + e*x])/(6*(b*d - a*e)^6*((a + b*x)^2)^(3/2))

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IntegrateAlgebraic [B]  time = 178.09, size = 9902, normalized size = 30.66 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)/((d + e*x)^3*(a^2 + 2*a*b*x + b^2*x^2)^(5/2)),x]

[Out]

Result too large to show

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fricas [B]  time = 0.48, size = 1151, normalized size = 3.56 \begin {gather*} -\frac {2 \, b^{5} d^{5} - 15 \, a b^{4} d^{4} e + 60 \, a^{2} b^{3} d^{3} e^{2} - 20 \, a^{3} b^{2} d^{2} e^{3} - 30 \, a^{4} b d e^{4} + 3 \, a^{5} e^{5} + 60 \, {\left (b^{5} d e^{4} - a b^{4} e^{5}\right )} x^{4} + 30 \, {\left (3 \, b^{5} d^{2} e^{3} + 2 \, a b^{4} d e^{4} - 5 \, a^{2} b^{3} e^{5}\right )} x^{3} + 10 \, {\left (2 \, b^{5} d^{3} e^{2} + 21 \, a b^{4} d^{2} e^{3} - 12 \, a^{2} b^{3} d e^{4} - 11 \, a^{3} b^{2} e^{5}\right )} x^{2} - 5 \, {\left (b^{5} d^{4} e - 12 \, a b^{4} d^{3} e^{2} - 24 \, a^{2} b^{3} d^{2} e^{3} + 32 \, a^{3} b^{2} d e^{4} + 3 \, a^{4} b e^{5}\right )} x + 60 \, {\left (b^{5} e^{5} x^{5} + a^{3} b^{2} d^{2} e^{3} + {\left (2 \, b^{5} d e^{4} + 3 \, a b^{4} e^{5}\right )} x^{4} + {\left (b^{5} d^{2} e^{3} + 6 \, a b^{4} d e^{4} + 3 \, a^{2} b^{3} e^{5}\right )} x^{3} + {\left (3 \, a b^{4} d^{2} e^{3} + 6 \, a^{2} b^{3} d e^{4} + a^{3} b^{2} e^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{3} d^{2} e^{3} + 2 \, a^{3} b^{2} d e^{4}\right )} x\right )} \log \left (b x + a\right ) - 60 \, {\left (b^{5} e^{5} x^{5} + a^{3} b^{2} d^{2} e^{3} + {\left (2 \, b^{5} d e^{4} + 3 \, a b^{4} e^{5}\right )} x^{4} + {\left (b^{5} d^{2} e^{3} + 6 \, a b^{4} d e^{4} + 3 \, a^{2} b^{3} e^{5}\right )} x^{3} + {\left (3 \, a b^{4} d^{2} e^{3} + 6 \, a^{2} b^{3} d e^{4} + a^{3} b^{2} e^{5}\right )} x^{2} + {\left (3 \, a^{2} b^{3} d^{2} e^{3} + 2 \, a^{3} b^{2} d e^{4}\right )} x\right )} \log \left (e x + d\right )}{6 \, {\left (a^{3} b^{6} d^{8} - 6 \, a^{4} b^{5} d^{7} e + 15 \, a^{5} b^{4} d^{6} e^{2} - 20 \, a^{6} b^{3} d^{5} e^{3} + 15 \, a^{7} b^{2} d^{4} e^{4} - 6 \, a^{8} b d^{3} e^{5} + a^{9} d^{2} e^{6} + {\left (b^{9} d^{6} e^{2} - 6 \, a b^{8} d^{5} e^{3} + 15 \, a^{2} b^{7} d^{4} e^{4} - 20 \, a^{3} b^{6} d^{3} e^{5} + 15 \, a^{4} b^{5} d^{2} e^{6} - 6 \, a^{5} b^{4} d e^{7} + a^{6} b^{3} e^{8}\right )} x^{5} + {\left (2 \, b^{9} d^{7} e - 9 \, a b^{8} d^{6} e^{2} + 12 \, a^{2} b^{7} d^{5} e^{3} + 5 \, a^{3} b^{6} d^{4} e^{4} - 30 \, a^{4} b^{5} d^{3} e^{5} + 33 \, a^{5} b^{4} d^{2} e^{6} - 16 \, a^{6} b^{3} d e^{7} + 3 \, a^{7} b^{2} e^{8}\right )} x^{4} + {\left (b^{9} d^{8} - 18 \, a^{2} b^{7} d^{6} e^{2} + 52 \, a^{3} b^{6} d^{5} e^{3} - 60 \, a^{4} b^{5} d^{4} e^{4} + 24 \, a^{5} b^{4} d^{3} e^{5} + 10 \, a^{6} b^{3} d^{2} e^{6} - 12 \, a^{7} b^{2} d e^{7} + 3 \, a^{8} b e^{8}\right )} x^{3} + {\left (3 \, a b^{8} d^{8} - 12 \, a^{2} b^{7} d^{7} e + 10 \, a^{3} b^{6} d^{6} e^{2} + 24 \, a^{4} b^{5} d^{5} e^{3} - 60 \, a^{5} b^{4} d^{4} e^{4} + 52 \, a^{6} b^{3} d^{3} e^{5} - 18 \, a^{7} b^{2} d^{2} e^{6} + a^{9} e^{8}\right )} x^{2} + {\left (3 \, a^{2} b^{7} d^{8} - 16 \, a^{3} b^{6} d^{7} e + 33 \, a^{4} b^{5} d^{6} e^{2} - 30 \, a^{5} b^{4} d^{5} e^{3} + 5 \, a^{6} b^{3} d^{4} e^{4} + 12 \, a^{7} b^{2} d^{3} e^{5} - 9 \, a^{8} b d^{2} e^{6} + 2 \, a^{9} d e^{7}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="fricas")

