3.2035 \(\int \frac{(a+b x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{5/2}} \, dx\)

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

[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])

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Rubi [A]  time = 0.10893, antiderivative size = 154, normalized size of antiderivative = 1., 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} \[ -\frac{2 e^2 (b d-a e)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e (b d-a e)^2}{2 b^4 (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^3}{3 b \left (a^2+2 a b x+b^2 x^2\right )^{3/2}}+\frac{e^3 (a+b x) \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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 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]

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 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])

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*}

Mathematica [A]  time = 0.0506494, size = 91, normalized size = 0.59 \[ \frac{6 e^3 (a+b x)^3 \log (a+b x)-(b d-a e) \left (11 a^2 e^2+a b e (5 d+27 e x)+b^2 \left (2 d^2+9 d e x+18 e^2 x^2\right )\right )}{6 b^4 \left ((a+b x)^2\right )^{3/2}} \]

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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Maple [A]  time = 0.011, size = 179, normalized size = 1.2 \begin{align*}{\frac{ \left ( 6\,\ln \left ( bx+a \right ){x}^{3}{b}^{3}{e}^{3}+18\,\ln \left ( bx+a \right ){x}^{2}a{b}^{2}{e}^{3}+18\,\ln \left ( bx+a \right ) x{a}^{2}b{e}^{3}+18\,{x}^{2}a{b}^{2}{e}^{3}-18\,{x}^{2}{b}^{3}d{e}^{2}+6\,\ln \left ( bx+a \right ){a}^{3}{e}^{3}+27\,x{a}^{2}b{e}^{3}-18\,xa{b}^{2}d{e}^{2}-9\,x{b}^{3}{d}^{2}e+11\,{e}^{3}{a}^{3}-6\,d{e}^{2}{a}^{2}b-3\,a{d}^{2}e{b}^{2}-2\,{d}^{3}{b}^{3} \right ) \left ( bx+a \right ) ^{2}}{6\,{b}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{5}{2}}}} \end{align*}

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

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Maxima [B]  time = 1.10006, size = 849, normalized size = 5.51 \begin{align*} \text{result too large to display} \end{align*}

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) + 3*a^3*b/((b^2)^(9/2)*(x + a/b)^4) - 8*a^2/((b^2)^(7/2)*(x + a/b)^3)
+ 6*a/((b^2)^(5/2)*b*(x + a/b)^2) - 6*a^3/((b^2)^(5/2)*b^3*(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) + 3*a^3*b/((b^2)^(9/2)*(x + a/b)^4) - 8*a^
2/((b^2)^(7/2)*(x + a/b)^3) + 6*a/((b^2)^(5/2)*b*(x + a/b)^2) - 6*a^3/((b^2)^(5/2)*b^3*(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^2)^(5/2)*b*(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^2)^(5/2)*b*(x + a/b)^4)) - 1/4*b*d^2*e*(3*a^2*b^2/((b^2)^(9/2)*(x + a/b)^4)
- 8*a*b/((b^2)^(7/2)*(x + a/b)^3) + 6/((b^2)^(5/2)*(x + a/b)^2)) - 1/4*a*d*e^2*(3*a^2*b^2/((b^2)^(9/2)*(x + a
/b)^4) - 8*a*b/((b^2)^(7/2)*(x + a/b)^3) + 6/((b^2)^(5/2)*(x + a/b)^2)) - 1/4*a*d^3/((b^2)^(5/2)*(x + a/b)^4)

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Fricas [A]  time = 1.52805, size = 360, normalized size = 2.34 \begin{align*} -\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{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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{align*}

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)