3.1588 \(\int \frac{(d+e x)^3}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=173 \[ \frac{e x (a+b x) (b d-a e)^2}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (d+e x)^2 (b d-a e)}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (d+e x)^3}{3 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (b d-a e)^3 \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(e*(b*d - a*e)^2*x*(a + b*x))/(b^3*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((b*d - a*e)*(a + b*x)*(d + e*x)^2)/(2*b^2
*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((a + b*x)*(d + e*x)^3)/(3*b*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) + ((b*d - a*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 = 0.0619336, antiderivative size = 173, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {646, 43} \[ \frac{e x (a+b x) (b d-a e)^2}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (d+e x)^2 (b d-a e)}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (d+e x)^3}{3 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (b d-a e)^3 \log (a+b x)}{b^4 \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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{(d+e x)^3}{\sqrt{a^2+2 a b x+b^2 x^2}} \, dx &=\frac{\left (a b+b^2 x\right ) \int \frac{(d+e x)^3}{a b+b^2 x} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (\frac{e (b d-a e)^2}{b^4}+\frac{(b d-a e)^3}{b^3 \left (a b+b^2 x\right )}+\frac{e (b d-a e) (d+e x)}{b^3}+\frac{e (d+e x)^2}{b^2}\right ) \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{e (b d-a e)^2 x (a+b x)}{b^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(b d-a e) (a+b x) (d+e x)^2}{2 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(a+b x) (d+e x)^3}{3 b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{(b d-a 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 = 0.0526176, size = 90, normalized size = 0.52 \[ \frac{(a+b x) \left (b e x \left (6 a^2 e^2-3 a b e (6 d+e x)+b^2 \left (18 d^2+9 d e x+2 e^2 x^2\right )\right )+6 (b d-a e)^3 \log (a+b x)\right )}{6 b^4 \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)*(b*e*x*(6*a^2*e^2 - 3*a*b*e*(6*d + e*x) + b^2*(18*d^2 + 9*d*e*x + 2*e^2*x^2)) + 6*(b*d - a*e)^3*Log
[a + b*x]))/(6*b^4*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.154, size = 147, normalized size = 0.9 \begin{align*} -{\frac{ \left ( bx+a \right ) \left ( -2\,{x}^{3}{b}^{3}{e}^{3}+3\,{x}^{2}a{b}^{2}{e}^{3}-9\,{x}^{2}{b}^{3}d{e}^{2}+6\,\ln \left ( bx+a \right ){a}^{3}{e}^{3}-18\,\ln \left ( bx+a \right ){a}^{2}bd{e}^{2}+18\,\ln \left ( bx+a \right ) a{b}^{2}{d}^{2}e-6\,\ln \left ( bx+a \right ){b}^{3}{d}^{3}-6\,x{a}^{2}b{e}^{3}+18\,xa{b}^{2}d{e}^{2}-18\,x{b}^{3}{d}^{2}e \right ) }{6\,{b}^{4}}{\frac{1}{\sqrt{ \left ( bx+a \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^3/((b*x+a)^2)^(1/2),x)

[Out]

-1/6*(b*x+a)*(-2*x^3*b^3*e^3+3*x^2*a*b^2*e^3-9*x^2*b^3*d*e^2+6*ln(b*x+a)*a^3*e^3-18*ln(b*x+a)*a^2*b*d*e^2+18*l
n(b*x+a)*a*b^2*d^2*e-6*ln(b*x+a)*b^3*d^3-6*x*a^2*b*e^3+18*x*a*b^2*d*e^2-18*x*b^3*d^2*e)/((b*x+a)^2)^(1/2)/b^4

