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

Optimal. Leaf size=166 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 \log (d+e x)}{e^4 (a+b x)}+\frac {b x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x)}-\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^2}+\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e} \]

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Rubi [A]  time = 0.07, antiderivative size = 166, 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 {b x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x)}-\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{2 e^2}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3 \log (d+e x)}{e^4 (a+b x)}+\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{d+e x} \, dx &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \frac {\left (a b+b^2 x\right )^3}{d+e x} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\sqrt {a^2+2 a b x+b^2 x^2} \int \left (\frac {b^4 (b d-a e)^2}{e^3}-\frac {b^3 (b d-a e) \left (a b+b^2 x\right )}{e^2}+\frac {b^2 \left (a b+b^2 x\right )^2}{e}-\frac {b^3 (b d-a e)^3}{e^3 (d+e x)}\right ) \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {b (b d-a e)^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x)}-\frac {(b d-a e) (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^2}+\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e}-\frac {(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^4 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 92, normalized size = 0.55 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (b e x \left (18 a^2 e^2+9 a b e (e x-2 d)+b^2 \left (6 d^2-3 d e x+2 e^2 x^2\right )\right )-6 (b d-a e)^3 \log (d+e x)\right )}{6 e^4 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 1.68, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a^2+2 a b x+b^2 x^2\right )^{3/2}}{d+e x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.41, size = 115, normalized size = 0.69 \begin {gather*} \frac {2 \, b^{3} e^{3} x^{3} - 3 \, {\left (b^{3} d e^{2} - 3 \, a b^{2} e^{3}\right )} x^{2} + 6 \, {\left (b^{3} d^{2} e - 3 \, a b^{2} d e^{2} + 3 \, 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 (e x + d\right )}{6 \, e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.18, size = 173, normalized size = 1.04 \begin {gather*} -{\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 )} e^{\left (-4\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{6} \, {\left (2 \, b^{3} x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) - 3 \, b^{3} d x^{2} e \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{3} d^{2} x \mathrm {sgn}\left (b x + a\right ) + 9 \, a b^{2} x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 18 \, a b^{2} d x e \mathrm {sgn}\left (b x + a\right ) + 18 \, a^{2} b x e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 149, normalized size = 0.90 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (2 b^{3} e^{3} x^{3}+9 a \,b^{2} e^{3} x^{2}-3 b^{3} d \,e^{2} x^{2}+6 a^{3} e^{3} \ln \left (e x +d \right )-18 a^{2} b d \,e^{2} \ln \left (e x +d \right )+18 a^{2} b \,e^{3} x +18 a \,b^{2} d^{2} e \ln \left (e x +d \right )-18 a \,b^{2} d \,e^{2} x -6 b^{3} d^{3} \ln \left (e x +d \right )+6 b^{3} d^{2} e x \right )}{6 \left (b x +a \right )^{3} e^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*e-b*d>0)', see `assume?` for
more details)Is a*e-b*d zero or nonzero?

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (\left (a + b x\right )^{2}\right )^{\frac {3}{2}}}{d + e x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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