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

Optimal. Leaf size=154 \[ -\frac {(d+e x)^3}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {2 e^2 (b d-a e)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (b d-a e)^2}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.11, antiderivative size = 154, 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 = {768, 646, 43} \begin {gather*} -\frac {2 e^2 (b d-a e)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (b d-a e)^2}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^3}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {e^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}} \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]

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

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 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 {(d+e x)^3}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {e \int \frac {(d+e x)^2}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx}{b}\\ &=-\frac {(d+e x)^3}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {\left (b e \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^2}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^3}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac {\left (b e \left (a b+b^2 x\right )\right ) \int \left (\frac {(b d-a e)^2}{b^5 (a+b x)^3}+\frac {2 e (b d-a e)}{b^5 (a+b x)^2}+\frac {e^2}{b^5 (a+b x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^3}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}-\frac {2 e^2 (b d-a e)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {e (b d-a e)^2}{2 b^4 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {e^3 (a+b x) \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 91, normalized size = 0.59 \begin {gather*} \frac {6 e^3 (a+b x)^3 \log (a+b x)-(b d-a e) \left (11 a^2 e^2+a b e (5 d+27 e x)+b^2 \left (2 d^2+9 d e x+18 e^2 x^2\right )\right )}{6 b^4 \left ((a+b x)^2\right )^{3/2}} \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]

(-((b*d - a*e)*(11*a^2*e^2 + a*b*e*(5*d + 27*e*x) + b^2*(2*d^2 + 9*d*e*x + 18*e^2*x^2))) + 6*e^3*(a + b*x)^3*L
og[a + b*x])/(6*b^4*((a + b*x)^2)^(3/2))

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IntegrateAlgebraic [B]  time = 3.21, size = 2559, normalized size = 16.62 \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]

