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

Optimal. Leaf size=122 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x)}{e^3 (a+b x)}-\frac {b x \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}{e^2 (a+b x)}+\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e} \]

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 122, 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)}{e^2 (a+b x)}+\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2 \log (d+e x)}{e^3 (a+b x)}+\frac {(a+b x) \sqrt {a^2+2 a b x+b^2 x^2}}{2 e} \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),x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 61, normalized size = 0.50 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (b e x (4 a e-2 b d+b e x)+2 (b d-a e)^2 \log (d+e x)\right )}{2 e^3 (a+b x)} \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),x]

[Out]

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

________________________________________________________________________________________

IntegrateAlgebraic [F]  time = 1.10, 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} \, 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),x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.42, size = 62, normalized size = 0.51 \begin {gather*} \frac {b^{2} e^{2} x^{2} - 2 \, {\left (b^{2} d e - 2 \, a b e^{2}\right )} x + 2 \, {\left (b^{2} d^{2} - 2 \, a b d e + a^{2} e^{2}\right )} \log \left (e x + d\right )}{2 \, e^{3}} \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),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.16, size = 97, normalized size = 0.80 \begin {gather*} {\left (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 (-3\right )} \log \left ({\left | x e + d \right |}\right ) + \frac {1}{2} \, {\left (b^{2} x^{2} e \mathrm {sgn}\left (b x + a\right ) - 2 \, b^{2} d x \mathrm {sgn}\left (b x + a\right ) + 4 \, a b x e \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-2\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),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [C]  time = 0.08, size = 102, normalized size = 0.84 \begin {gather*} \frac {\left (b^{2} e^{2} x^{2}+2 a^{2} e^{2} \ln \left (b e x +b d \right )-4 a b d e \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 )-2 b^{2} d e x +3 a^{2} e^{2}-2 a b d e \right ) \mathrm {csgn}\left (b x +a \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),x)

[Out]

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

________________________________________________________________________________________

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),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?

________________________________________________________________________________________

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 )}{d+e\,x} \,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),x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 0.24, size = 44, normalized size = 0.36 \begin {gather*} \frac {b^{2} x^{2}}{2 e} + x \left (\frac {2 a b}{e} - \frac {b^{2} d}{e^{2}}\right ) + \frac {\left (a e - b d\right )^{2} \log {\left (d + e x \right )}}{e^{3}} \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),x)

[Out]

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

________________________________________________________________________________________