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

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

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Rubi [A]  time = 0.06, antiderivative size = 173, normalized size of antiderivative = 1.00, 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} \begin {gather*} \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}} \end {gather*}

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 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]

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*}

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Mathematica [A]  time = 0.05, size = 90, normalized size = 0.52 \begin {gather*} \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}} \end {gather*}

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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IntegrateAlgebraic [B]  time = 0.84, size = 450, normalized size = 2.60 \begin {gather*} \frac {-6 a^2 e^3 x+18 a b d e^2 x+3 a b e^3 x^2-18 b^2 d^2 e x-9 b^2 d e^2 x^2-2 b^2 e^3 x^3}{12 \left (b^2\right )^{3/2}}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} \left (11 a^2 e^3-27 a b d e^2-5 a b e^3 x+18 b^2 d^2 e+9 b^2 d e^2 x+2 b^2 e^3 x^2\right )}{12 b^4}+\frac {\left (a^3 \sqrt {b^2} e^3+a^3 b e^3-3 a^2 b^2 d e^2-3 a^2 \sqrt {b^2} b d e^2+3 a b^3 d^2 e+3 a \left (b^2\right )^{3/2} d^2 e+b^4 \left (-d^3\right )-\sqrt {b^2} b^3 d^3\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )}{2 b^4 \sqrt {b^2}}+\frac {\left (-a^3 \sqrt {b^2} e^3+a^3 b e^3-3 a^2 b^2 d e^2+3 a^2 \sqrt {b^2} b d e^2+3 a b^3 d^2 e-3 a \left (b^2\right )^{3/2} d^2 e+b^4 \left (-d^3\right )+\sqrt {b^2} b^3 d^3\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}{2 b^4 \sqrt {b^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a^2 + 2*a*b*x + b^2*x^2]*(18*b^2*d^2*e - 27*a*b*d*e^2 + 11*a^2*e^3 + 9*b^2*d*e^2*x - 5*a*b*e^3*x + 2*b^2
*e^3*x^2))/(12*b^4) + (-18*b^2*d^2*e*x + 18*a*b*d*e^2*x - 6*a^2*e^3*x - 9*b^2*d*e^2*x^2 + 3*a*b*e^3*x^2 - 2*b^
2*e^3*x^3)/(12*(b^2)^(3/2)) + ((-(b^4*d^3) - b^3*Sqrt[b^2]*d^3 + 3*a*b^3*d^2*e + 3*a*(b^2)^(3/2)*d^2*e - 3*a^2
*b^2*d*e^2 - 3*a^2*b*Sqrt[b^2]*d*e^2 + a^3*b*e^3 + a^3*Sqrt[b^2]*e^3)*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*
x + b^2*x^2]])/(2*b^4*Sqrt[b^2]) + ((-(b^4*d^3) + b^3*Sqrt[b^2]*d^3 + 3*a*b^3*d^2*e - 3*a*(b^2)^(3/2)*d^2*e -
3*a^2*b^2*d*e^2 + 3*a^2*b*Sqrt[b^2]*d*e^2 + a^3*b*e^3 - a^3*Sqrt[b^2]*e^3)*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*
a*b*x + b^2*x^2]])/(2*b^4*Sqrt[b^2])

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fricas [A]  time = 0.40, size = 116, normalized size = 0.67 \begin {gather*} \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 {gather*}

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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giac [A]  time = 0.16, size = 170, normalized size = 0.98 \begin {gather*} \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 {gather*}

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

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maple [A]  time = 0.05, size = 147, normalized size = 0.85 \begin {gather*} -\frac {\left (b x +a \right ) \left (-2 b^{3} e^{3} x^{3}+3 a \,b^{2} e^{3} x^{2}-9 b^{3} d \,e^{2} x^{2}+6 a^{3} e^{3} \ln \left (b x +a \right )-18 a^{2} b d \,e^{2} \ln \left (b x +a \right )-6 a^{2} b \,e^{3} x +18 a \,b^{2} d^{2} e \ln \left (b x +a \right )+18 a \,b^{2} d \,e^{2} x -6 b^{3} d^{3} \ln \left (b x +a \right )-18 b^{3} d^{2} e x \right )}{6 \sqrt {\left (b x +a \right )^{2}}\, b^{4}} \end {gather*}

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

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maxima [A]  time = 1.11, size = 205, normalized size = 1.18 \begin {gather*} \frac {3 \, d e^{2} x^{2}}{2 \, b} - \frac {5 \, a e^{3} x^{2}}{6 \, b^{2}} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2}} e^{3} x^{2}}{3 \, b^{2}} - \frac {3 \, a d e^{2} x}{b^{2}} + \frac {5 \, a^{2} e^{3} x}{3 \, b^{3}} + \frac {d^{3} \log \left (x + \frac {a}{b}\right )}{b} - \frac {3 \, a d^{2} e \log \left (x + \frac {a}{b}\right )}{b^{2}} + \frac {3 \, a^{2} d e^{2} \log \left (x + \frac {a}{b}\right )}{b^{3}} - \frac {a^{3} e^{3} \log \left (x + \frac {a}{b}\right )}{b^{4}} + \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 {gather*}

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]

3/2*d*e^2*x^2/b - 5/6*a*e^3*x^2/b^2 + 1/3*sqrt(b^2*x^2 + 2*a*b*x + a^2)*e^3*x^2/b^2 - 3*a*d*e^2*x/b^2 + 5/3*a^
2*e^3*x/b^3 + d^3*log(x + a/b)/b - 3*a*d^2*e*log(x + a/b)/b^2 + 3*a^2*d*e^2*log(x + a/b)/b^3 - a^3*e^3*log(x +
a/b)/b^4 + 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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^3}{\sqrt {{\left (a+b\,x\right )}^2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.33, size = 83, normalized size = 0.48 \begin {gather*} x^{2} \left (- \frac {a e^{3}}{2 b^{2}} + \frac {3 d e^{2}}{2 b}\right ) + x \left (\frac {a^{2} e^{3}}{b^{3}} - \frac {3 a d e^{2}}{b^{2}} + \frac {3 d^{2} e}{b}\right ) + \frac {e^{3} x^{3}}{3 b} - \frac {\left (a e - b d\right )^{3} \log {\left (a + b x \right )}}{b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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