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

Optimal. Leaf size=140 \[ \frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^3 (a+b x) (d+e x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^3 (a+b x) (d+e x)^2}+\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)} \]

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Rubi [A]  time = 0.08, antiderivative size = 140, 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 {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^3 (a+b x) (d+e x)}-\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{2 e^3 (a+b x) (d+e x)^2}+\frac {b^2 \sqrt {a^2+2 a b x+b^2 x^2} \log (d+e x)}{e^3 (a+b x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.03, size = 73, normalized size = 0.52 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left ((b d-a e) (a e+3 b d+4 b e x)+2 b^2 (d+e x)^2 \log (d+e x)\right )}{2 e^3 (a+b x) (d+e x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [F]  time = 2.13, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{(d+e x)^3} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [A]  time = 0.43, size = 100, normalized size = 0.71 \begin {gather*} \frac {3 \, b^{2} d^{2} - 2 \, a b d e - a^{2} e^{2} + 4 \, {\left (b^{2} d e - a b e^{2}\right )} x + 2 \, {\left (b^{2} e^{2} x^{2} + 2 \, b^{2} d e x + b^{2} d^{2}\right )} \log \left (e x + d\right )}{2 \, {\left (e^{5} x^{2} + 2 \, d e^{4} x + d^{2} e^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.16, size = 105, normalized size = 0.75 \begin {gather*} b^{2} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) \mathrm {sgn}\left (b x + a\right ) + \frac {{\left (4 \, {\left (b^{2} d \mathrm {sgn}\left (b x + a\right ) - a b e \mathrm {sgn}\left (b x + a\right )\right )} x + {\left (3 \, b^{2} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b d e \mathrm {sgn}\left (b x + a\right ) - a^{2} e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )}\right )} e^{\left (-2\right )}}{2 \, {\left (x e + d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b^2*e^(-3)*log(abs(x*e + d))*sgn(b*x + a) + 1/2*(4*(b^2*d*sgn(b*x + a) - a*b*e*sgn(b*x + a))*x + (3*b^2*d^2*sg
n(b*x + a) - 2*a*b*d*e*sgn(b*x + a) - a^2*e^2*sgn(b*x + a))*e^(-1))*e^(-2)/(x*e + d)^2

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maple [C]  time = 0.07, size = 112, normalized size = 0.80 \begin {gather*} \frac {\left (2 b^{2} e^{2} x^{2} \ln \left (b e x +b d \right )+4 b^{2} d e x \ln \left (b e x +b d \right )-4 a b \,e^{2} x +2 b^{2} d^{2} \ln \left (b e x +b d \right )+4 b^{2} d e x -a^{2} e^{2}-2 a b d e +3 b^{2} d^{2}\right ) \mathrm {csgn}\left (b x +a \right )}{2 \left (e x +d \right )^{2} e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*x+a)^2)^(1/2)/(e*x+d)^3,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 {\sqrt {{\left (a+b\,x\right )}^2}\,\left (a+b\,x\right )}{{\left (d+e\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 0.48, size = 80, normalized size = 0.57 \begin {gather*} \frac {b^{2} \log {\left (d + e x \right )}}{e^{3}} + \frac {- a^{2} e^{2} - 2 a b d e + 3 b^{2} d^{2} + x \left (- 4 a b e^{2} + 4 b^{2} d e\right )}{2 d^{2} e^{3} + 4 d e^{4} x + 2 e^{5} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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