(2*Sqrt[b^2]*d*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(-6*a^3*d*e - 3*a*b^2*d^2*x + 6*a^2*b*d*e*x - 18*a^3*e^2*x - 6*a*
b^2*d*e*x^2 - 27*a^2*b*e^2*x^2 + 9*b^3*d*e*x^3 - 27*a*b^2*e^2*x^3) + 2*d*(2*a^3*b^2*d^2 - 6*a^4*b*d*e + 6*a^5*
e^2 + 3*a^2*b^3*d^2*x + 18*a^4*b*e^2*x + 3*a*b^4*d^2*x^2 + 45*a^3*b^2*e^2*x^2 - 3*a*b^4*d*e*x^3 + 54*a^2*b^3*e
^2*x^3 - 9*b^5*d*e*x^4 + 27*a*b^4*e^2*x^4))/(3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(4*a^2*b^6 + 8*a*b^7*x + 4*b^
8*x^2) + 3*Sqrt[b^2]*x^3*(-4*a^3*b^5 - 12*a^2*b^6*x - 12*a*b^7*x^2 - 4*b^8*x^3)) + ((-40*a^5*Sqrt[b^2]*e^3*x)/
b^4 - (136*a^4*Sqrt[b^2]*e^3*x^2)/b^3 - (752*a^3*(b^2)^(3/2)*e^3*x^3)/(3*b^4) - (240*a^2*Sqrt[b^2]*e^3*x^4)/b
- 96*a*Sqrt[b^2]*e^3*x^5 + (32*a^5*e^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b^4) + (88*a^4*e^3*x*Sqrt[a^2 + 2*a*b
*x + b^2*x^2])/(3*b^3) + (320*a^3*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b^2) + (144*a^2*e^3*x^3*Sqrt[a^2 +
2*a*b*x + b^2*x^2])/b + 96*a*e^3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2] + (32*a^3*e^3*x^3*ArcTanh[(-(Sqrt[b^2]*x)
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b + 96*a^2*e^3*x^4*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2
])/a] + 96*a*b*e^3*x^5*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] + 32*b^2*e^3*x^6*ArcTanh[(-
(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - (32*a^2*(b^2)^(3/2)*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*
ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^4 - (64*a*Sqrt[b^2]*e^3*x^4*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b - 32*Sqrt[b^2]*e^3*x^5*Sqrt[a^2 + 2*
a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/((-a - Sqrt[b^2]*x + Sqrt[a^2 +
2*a*b*x + b^2*x^2])^3*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^3) + ((32*a^6*e^3)/(3*b^3*Sqrt[b^2]) +
(16*a^2*b*d^3*x)/Sqrt[b^2] + (96*a^4*d*e^2*x)/(b*Sqrt[b^2]) + (40*a^5*e^3*x)/(b^2)^(3/2) + 16*a*Sqrt[b^2]*d^3
*x^2 + (240*a^3*d*e^2*x^2)/Sqrt[b^2] + (136*a^4*e^3*x^2)/(b*Sqrt[b^2]) + (32*b^3*d^3*x^3)/(3*Sqrt[b^2]) + (320
*a^2*b*d*e^2*x^3)/Sqrt[b^2] + (192*a^3*e^3*x^3)/Sqrt[b^2] + 240*a*Sqrt[b^2]*d*e^2*x^4 + (96*a^2*b*e^3*x^4)/Sqr
t[b^2] + (96*b^3*d*e^2*x^5)/Sqrt[b^2] - (32*a^2*d^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*b) - (32*a^4*d*e^2*Sqrt[
a^2 + 2*a*b*x + b^2*x^2])/b^3 - (16*a*d^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/3 - (64*a^3*d*e^2*x*Sqrt[a^2 + 2*a*
b*x + b^2*x^2])/b^2 - (40*a^4*e^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 - (32*b*d^3*x^2*Sqrt[a^2 + 2*a*b*x + b^
2*x^2])/3 - (176*a^2*d*e^2*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - (96*a^3*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2
])/b^2 - 144*a*d*e^2*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (96*a^2*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - 96
*b*d*e^2*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2] - (16*a^3*e^3*x^3*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x
^2]])/Sqrt[b^2] - (48*a^2*b*e^3*x^4*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - 48*a*Sq
rt[b^2]*e^3*x^5*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (16*b^3*e^3*x^6*Log[-a - Sqrt[b^2]*x +
Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (16*a^2*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*
x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/b + 32*a*e^3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt
[a^2 + 2*a*b*x + b^2*x^2]] + 16*b*e^3*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*
b*x + b^2*x^2]] - (16*a^3*e^3*x^3*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (48*a^2*b*
e^3*x^4*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - 48*a*Sqrt[b^2]*e^3*x^5*Log[a - Sqrt[
b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] - (16*b^3*e^3*x^6*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]]
)/Sqrt[b^2] + (16*a^2*e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2
]])/b + 32*a*e^3*x^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]] + 16*b
*e^3*x^5*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/((-a - Sqrt[b^2]*
x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^3*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2])^3)

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fricas [A]  time = 0.49, size = 176, normalized size = 1.14 \begin {gather*} -\frac {2 \, b^{3} d^{3} + 3 \, a b^{2} d^{2} e + 6 \, a^{2} b d e^{2} - 11 \, a^{3} e^{3} + 18 \, {\left (b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 9 \, {\left (b^{3} d^{2} e + 2 \, a b^{2} d e^{2} - 3 \, a^{2} b e^{3}\right )} x - 6 \, {\left (b^{3} e^{3} x^{3} + 3 \, a b^{2} e^{3} x^{2} + 3 \, a^{2} b e^{3} x + a^{3} e^{3}\right )} \log \left (b x + a\right )}{6 \, {\left (b^{7} x^{3} + 3 \, a b^{6} x^{2} + 3 \, a^{2} b^{5} x + a^{3} b^{4}\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^3*d^3 + 3*a*b^2*d^2*e + 6*a^2*b*d*e^2 - 11*a^3*e^3 + 18*(b^3*d*e^2 - a*b^2*e^3)*x^2 + 9*(b^3*d^2*e +
2*a*b^2*d*e^2 - 3*a^2*b*e^3)*x - 6*(b^3*e^3*x^3 + 3*a*b^2*e^3*x^2 + 3*a^2*b*e^3*x + a^3*e^3)*log(b*x + a))/(b
^7*x^3 + 3*a*b^6*x^2 + 3*a^2*b^5*x + a^3*b^4)

