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

Optimal. Leaf size=162 \[ -\frac {(d+e x)^3}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (d+e x)^2}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (b d-a e)^2 \log (a+b x)}{b^4 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e^2 x (a+b x) (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}} \]

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Rubi [A]  time = 0.09, antiderivative size = 162, 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 {3 e^2 x (a+b x) (b d-a e)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^3}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (d+e x)^2}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (b d-a e)^2 \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)^(3/2),x]

[Out]

(3*e^2*(b*d - a*e)*x*(a + b*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*e*(a + b*x)*(d + e*x)^2)/(2*b^2*Sqrt[
a^2 + 2*a*b*x + b^2*x^2]) - (d + e*x)^3/(b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + (3*e*(b*d - a*e)^2*(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 )^{3/2}} \, dx &=-\frac {(d+e x)^3}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(3 e) \int \frac {(d+e x)^2}{\sqrt {a^2+2 a b x+b^2 x^2}} \, dx}{b}\\ &=-\frac {(d+e x)^3}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (3 e \left (a b+b^2 x\right )\right ) \int \frac {(d+e x)^2}{a b+b^2 x} \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(d+e x)^3}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {\left (3 e \left (a b+b^2 x\right )\right ) \int \left (\frac {e (b d-a e)}{b^3}+\frac {(b d-a e)^2}{b^2 \left (a b+b^2 x\right )}+\frac {e (d+e x)}{b^2}\right ) \, dx}{b \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {3 e^2 (b d-a e) x (a+b x)}{b^3 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (a+b x) (d+e x)^2}{2 b^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(d+e x)^3}{b \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {3 e (b d-a e)^2 (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 = 120, normalized size = 0.74 \begin {gather*} \frac {2 a^3 e^3-2 a^2 b e^2 (3 d+2 e x)+3 a b^2 e \left (2 d^2+2 d e x-e^2 x^2\right )+6 e (a+b x) (b d-a e)^2 \log (a+b x)+b^3 \left (-2 d^3+6 d e^2 x^2+e^3 x^3\right )}{2 b^4 \sqrt {(a+b x)^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)^(3/2),x]

[Out]

(2*a^3*e^3 - 2*a^2*b*e^2*(3*d + 2*e*x) + 3*a*b^2*e*(2*d^2 + 2*d*e*x - e^2*x^2) + b^3*(-2*d^3 + 6*d*e^2*x^2 + e
^3*x^3) + 6*e*(b*d - a*e)^2*(a + b*x)*Log[a + b*x])/(2*b^4*Sqrt[(a + b*x)^2])

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IntegrateAlgebraic [B]  time = 1.91, size = 2469, normalized size = 15.24 \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)^(3/2),x]

[Out]

-((a*d^3)/(b*Sqrt[b^2]*x*(a + b*x) - b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])) + ((a^3*Sqrt[b^2]*e^3*x)/(2*b^4) +
(15*a^2*Sqrt[b^2]*e^3*x^2)/(2*b^3) + (5*a*(b^2)^(3/2)*e^3*x^3)/b^4 + (2*a^3*e^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/b^4 - (5*a^2*e^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b^3) - (5*a*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 +
(6*a^3*e^3*x*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^3 + (6*a^2*e^3*x^2*ArcTanh[(-(Sqr
t[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^2 - (6*a^2*Sqrt[b^2]*e^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcT
anh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^4)/((-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])) + ((6*a^2*d^2*e)/(b*Sqrt[b^2]) + (2*b*d^3*x)/Sqrt[b^2]
- (2*d^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b - (3*a*d^2*e*x*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]]
)/Sqrt[b^2] - (3*b*d^2*e*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] + (3*d^2*e*x*Sqr
t[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 - (3*a*d^2*e*x*Log[a - Sqr
t[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (3*b*d^2*e*x^2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]])/Sqrt[b^2] + (3*d^2*e*x*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)/((-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])) + ((-6*a^3*d*e^2)/(b^2)^(3/2) - (6*a*d^2*e*x)/Sqrt[b^2] + (6*a*d^2*e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2
+ (6*a*d^2*e*x*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b + 6*d^2*e*x^2*ArcTanh[(-(Sqrt[b
^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a] - (6*d^2*e*x*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])/Sqrt[b^2] + (6*a^2*b*d*e^2*x*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x +
b^2*x^2]])/(b^2)^(3/2) + (6*a*d*e^2*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (6*
a*d*e^2*x*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^2 + (6*a^2*b*
d*e^2*x*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) + (6*a*d*e^2*x^2*Log[a - Sqrt[b^2]*x
+ Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/Sqrt[b^2] - (6*a*d*e^2*x*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^2)/((-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])) + ((2*a^4*e^3)/(b^3*Sqrt[b^2]) + (3*a^2*d*e^2*x)/(b*Sqrt[b^2]) - (a^3*e^3*x)/(
4*(b^2)^(3/2)) - (9*a*d*e^2*x^2)/Sqrt[b^2] - (5*a^2*e^3*x^2)/(4*b*Sqrt[b^2]) - (6*b*d*e^2*x^3)/Sqrt[b^2] - (2*
a*e^3*x^3)/Sqrt[b^2] - (b*e^3*x^4)/Sqrt[b^2] - (6*a^2*d*e^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^3 + (3*a*d*e^2*x*
Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (a^2*e^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(4*b^3) + (6*d*e^2*x^2*Sqrt[a^2
+ 2*a*b*x + b^2*x^2])/b + (a*e^3*x^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/b^2 + (e^3*x^3*Sqrt[a^2 + 2*a*b*x + b^2*x
^2])/b - (12*a^2*d*e^2*x*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b^2 - (12*a*d*e^2*x^2*Ar
cTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/b + (12*a*d*e^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Arc
Tanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(b*Sqrt[b^2]) - (3*a^3*e^3*x*Log[-a - Sqrt[b^2]*x +
Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (3*a^2*e^3*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x
^2]])/(b*Sqrt[b^2]) + (3*a^2*e^3*x*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^3 - (3*a^3*e^3*x*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b^2)^(3/2) - (3*a^2*e^3*x^
2*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(b*Sqrt[b^2]) + (3*a^2*e^3*x*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^3)/((-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]))