[Out]

-1/6*(2*b^5*d^5 - 15*a*b^4*d^4*e + 60*a^2*b^3*d^3*e^2 - 20*a^3*b^2*d^2*e^3 - 30*a^4*b*d*e^4 + 3*a^5*e^5 + 60*(
b^5*d*e^4 - a*b^4*e^5)*x^4 + 30*(3*b^5*d^2*e^3 + 2*a*b^4*d*e^4 - 5*a^2*b^3*e^5)*x^3 + 10*(2*b^5*d^3*e^2 + 21*a
*b^4*d^2*e^3 - 12*a^2*b^3*d*e^4 - 11*a^3*b^2*e^5)*x^2 - 5*(b^5*d^4*e - 12*a*b^4*d^3*e^2 - 24*a^2*b^3*d^2*e^3 +
32*a^3*b^2*d*e^4 + 3*a^4*b*e^5)*x + 60*(b^5*e^5*x^5 + a^3*b^2*d^2*e^3 + (2*b^5*d*e^4 + 3*a*b^4*e^5)*x^4 + (b^
5*d^2*e^3 + 6*a*b^4*d*e^4 + 3*a^2*b^3*e^5)*x^3 + (3*a*b^4*d^2*e^3 + 6*a^2*b^3*d*e^4 + a^3*b^2*e^5)*x^2 + (3*a^
2*b^3*d^2*e^3 + 2*a^3*b^2*d*e^4)*x)*log(b*x + a) - 60*(b^5*e^5*x^5 + a^3*b^2*d^2*e^3 + (2*b^5*d*e^4 + 3*a*b^4*
e^5)*x^4 + (b^5*d^2*e^3 + 6*a*b^4*d*e^4 + 3*a^2*b^3*e^5)*x^3 + (3*a*b^4*d^2*e^3 + 6*a^2*b^3*d*e^4 + a^3*b^2*e^
5)*x^2 + (3*a^2*b^3*d^2*e^3 + 2*a^3*b^2*d*e^4)*x)*log(e*x + d))/(a^3*b^6*d^8 - 6*a^4*b^5*d^7*e + 15*a^5*b^4*d^
6*e^2 - 20*a^6*b^3*d^5*e^3 + 15*a^7*b^2*d^4*e^4 - 6*a^8*b*d^3*e^5 + a^9*d^2*e^6 + (b^9*d^6*e^2 - 6*a*b^8*d^5*e
^3 + 15*a^2*b^7*d^4*e^4 - 20*a^3*b^6*d^3*e^5 + 15*a^4*b^5*d^2*e^6 - 6*a^5*b^4*d*e^7 + a^6*b^3*e^8)*x^5 + (2*b^
9*d^7*e - 9*a*b^8*d^6*e^2 + 12*a^2*b^7*d^5*e^3 + 5*a^3*b^6*d^4*e^4 - 30*a^4*b^5*d^3*e^5 + 33*a^5*b^4*d^2*e^6 -
16*a^6*b^3*d*e^7 + 3*a^7*b^2*e^8)*x^4 + (b^9*d^8 - 18*a^2*b^7*d^6*e^2 + 52*a^3*b^6*d^5*e^3 - 60*a^4*b^5*d^4*e
^4 + 24*a^5*b^4*d^3*e^5 + 10*a^6*b^3*d^2*e^6 - 12*a^7*b^2*d*e^7 + 3*a^8*b*e^8)*x^3 + (3*a*b^8*d^8 - 12*a^2*b^7
*d^7*e + 10*a^3*b^6*d^6*e^2 + 24*a^4*b^5*d^5*e^3 - 60*a^5*b^4*d^4*e^4 + 52*a^6*b^3*d^3*e^5 - 18*a^7*b^2*d^2*e^
6 + a^9*e^8)*x^2 + (3*a^2*b^7*d^8 - 16*a^3*b^6*d^7*e + 33*a^4*b^5*d^6*e^2 - 30*a^5*b^4*d^5*e^3 + 5*a^6*b^3*d^4
*e^4 + 12*a^7*b^2*d^3*e^5 - 9*a^8*b*d^2*e^6 + 2*a^9*d*e^7)*x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {b x + a}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {5}{2}} {\left (e x + d\right )}^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)/(e*x+d)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x, algorithm="giac")

