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

Optimal. Leaf size=210 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4 \log (d+e x)}{e^5 (a+b x)}-\frac {b x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^4 (a+b x)}+\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^3}-\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^2}+\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e} \]

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Rubi [A]  time = 0.10, antiderivative size = 210, normalized size of antiderivative = 1.00, 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 = {770, 21, 43} \begin {gather*} -\frac {b x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}{e^4 (a+b x)}+\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^3}-\frac {(a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{3 e^2}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^4 \log (d+e x)}{e^5 (a+b x)}+\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*(b*d - a*e)^3*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^4*(a + b*x))) + ((b*d - a*e)^2*(a + b*x)*Sqrt[a^2 + 2*a
*b*x + b^2*x^2])/(2*e^3) - ((b*d - a*e)*(a + b*x)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(3*e^2) + ((a + b*x)^3*Sqrt
[a^2 + 2*a*b*x + b^2*x^2])/(4*e) + ((b*d - a*e)^4*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[d + e*x])/(e^5*(a + b*x))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, 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 770

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dis
t[(a + b*x + c*x^2)^FracPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(f + g*x)*(b/2 + c
*x)^(2*p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && EqQ[b^2 - 4*a*c, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x) \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 {(a+b x) \left (a b+b^2 x\right )^3}{d+e x} \, dx}{b^2 \left (a b+b^2 x\right )}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \frac {(a+b x)^4}{d+e x} \, dx}{a b+b^2 x}\\ &=\frac {\left (b \sqrt {a^2+2 a b x+b^2 x^2}\right ) \int \left (-\frac {b (b d-a e)^3}{e^4}+\frac {b (b d-a e)^2 (a+b x)}{e^3}-\frac {b (b d-a e) (a+b x)^2}{e^2}+\frac {b (a+b x)^3}{e}+\frac {(-b d+a e)^4}{e^4 (d+e x)}\right ) \, dx}{a b+b^2 x}\\ &=-\frac {b (b d-a e)^3 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^4 (a+b x)}+\frac {(b d-a e)^2 (a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e^3}-\frac {(b d-a e) (a+b x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{3 e^2}+\frac {(a+b x)^3 \sqrt {a^2+2 a b x+b^2 x^2}}{4 e}+\frac {(b d-a e)^4 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^5 (a+b x)}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 133, normalized size = 0.63 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (b e x \left (48 a^3 e^3+36 a^2 b e^2 (e x-2 d)+8 a b^2 e \left (6 d^2-3 d e x+2 e^2 x^2\right )+b^3 \left (-12 d^3+6 d^2 e x-4 d e^2 x^2+3 e^3 x^3\right )\right )+12 (b d-a e)^4 \log (d+e x)\right )}{12 e^5 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)*(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*(48*a^3*e^3 + 36*a^2*b*e^2*(-2*d + e*x) + 8*a*b^2*e*(6*d^2 - 3*d*e*x + 2*e^2*x^2) +
b^3*(-12*d^3 + 6*d^2*e*x - 4*d*e^2*x^2 + 3*e^3*x^3)) + 12*(b*d - a*e)^4*Log[d + e*x]))/(12*e^5*(a + b*x))

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IntegrateAlgebraic [F]  time = 1.88, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x) \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 + b*x)*(a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(d + e*x),x]

[Out]

Defer[IntegrateAlgebraic][((a + b*x)*(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 = 179, normalized size = 0.85 \begin {gather*} \frac {3 \, b^{4} e^{4} x^{4} - 4 \, {\left (b^{4} d e^{3} - 4 \, a b^{3} e^{4}\right )} x^{3} + 6 \, {\left (b^{4} d^{2} e^{2} - 4 \, a b^{3} d e^{3} + 6 \, a^{2} b^{2} e^{4}\right )} x^{2} - 12 \, {\left (b^{4} d^{3} e - 4 \, a b^{3} d^{2} e^{2} + 6 \, a^{2} b^{2} d e^{3} - 4 \, a^{3} b e^{4}\right )} x + 12 \, {\left (b^{4} d^{4} - 4 \, a b^{3} d^{3} e + 6 \, a^{2} b^{2} d^{2} e^{2} - 4 \, a^{3} b d e^{3} + a^{4} e^{4}\right )} \log \left (e x + d\right )}{12 \, e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^4*e^4*x^4 - 4*(b^4*d*e^3 - 4*a*b^3*e^4)*x^3 + 6*(b^4*d^2*e^2 - 4*a*b^3*d*e^3 + 6*a^2*b^2*e^4)*x^2 -
12*(b^4*d^3*e - 4*a*b^3*d^2*e^2 + 6*a^2*b^2*d*e^3 - 4*a^3*b*e^4)*x + 12*(b^4*d^4 - 4*a*b^3*d^3*e + 6*a^2*b^2*d
^2*e^2 - 4*a^3*b*d*e^3 + a^4*e^4)*log(e*x + d))/e^5