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fricas [A]  time = 0.45, size = 173, normalized size = 1.07 \begin {gather*} \frac {b^{3} e^{3} x^{3} - 2 \, b^{3} d^{3} + 6 \, a b^{2} d^{2} e - 6 \, a^{2} b d e^{2} + 2 \, a^{3} e^{3} + 3 \, {\left (2 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 2 \, {\left (3 \, a b^{2} d e^{2} - 2 \, a^{2} b e^{3}\right )} x + 6 \, {\left (a b^{2} d^{2} e - 2 \, a^{2} b d e^{2} + a^{3} e^{3} + {\left (b^{3} d^{2} e - 2 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x\right )} \log \left (b x + a\right )}{2 \, {\left (b^{5} x + a 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)^(3/2),x, algorithm="fricas")

[Out]

1/2*(b^3*e^3*x^3 - 2*b^3*d^3 + 6*a*b^2*d^2*e - 6*a^2*b*d*e^2 + 2*a^3*e^3 + 3*(2*b^3*d*e^2 - a*b^2*e^3)*x^2 + 2
*(3*a*b^2*d*e^2 - 2*a^2*b*e^3)*x + 6*(a*b^2*d^2*e - 2*a^2*b*d*e^2 + a^3*e^3 + (b^3*d^2*e - 2*a*b^2*d*e^2 + a^2
*b*e^3)*x)*log(b*x + a))/(b^5*x + a*b^4)

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giac [A]  time = 0.34, size = 185, normalized size = 1.14 \begin {gather*} \frac {1}{2} \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} {\left (\frac {x e^{3}}{b^{3}} + \frac {6 \, b^{7} d e^{2} - 5 \, a b^{6} e^{3}}{b^{10}}\right )} - \frac {{\left (b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}\right )} \log \left ({\left | -3 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}^{2} a b - a^{3} b - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )}^{3} {\left | b \right |} - 3 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}}\right )} a^{2} {\left | b \right |} \right |}\right )}{b^{3} {\left | b \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)^(3/2),x, algorithm="giac")

[Out]

1/2*sqrt(b^2*x^2 + 2*a*b*x + a^2)*(x*e^3/b^3 + (6*b^7*d*e^2 - 5*a*b^6*e^3)/b^10) - (b^2*d^2*e - 2*a*b*d*e^2 +
a^2*e^3)*log(abs(-3*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2))^2*a*b - a^3*b - (x*abs(b) - sqrt(b^2*x^2 + 2*a*
b*x + a^2))^3*abs(b) - 3*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2))*a^2*abs(b)))/(b^3*abs(b))