[Out]

integrate((b*x + a)/((b^2*x^2 + 2*a*b*x + a^2)^(5/2)*(e*x + d)^3), x)

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maple [B]  time = 0.08, size = 753, normalized size = 2.33 \begin {gather*} -\frac {\left (60 b^{5} e^{5} x^{5} \ln \left (b x +a \right )-60 b^{5} e^{5} x^{5} \ln \left (e x +d \right )+180 a \,b^{4} e^{5} x^{4} \ln \left (b x +a \right )-180 a \,b^{4} e^{5} x^{4} \ln \left (e x +d \right )+120 b^{5} d \,e^{4} x^{4} \ln \left (b x +a \right )-120 b^{5} d \,e^{4} x^{4} \ln \left (e x +d \right )+180 a^{2} b^{3} e^{5} x^{3} \ln \left (b x +a \right )-180 a^{2} b^{3} e^{5} x^{3} \ln \left (e x +d \right )+360 a \,b^{4} d \,e^{4} x^{3} \ln \left (b x +a \right )-360 a \,b^{4} d \,e^{4} x^{3} \ln \left (e x +d \right )-60 a \,b^{4} e^{5} x^{4}+60 b^{5} d^{2} e^{3} x^{3} \ln \left (b x +a \right )-60 b^{5} d^{2} e^{3} x^{3} \ln \left (e x +d \right )+60 b^{5} d \,e^{4} x^{4}+60 a^{3} b^{2} e^{5} x^{2} \ln \left (b x +a \right )-60 a^{3} b^{2} e^{5} x^{2} \ln \left (e x +d \right )+360 a^{2} b^{3} d \,e^{4} x^{2} \ln \left (b x +a \right )-360 a^{2} b^{3} d \,e^{4} x^{2} \ln \left (e x +d \right )-150 a^{2} b^{3} e^{5} x^{3}+180 a \,b^{4} d^{2} e^{3} x^{2} \ln \left (b x +a \right )-180 a \,b^{4} d^{2} e^{3} x^{2} \ln \left (e x +d \right )+60 a \,b^{4} d \,e^{4} x^{3}+90 b^{5} d^{2} e^{3} x^{3}+120 a^{3} b^{2} d \,e^{4} x \ln \left (b x +a \right )-120 a^{3} b^{2} d \,e^{4} x \ln \left (e x +d \right )-110 a^{3} b^{2} e^{5} x^{2}+180 a^{2} b^{3} d^{2} e^{3} x \ln \left (b x +a \right )-180 a^{2} b^{3} d^{2} e^{3} x \ln \left (e x +d \right )-120 a^{2} b^{3} d \,e^{4} x^{2}+210 a \,b^{4} d^{2} e^{3} x^{2}+20 b^{5} d^{3} e^{2} x^{2}-15 a^{4} b \,e^{5} x +60 a^{3} b^{2} d^{2} e^{3} \ln \left (b x +a \right )-60 a^{3} b^{2} d^{2} e^{3} \ln \left (e x +d \right )-160 a^{3} b^{2} d \,e^{4} x +120 a^{2} b^{3} d^{2} e^{3} x +60 a \,b^{4} d^{3} e^{2} x -5 b^{5} d^{4} e x +3 a^{5} e^{5}-30 a^{4} b d \,e^{4}-20 a^{3} b^{2} d^{2} e^{3}+60 a^{2} b^{3} d^{3} e^{2}-15 a \,b^{4} d^{4} e +2 b^{5} d^{5}\right ) \left (b x +a \right )^{2}}{6 \left (e x +d \right )^{2} \left (a e -b d \right )^{6} \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)^3/(b^2*x^2+2*a*b*x+a^2)^(5/2),x)