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

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maple [A]  time = 0.06, size = 179, normalized size = 1.16 \begin {gather*} \frac {\left (6 b^{3} e^{3} x^{3} \ln \left (b x +a \right )+18 a \,b^{2} e^{3} x^{2} \ln \left (b x +a \right )+18 a^{2} b \,e^{3} x \ln \left (b x +a \right )+18 a \,b^{2} e^{3} x^{2}-18 b^{3} d \,e^{2} x^{2}+6 a^{3} e^{3} \ln \left (b x +a \right )+27 a^{2} b \,e^{3} x -18 a \,b^{2} d \,e^{2} x -9 b^{3} d^{2} e x +11 a^{3} e^{3}-6 a^{2} b d \,e^{2}-3 a \,b^{2} d^{2} e -2 b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{6 \left (\left (b x +a \right )^{2}\right )^{\frac {5}{2}} b^{4}} \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*(6*b^3*e^3*x^3*ln(b*x+a)+18*a*b^2*e^3*x^2*ln(b*x+a)+18*a^2*b*e^3*x*ln(b*x+a)+18*a*b^2*e^3*x^2-18*b^3*d*e^2
*x^2+6*a^3*e^3*ln(b*x+a)+27*a^2*b*e^3*x-18*a*b^2*d*e^2*x-9*b^3*d^2*e*x+11*a^3*e^3-6*a^2*b*d*e^2-3*a*b^2*d^2*e-
2*b^3*d^3)*(b*x+a)^2/b^4/((b*x+a)^2)^(5/2)

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maxima [B]  time = 0.78, size = 533, normalized size = 3.46 \begin {gather*} \frac {1}{12} \, b e^{3} {\left (\frac {48 \, a b^{3} x^{3} + 108 \, a^{2} b^{2} x^{2} + 88 \, a^{3} b x + 25 \, a^{4}}{b^{9} x^{4} + 4 \, a b^{8} x^{3} + 6 \, a^{2} b^{7} x^{2} + 4 \, a^{3} b^{6} x + a^{4} b^{5}} + \frac {12 \, \log \left (b x + a\right )}{b^{5}}\right )} - \frac {1}{4} \, b d e^{2} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{12} \, a e^{3} {\left (\frac {12 \, x^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} + \frac {8 \, a^{2}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{4}} + \frac {6 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{3}} - \frac {3 \, a^{3}}{b^{8} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{12} \, b d^{3} {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{4} \, a d^{2} e {\left (\frac {4}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac {3}{2}} b^{2}} - \frac {3 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{4} \, b d^{2} e {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {1}{4} \, a d e^{2} {\left (\frac {6}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {8 \, a}{b^{6} {\left (x + \frac {a}{b}\right )}^{3}} + \frac {3 \, a^{2}}{b^{7} {\left (x + \frac {a}{b}\right )}^{4}}\right )} - \frac {a d^{3}}{4 \, b^{5} {\left (x + \frac {a}{b}\right )}^{4}} \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]

1/12*b*e^3*((48*a*b^3*x^3 + 108*a^2*b^2*x^2 + 88*a^3*b*x + 25*a^4)/(b^9*x^4 + 4*a*b^8*x^3 + 6*a^2*b^7*x^2 + 4*
a^3*b^6*x + a^4*b^5) + 12*log(b*x + a)/b^5) - 1/4*b*d*e^2*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^
2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) - 3*a^3/(b^8*(x + a/
b)^4)) - 1/12*a*e^3*(12*x^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) + 8*a^2/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^4
) + 6*a/(b^6*(x + a/b)^2) - 8*a^2/(b^7*(x + a/b)^3) - 3*a^3/(b^8*(x + a/b)^4)) - 1/12*b*d^3*(4/((b^2*x^2 + 2*a
*b*x + a^2)^(3/2)*b^2) - 3*a/(b^6*(x + a/b)^4)) - 1/4*a*d^2*e*(4/((b^2*x^2 + 2*a*b*x + a^2)^(3/2)*b^2) - 3*a/(
b^6*(x + a/b)^4)) - 1/4*b*d^2*e*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/(b^7*(x + a/b)^4)) - 1/4*
a*d*e^2*(6/(b^5*(x + a/b)^2) - 8*a/(b^6*(x + a/b)^3) + 3*a^2/(b^7*(x + a/b)^4)) - 1/4*a*d^3/(b^5*(x + a/b)^4)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\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]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b x\right ) \left (d + e x\right )^{3}}{\left (\left (a + b x\right )^{2}\right )^{\frac {5}{2}}}\, dx \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]

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

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