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giac [A]  time = 0.17, size = 266, normalized size = 1.27 \begin {gather*} {\left (b^{4} d^{4} \mathrm {sgn}\left (b x + a\right ) - 4 \, a b^{3} d^{3} e \mathrm {sgn}\left (b x + a\right ) + 6 \, a^{2} b^{2} d^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) - 4 \, a^{3} b d e^{3} \mathrm {sgn}\left (b x + a\right ) + a^{4} e^{4} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-5\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{12} \, {\left (3 \, b^{4} x^{4} e^{3} \mathrm {sgn}\left (b x + a\right ) - 4 \, b^{4} d x^{3} e^{2} \mathrm {sgn}\left (b x + a\right ) + 6 \, b^{4} d^{2} x^{2} e \mathrm {sgn}\left (b x + a\right ) - 12 \, b^{4} d^{3} x \mathrm {sgn}\left (b x + a\right ) + 16 \, a b^{3} x^{3} e^{3} \mathrm {sgn}\left (b x + a\right ) - 24 \, a b^{3} d x^{2} e^{2} \mathrm {sgn}\left (b x + a\right ) + 48 \, a b^{3} d^{2} x e \mathrm {sgn}\left (b x + a\right ) + 36 \, a^{2} b^{2} x^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) - 72 \, a^{2} b^{2} d x e^{2} \mathrm {sgn}\left (b x + a\right ) + 48 \, a^{3} b x e^{3} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b^4*d^4*sgn(b*x + a) - 4*a*b^3*d^3*e*sgn(b*x + a) + 6*a^2*b^2*d^2*e^2*sgn(b*x + a) - 4*a^3*b*d*e^3*sgn(b*x +
a) + a^4*e^4*sgn(b*x + a))*e^(-5)*log(abs(x*e + d)) + 1/12*(3*b^4*x^4*e^3*sgn(b*x + a) - 4*b^4*d*x^3*e^2*sgn(b
*x + a) + 6*b^4*d^2*x^2*e*sgn(b*x + a) - 12*b^4*d^3*x*sgn(b*x + a) + 16*a*b^3*x^3*e^3*sgn(b*x + a) - 24*a*b^3*
d*x^2*e^2*sgn(b*x + a) + 48*a*b^3*d^2*x*e*sgn(b*x + a) + 36*a^2*b^2*x^2*e^3*sgn(b*x + a) - 72*a^2*b^2*d*x*e^2*
sgn(b*x + a) + 48*a^3*b*x*e^3*sgn(b*x + a))*e^(-4)

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maple [A]  time = 0.06, size = 225, normalized size = 1.07 \begin {gather*} \frac {\left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}} \left (3 b^{4} e^{4} x^{4}+16 a \,b^{3} e^{4} x^{3}-4 b^{4} d \,e^{3} x^{3}+36 a^{2} b^{2} e^{4} x^{2}-24 a \,b^{3} d \,e^{3} x^{2}+6 b^{4} d^{2} e^{2} x^{2}+12 a^{4} e^{4} \ln \left (e x +d \right )-48 a^{3} b d \,e^{3} \ln \left (e x +d \right )+48 a^{3} b \,e^{4} x +72 a^{2} b^{2} d^{2} e^{2} \ln \left (e x +d \right )-72 a^{2} b^{2} d \,e^{3} x -48 a \,b^{3} d^{3} e \ln \left (e x +d \right )+48 a \,b^{3} d^{2} e^{2} x +12 b^{4} d^{4} \ln \left (e x +d \right )-12 b^{4} d^{3} e x \right )}{12 \left (b x +a \right )^{3} e^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*((b*x+a)^2)^(3/2)*(3*x^4*b^4*e^4+16*x^3*a*b^3*e^4-4*x^3*b^4*d*e^3+36*x^2*a^2*b^2*e^4-24*x^2*a*b^3*d*e^3+6
*x^2*b^4*d^2*e^2+12*ln(e*x+d)*a^4*e^4-48*ln(e*x+d)*a^3*b*d*e^3+72*ln(e*x+d)*a^2*b^2*d^2*e^2-48*ln(e*x+d)*a*b^3
*d^3*e+12*ln(e*x+d)*b^4*d^4+48*x*a^3*b*e^4-72*x*a^2*b^2*d*e^3+48*x*a*b^3*d^2*e^2-12*x*b^4*d^3*e)/(b*x+a)^3/e^5

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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*x+a)*(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 = -0.00 \begin {gather*} \int \frac {\left (a+b\,x\right )\,{\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 + b*x)*(a^2 + b^2*x^2 + 2*a*b*x)^(3/2))/(d + e*x),x)

[Out]

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

[Out]

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

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