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

Optimal. Leaf size=132 \[ -\frac {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x) (d+e x)}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^3 (a+b x)}+\frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)} \]

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Rubi [A]  time = 0.08, antiderivative size = 132, 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 {\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}{e^3 (a+b x) (d+e x)}-\frac {2 b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e) \log (d+e x)}{e^3 (a+b x)}+\frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (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)^2,x]

[Out]

(b^2*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^2*(a + b*x)) - ((b*d - a*e)^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(e^3*(a
+ b*x)*(d + e*x)) - (2*b*(b*d - a*e)*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)^2} \, 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)^2} \, 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)^2} \, 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^2}{e^2}+\frac {(-b d+a e)^2}{e^2 (d+e x)^2}-\frac {2 b (b d-a e)}{e^2 (d+e x)}\right ) \, dx}{a b+b^2 x}\\ &=\frac {b^2 x \sqrt {a^2+2 a b x+b^2 x^2}}{e^2 (a+b x)}-\frac {(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}{e^3 (a+b x) (d+e x)}-\frac {2 b (b d-a e) \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 = 0.05, size = 87, normalized size = 0.66 \begin {gather*} \frac {\sqrt {(a+b x)^2} \left (-a^2 e^2-2 b (d+e x) (b d-a e) \log (d+e x)+2 a b d e+b^2 \left (-d^2+d e x+e^2 x^2\right )\right )}{e^3 (a+b x) (d+e 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)^2,x]

[Out]

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

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IntegrateAlgebraic [F]  time = 1.49, 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)^2} \, 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)^2,x]

[Out]

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

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fricas [A]  time = 0.42, size = 92, normalized size = 0.70 \begin {gather*} \frac {b^{2} e^{2} x^{2} + b^{2} d e x - b^{2} d^{2} + 2 \, a b d e - a^{2} e^{2} - 2 \, {\left (b^{2} d^{2} - a b d e + {\left (b^{2} d e - a b e^{2}\right )} x\right )} \log \left (e x + d\right )}{e^{4} x + d 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)^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.15, size = 101, normalized size = 0.77 \begin {gather*} b^{2} x e^{\left (-2\right )} \mathrm {sgn}\left (b x + a\right ) - 2 \, {\left (b^{2} d \mathrm {sgn}\left (b x + a\right ) - a b e \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-3\right )} \log \left ({\left | x e + d \right |}\right ) - \frac {{\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 )}}{x e + d} \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)^2,x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.07, size = 131, normalized size = 0.99 \begin {gather*} \frac {\left (2 a b \,e^{2} x \ln \left (b e x +b d \right )-2 b^{2} d e x \ln \left (b e x +b d \right )+b^{2} e^{2} x^{2}+2 a b d e \ln \left (b e x +b d \right )+a b \,e^{2} x -2 b^{2} d^{2} \ln \left (b e x +b d \right )+b^{2} d e x -a^{2} e^{2}+3 a b d e -b^{2} d^{2}\right ) \mathrm {csgn}\left (b x +a \right )}{\left (e x +d \right ) 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)^2,x)

[Out]

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

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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)^2,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 )}^2} \,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)^2,x)

[Out]

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

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sympy [A]  time = 0.37, size = 60, normalized size = 0.45 \begin {gather*} \frac {b^{2} x}{e^{2}} + \frac {2 b \left (a e - b d\right ) \log {\left (d + e x \right )}}{e^{3}} + \frac {- a^{2} e^{2} + 2 a b d e - b^{2} d^{2}}{d e^{3} + e^{4} x} \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)**2,x)

[Out]

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

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