[Out]

-1/6*(-30*a^4*b*d*e^4-20*a^3*b^2*d^2*e^3+60*a^2*b^3*d^3*e^2-15*a*b^4*d^4*e+180*a*b^4*d^2*e^3*x^2*ln(b*x+a)+120
*a^3*b^2*d*e^4*x*ln(b*x+a)+360*a^2*b^3*d*e^4*x^2*ln(b*x+a)+360*a*b^4*d*e^4*x^3*ln(b*x+a)+180*a^2*b^3*d^2*e^3*x
*ln(b*x+a)+3*a^5*e^5+90*x^3*b^5*d^2*e^3+20*b^5*d^3*e^2*x^2-110*a^3*b^2*e^5*x^2-150*a^2*b^3*e^5*x^3-60*a*b^4*e^
5*x^4+60*b^5*d*e^4*x^4-5*b^5*d^4*e*x-15*a^4*b*e^5*x+2*b^5*d^5-120*ln(e*x+d)*x*a^3*b^2*d*e^4-180*ln(e*x+d)*x*a^
2*b^3*d^2*e^3-360*ln(e*x+d)*x^2*a^2*b^3*d*e^4-180*ln(e*x+d)*x^2*a*b^4*d^2*e^3-360*ln(e*x+d)*x^3*a*b^4*d*e^4+12
0*a^2*b^3*d^2*e^3*x+60*a*b^4*d^3*e^2*x+60*ln(b*x+a)*x^5*b^5*e^5-60*ln(e*x+d)*x^5*b^5*e^5-180*ln(e*x+d)*x^3*a^2
*b^3*e^5-60*ln(e*x+d)*x^3*b^5*d^2*e^3-60*ln(e*x+d)*x^2*a^3*b^2*e^5-60*ln(e*x+d)*a^3*b^2*d^2*e^3+180*ln(b*x+a)*
x^4*a*b^4*e^5+120*ln(b*x+a)*x^4*b^5*d*e^4-180*ln(e*x+d)*x^4*a*b^4*e^5-120*ln(e*x+d)*x^4*b^5*d*e^4-120*a^2*b^3*
d*e^4*x^2-160*a^3*b^2*d*e^4*x+60*a*b^4*d*e^4*x^3+180*a^2*b^3*e^5*x^3*ln(b*x+a)+60*b^5*d^2*e^3*x^3*ln(b*x+a)+60
*a^3*b^2*e^5*x^2*ln(b*x+a)+60*a^3*b^2*d^2*e^3*ln(b*x+a)+210*a*b^4*d^2*e^3*x^2)*(b*x+a)^2/(e*x+d)^2/(a*e-b*d)^6
/((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)^3/(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?

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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 )}^3\,{\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)^3*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)),x)

[Out]

int((a + b*x)/((d + e*x)^3*(a^2 + b^2*x^2 + 2*a*b*x)^(5/2)), x)

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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)**3/(b**2*x**2+2*a*b*x+a**2)**(5/2),x)

[Out]

Timed out

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