3.2026 \(\int \frac{(a+b x) (d+e x)^3}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

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

[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])

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Rubi [A]  time = 0.0872255, antiderivative size = 162, 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} \[ \frac{3 e^2 x (a+b x) (b d-a e)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^3}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e (a+b x) (d+e x)^2}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e (a+b x) (b d-a e)^2 \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

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 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 )^{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*}

Mathematica [A]  time = 0.0631258, size = 120, normalized size = 0.74 \[ \frac{-2 a^2 b e^2 (3 d+2 e x)+2 a^3 e^3+3 a b^2 e \left (2 d^2+2 d e x-e^2 x^2\right )+6 e (a+b x) (b d-a e)^2 \log (a+b x)+b^3 \left (-2 d^3+6 d e^2 x^2+e^3 x^3\right )}{2 b^4 \sqrt{(a+b x)^2}} \]

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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Maple [A]  time = 0.013, size = 209, normalized size = 1.3 \begin{align*}{\frac{ \left ({x}^{3}{b}^{3}{e}^{3}+6\,\ln \left ( bx+a \right ) x{a}^{2}b{e}^{3}-12\,\ln \left ( bx+a \right ) xa{b}^{2}d{e}^{2}+6\,\ln \left ( bx+a \right ) x{b}^{3}{d}^{2}e-3\,{x}^{2}a{b}^{2}{e}^{3}+6\,{x}^{2}{b}^{3}d{e}^{2}+6\,\ln \left ( bx+a \right ){a}^{3}{e}^{3}-12\,\ln \left ( bx+a \right ){a}^{2}bd{e}^{2}+6\,\ln \left ( bx+a \right ) a{b}^{2}{d}^{2}e-4\,x{a}^{2}b{e}^{3}+6\,xa{b}^{2}d{e}^{2}+2\,{e}^{3}{a}^{3}-6\,d{e}^{2}{a}^{2}b+6\,a{d}^{2}e{b}^{2}-2\,{d}^{3}{b}^{3} \right ) \left ( bx+a \right ) ^{2}}{2\,{b}^{4}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{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)^(3/2),x)

[Out]

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

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Maxima [B]  time = 0.994657, size = 780, normalized size = 4.81 \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)^(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^2)^(3/2)*b) + 9*a^4*b*e^3/((b^2)^(7/2)*(x + a/b)^2) + 12*a^3*e^3*x/((b^2)^(5/2)*(x + a/b)^2)
- 5*a^3*e^3/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^4) - 1/2*a*d^3/((b^2)^(3/2)*(x + a/b)^2) + 5/2*a^4*e^3/((b^2)^(3/
2)*b^3*(x + a/b)^2) + (3*b*d*e^2 + a*e^3)*x^2/(sqrt(b^2*x^2 + 2*a*b*x + a^2)*b^2) + 3*(b*d^2*e + a*d*e^2)*log(
x + a/b)/(b^2)^(3/2) - 3*(3*b*d*e^2 + a*e^3)*a*log(x + a/b)/((b^2)^(3/2)*b) - 9/2*(3*b*d*e^2 + a*e^3)*a^3*b/((
b^2)^(7/2)*(x + a/b)^2) + 9/2*(b*d^2*e + a*d*e^2)*a^2*b^2/((b^2)^(7/2)*(x + a/b)^2) - 6*(3*b*d*e^2 + a*e^3)*a^
2*x/((b^2)^(5/2)*(x + a/b)^2) + 6*(b*d^2*e + a*d*e^2)*a*b*x/((b^2)^(5/2)*(x + a/b)^2) + 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) - (3*b*d*e^2
+ a*e^3)*a^3/((b^2)^(3/2)*b^3*(x + a/b)^2) + 1/2*(b*d^3 + 3*a*d^2*e)*a/((b^2)^(3/2)*b*(x + a/b)^2)

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Fricas [A]  time = 1.60269, size = 354, normalized size = 2.19 \begin{align*} \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{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)^(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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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{3}{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)**(3/2),x)

[Out]

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

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Giac [A]  time = 1.21603, size = 250, normalized size = 1.54 \begin{align*} \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{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)^(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))