3.2031 \(\int \frac{a+b x}{(d+e x)^2 (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=169 \[ -\frac{b}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{e (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}-\frac{2 b e (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{2 b e (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

[Out]

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

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Rubi [A]  time = 0.111761, antiderivative size = 169, normalized size of antiderivative = 1., 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, 44} \[ -\frac{b}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac{e (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}-\frac{2 b e (a+b x) \log (a+b x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3}+\frac{2 b e (a+b x) \log (d+e x)}{\sqrt{a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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 44

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}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{a+b x}{(d+e x)^2 \left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{a+b x}{\left (a b+b^2 x\right )^3 (d+e x)^2} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \frac{1}{(a+b x)^2 (d+e x)^2} \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=\frac{\left (a b+b^2 x\right ) \int \left (\frac{b^2}{(b d-a e)^2 (a+b x)^2}-\frac{2 b^2 e}{(b d-a e)^3 (a+b x)}+\frac{e^2}{(b d-a e)^2 (d+e x)^2}+\frac{2 b e^2}{(b d-a e)^3 (d+e x)}\right ) \, dx}{b \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{b}{(b d-a e)^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{e (a+b x)}{(b d-a e)^2 (d+e x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{2 b e (a+b x) \log (a+b x)}{(b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{2 b e (a+b x) \log (d+e x)}{(b d-a e)^3 \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.0691426, size = 92, normalized size = 0.54 \[ \frac{-(b d-a e) (a e+b (d+2 e x))-2 b e (a+b x) (d+e x) \log (a+b x)+2 b e (a+b x) (d+e x) \log (d+e x)}{\sqrt{(a+b x)^2} (d+e x) (b d-a e)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.016, size = 182, normalized size = 1.1 \begin{align*} -{\frac{ \left ( 2\,\ln \left ( ex+d \right ){x}^{2}{b}^{2}{e}^{2}-2\,\ln \left ( bx+a \right ){x}^{2}{b}^{2}{e}^{2}+2\,\ln \left ( ex+d \right ) xab{e}^{2}+2\,\ln \left ( ex+d \right ) x{b}^{2}de-2\,\ln \left ( bx+a \right ) xab{e}^{2}-2\,\ln \left ( bx+a \right ) x{b}^{2}de+2\,\ln \left ( ex+d \right ) abde-2\,\ln \left ( bx+a \right ) abde+2\,xab{e}^{2}-2\,x{b}^{2}de+{a}^{2}{e}^{2}-{b}^{2}{d}^{2} \right ) \left ( bx+a \right ) ^{2}}{ \left ( ex+d \right ) \left ( ae-bd \right ) ^{3}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.80739, size = 486, normalized size = 2.88 \begin{align*} -\frac{b^{2} d^{2} - a^{2} e^{2} + 2 \,{\left (b^{2} d e - a b e^{2}\right )} x + 2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (b x + a\right ) - 2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )} \log \left (e x + d\right )}{a b^{3} d^{4} - 3 \, a^{2} b^{2} d^{3} e + 3 \, a^{3} b d^{2} e^{2} - a^{4} d e^{3} +{\left (b^{4} d^{3} e - 3 \, a b^{3} d^{2} e^{2} + 3 \, a^{2} b^{2} d e^{3} - a^{3} b e^{4}\right )} x^{2} +{\left (b^{4} d^{4} - 2 \, a b^{3} d^{3} e + 2 \, a^{3} b d e^{3} - a^{4} e^{4}\right )} x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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