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maple [A]  time = 0.07, size = 209, normalized size = 1.29 \begin {gather*} \frac {\left (b^{3} e^{3} x^{3}+6 a^{2} b \,e^{3} x \ln \left (b x +a \right )-12 a \,b^{2} d \,e^{2} x \ln \left (b x +a \right )-3 a \,b^{2} e^{3} x^{2}+6 b^{3} d^{2} e x \ln \left (b x +a \right )+6 b^{3} d \,e^{2} x^{2}+6 a^{3} e^{3} \ln \left (b x +a \right )-12 a^{2} b d \,e^{2} \ln \left (b x +a \right )-4 a^{2} b \,e^{3} x +6 a \,b^{2} d^{2} e \ln \left (b x +a \right )+6 a \,b^{2} d \,e^{2} x +2 a^{3} e^{3}-6 a^{2} b d \,e^{2}+6 a \,b^{2} d^{2} e -2 b^{3} d^{3}\right ) \left (b x +a \right )^{2}}{2 \left (\left (b x +a \right )^{2}\right )^{\frac {3}{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)^(3/2),x)

[Out]

1/2*(b^3*e^3*x^3+6*ln(b*x+a)*x*a^2*b*e^3-12*ln(b*x+a)*x*a*b^2*d*e^2+6*ln(b*x+a)*x*b^3*d^2*e-3*a*b^2*e^3*x^2+6*
b^3*d*e^2*x^2+6*a^3*e^3*ln(b*x+a)-12*a^2*b*d*e^2*ln(b*x+a)+6*a*b^2*d^2*e*ln(b*x+a)-4*a^2*b*e^3*x+6*a*b^2*d*e^2
*x+2*a^3*e^3-6*a^2*b*d*e^2+6*a*b^2*d^2*e-2*b^3*d^3)*(b*x+a)^2/b^4/((b*x+a)^2)^(3/2)

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maxima [B]  time = 0.63, size = 481, normalized size = 2.97 \begin {gather*} \frac {e^{3} x^{3}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b} - \frac {5 \, a e^{3} x^{2}}{2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {6 \, a^{2} e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} - \frac {5 \, a^{3} e^{3}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} + \frac {{\left (3 \, b d e^{2} + a e^{3}\right )} x^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} + \frac {12 \, a^{3} e^{3} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {a d^{3}}{2 \, b^{3} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {23 \, a^{4} e^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {3 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a \log \left (x + \frac {a}{b}\right )}{b^{4}} + \frac {3 \, {\left (b d^{2} e + a d e^{2}\right )} \log \left (x + \frac {a}{b}\right )}{b^{3}} + \frac {2 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a^{2}}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{4}} - \frac {b d^{3} + 3 \, a d^{2} e}{\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} b^{2}} - \frac {6 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a^{2} x}{b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {6 \, {\left (b d^{2} e + a d e^{2}\right )} a x}{b^{4} {\left (x + \frac {a}{b}\right )}^{2}} - \frac {11 \, {\left (3 \, b d e^{2} + a e^{3}\right )} a^{3}}{2 \, b^{6} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {9 \, {\left (b d^{2} e + a d e^{2}\right )} a^{2}}{2 \, b^{5} {\left (x + \frac {a}{b}\right )}^{2}} + \frac {{\left (b d^{3} + 3 \, a d^{2} e\right )} a}{2 \, b^{4} {\left (x + \frac {a}{b}\right )}^{2}} \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)^(3/2),x, algorithm="maxima")

[Out]

1/2*e^3*x^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b) - 5/2*a*e^3*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 6*a^2*e^3*
log(x + a/b)/b^4 - 5*a^3*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) + (3*b*d*e^2 + a*e^3)*x^2/(sqrt(b^2*x^2 + 2*a
*b*x + a^2)*b^2) + 12*a^3*e^3*x/(b^5*(x + a/b)^2) - 1/2*a*d^3/(b^3*(x + a/b)^2) + 23/2*a^4*e^3/(b^6*(x + a/b)^
2) - 3*(3*b*d*e^2 + a*e^3)*a*log(x + a/b)/b^4 + 3*(b*d^2*e + a*d*e^2)*log(x + a/b)/b^3 + 2*(3*b*d*e^2 + a*e^3)
*a^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - (b*d^3 + 3*a*d^2*e)/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) - 6*(3*b*d*
e^2 + a*e^3)*a^2*x/(b^5*(x + a/b)^2) + 6*(b*d^2*e + a*d*e^2)*a*x/(b^4*(x + a/b)^2) - 11/2*(3*b*d*e^2 + a*e^3)*
a^3/(b^6*(x + a/b)^2) + 9/2*(b*d^2*e + a*d*e^2)*a^2/(b^5*(x + a/b)^2) + 1/2*(b*d^3 + 3*a*d^2*e)*a/(b^4*(x + a/
b)^2)

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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 )}^{3/2}} \,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)^(3/2),x)

[Out]

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

[Out]

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

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