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Maxima [B]  time = 1.14639, size = 346, normalized size = 2. \begin{align*} \frac{3 \, a^{2} b^{2} d e^{2} \log \left (x + \frac{a}{b}\right )}{{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{5 \, a^{3} b e^{3} \log \left (x + \frac{a}{b}\right )}{3 \,{\left (b^{2}\right )}^{\frac{5}{2}}} - \frac{3 \, a b d e^{2} x}{{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{5 \, a^{2} e^{3} x}{3 \,{\left (b^{2}\right )}^{\frac{3}{2}}} + \frac{3 \, d e^{2} x^{2}}{2 \, \sqrt{b^{2}}} - \frac{5 \, a e^{3} x^{2}}{6 \, \sqrt{b^{2}} b} + \sqrt{\frac{1}{b^{2}}} d^{3} \log \left (x + \frac{a}{b}\right ) - \frac{3 \, a \sqrt{\frac{1}{b^{2}}} d^{2} e \log \left (x + \frac{a}{b}\right )}{b} + \frac{2 \, a^{3} \sqrt{\frac{1}{b^{2}}} e^{3} \log \left (x + \frac{a}{b}\right )}{3 \, b^{3}} + \frac{\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} e^{3} x^{2}}{3 \, b^{2}} + \frac{3 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} d^{2} e}{b^{2}} - \frac{2 \, \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2}} a^{2} e^{3}}{3 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.55755, size = 238, normalized size = 1.38 \begin{align*} \frac{2 \, b^{3} e^{3} x^{3} + 3 \,{\left (3 \, b^{3} d e^{2} - a b^{2} e^{3}\right )} x^{2} + 6 \,{\left (3 \, b^{3} d^{2} e - 3 \, a b^{2} d e^{2} + a^{2} b e^{3}\right )} x + 6 \,{\left (b^{3} d^{3} - 3 \, a b^{2} d^{2} e + 3 \, a^{2} b d e^{2} - a^{3} e^{3}\right )} \log \left (b x + a\right )}{6 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.549075, size = 82, normalized size = 0.47 \begin{align*} \frac{e^{3} x^{3}}{3 b} - \frac{x^{2} \left (a e^{3} - 3 b d e^{2}\right )}{2 b^{2}} + \frac{x \left (a^{2} e^{3} - 3 a b d e^{2} + 3 b^{2} d^{2} e\right )}{b^{3}} - \frac{\left (a e - b d\right )^{3} \log{\left (a + b x \right )}}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**3/((b*x+a)**2)**(1/2),x)

[Out]

e**3*x**3/(3*b) - x**2*(a*e**3 - 3*b*d*e**2)/(2*b**2) + x*(a**2*e**3 - 3*a*b*d*e**2 + 3*b**2*d**2*e)/b**3 - (a
*e - b*d)**3*log(a + b*x)/b**4

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Giac [A]  time = 1.15909, size = 230, normalized size = 1.33 \begin{align*} \frac{2 \, b^{2} x^{3} e^{3} \mathrm{sgn}\left (b x + a\right ) + 9 \, b^{2} d x^{2} e^{2} \mathrm{sgn}\left (b x + a\right ) + 18 \, b^{2} d^{2} x e \mathrm{sgn}\left (b x + a\right ) - 3 \, a b x^{2} e^{3} \mathrm{sgn}\left (b x + a\right ) - 18 \, a b d x e^{2} \mathrm{sgn}\left (b x + a\right ) + 6 \, a^{2} x e^{3} \mathrm{sgn}\left (b x + a\right )}{6 \, b^{3}} + \frac{{\left (b^{3} d^{3} \mathrm{sgn}\left (b x + a\right ) - 3 \, a b^{2} d^{2} e \mathrm{sgn}\left (b x + a\right ) + 3 \, a^{2} b d e^{2} \mathrm{sgn}\left (b x + a\right ) - a^{3} e^{3} \mathrm{sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

1/6*(2*b^2*x^3*e^3*sgn(b*x + a) + 9*b^2*d*x^2*e^2*sgn(b*x + a) + 18*b^2*d^2*x*e*sgn(b*x + a) - 3*a*b*x^2*e^3*s
gn(b*x + a) - 18*a*b*d*x*e^2*sgn(b*x + a) + 6*a^2*x*e^3*sgn(b*x + a))/b^3 + (b^3*d^3*sgn(b*x + a) - 3*a*b^2*d^
2*e*sgn(b*x + a) + 3*a^2*b*d*e^2*sgn(b*x + a) - a^3*e^3*sgn(b*x + a))*log(abs(b*x + a))/b^4