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

Optimal. Leaf size=120 \[ -\frac {1}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {e (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}+\frac {e (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

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Rubi [A]  time = 0.08, antiderivative size = 120, 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, 44} \begin {gather*} -\frac {1}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}-\frac {e (a+b x) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}+\frac {e (a+b x) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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

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Mathematica [A]  time = 0.04, size = 57, normalized size = 0.48 \begin {gather*} \frac {e (a+b x) \log (d+e x)-e (a+b x) \log (a+b x)+a e-b d}{\sqrt {(a+b x)^2} (b d-a e)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 2.47, size = 1732, normalized size = 14.43 \begin {gather*} \frac {b d \tanh ^{-1}\left (\frac {\sqrt {b^2} x}{a}-\frac {\sqrt {a^2+2 b x a+b^2 x^2}}{a}\right )}{a (b d-a e)^2}-\frac {b d \tanh ^{-1}\left (\frac {\sqrt {b^2} e x}{2 b d-a e}-\frac {e \sqrt {a^2+2 b x a+b^2 x^2}}{2 b d-a e}\right )}{a (b d-a e)^2}+\frac {\sqrt {b^2} d \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )}{2 a (b d-a e)^2}+\frac {\sqrt {b^2} d \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )}{2 a (b d-a e)^2}-\frac {\sqrt {b^2} d \log \left (2 b d-a e+\sqrt {b^2} e x-e \sqrt {a^2+2 b x a+b^2 x^2}\right )}{2 a (b d-a e)^2}-\frac {\sqrt {b^2} d \log \left (2 b d-a e-\sqrt {b^2} e x+e \sqrt {a^2+2 b x a+b^2 x^2}\right )}{2 a (b d-a e)^2}+\frac {-\frac {2 b^2 \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x^2}{a (a e-b d)}+\frac {2 b^2 \tanh ^{-1}\left (\frac {e \sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} e x}{2 b d-a e}\right ) x^2}{a (a e-b d)}+\frac {\left (b^2\right )^{3/2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x^2}{a b (a e-b d)}+\frac {\left (b^2\right )^{3/2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x^2}{a b (a e-b d)}-\frac {\left (b^2\right )^{3/2} \log \left (2 b d-a e+\sqrt {b^2} e x-e \sqrt {a^2+2 b x a+b^2 x^2}\right ) x^2}{a b (a e-b d)}-\frac {\left (b^2\right )^{3/2} \log \left (2 b d-a e-\sqrt {b^2} e x+e \sqrt {a^2+2 b x a+b^2 x^2}\right ) x^2}{a b (a e-b d)}+\frac {2 \sqrt {b^2} \sqrt {a^2+2 b x a+b^2 x^2} \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x}{a (a e-b d)}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} x}{a}\right ) x}{a e-b d}-\frac {2 \sqrt {b^2} \sqrt {a^2+2 b x a+b^2 x^2} \tanh ^{-1}\left (\frac {e \sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} e x}{2 b d-a e}\right ) x}{a (a e-b d)}+\frac {2 b \tanh ^{-1}\left (\frac {e \sqrt {a^2+2 b x a+b^2 x^2}-\sqrt {b^2} e x}{2 b d-a e}\right ) x}{a e-b d}-\frac {b \sqrt {a^2+2 b x a+b^2 x^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x}{a (a e-b d)}+\frac {\sqrt {b^2} \log \left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x}{a e-b d}-\frac {b \sqrt {a^2+2 b x a+b^2 x^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x}{a (a e-b d)}+\frac {\sqrt {b^2} \log \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) x}{a e-b d}+\frac {b \sqrt {a^2+2 b x a+b^2 x^2} \log \left (2 b d-a e+\sqrt {b^2} e x-e \sqrt {a^2+2 b x a+b^2 x^2}\right ) x}{a (a e-b d)}-\frac {\sqrt {b^2} \log \left (2 b d-a e+\sqrt {b^2} e x-e \sqrt {a^2+2 b x a+b^2 x^2}\right ) x}{a e-b d}+\frac {b \sqrt {a^2+2 b x a+b^2 x^2} \log \left (2 b d-a e-\sqrt {b^2} e x+e \sqrt {a^2+2 b x a+b^2 x^2}\right ) x}{a (a e-b d)}-\frac {\sqrt {b^2} \log \left (2 b d-a e-\sqrt {b^2} e x+e \sqrt {a^2+2 b x a+b^2 x^2}\right ) x}{a e-b d}-\frac {2 \sqrt {b^2} x}{a e-b d}+\frac {2 \sqrt {a^2+2 b x a+b^2 x^2}}{a e-b d}+\frac {2 a \sqrt {b^2}}{b (a e-b d)}}{\left (-a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right ) \left (a-\sqrt {b^2} x+\sqrt {a^2+2 b x a+b^2 x^2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*d*ArcTanh[(Sqrt[b^2]*x)/a - Sqrt[a^2 + 2*a*b*x + b^2*x^2]/a])/(a*(b*d - a*e)^2) - (b*d*ArcTanh[(Sqrt[b^2]*e
*x)/(2*b*d - a*e) - (e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(a*(b*d - a*e)^2) + (Sqrt[b^2]*d*Log[-a
- Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(2*a*(b*d - a*e)^2) + (Sqrt[b^2]*d*Log[a - Sqrt[b^2]*x + Sqrt[
a^2 + 2*a*b*x + b^2*x^2]])/(2*a*(b*d - a*e)^2) - (Sqrt[b^2]*d*Log[2*b*d - a*e + Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2
*a*b*x + b^2*x^2]])/(2*a*(b*d - a*e)^2) - (Sqrt[b^2]*d*Log[2*b*d - a*e - Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]])/(2*a*(b*d - a*e)^2) + ((2*a*Sqrt[b^2])/(b*(-(b*d) + a*e)) - (2*Sqrt[b^2]*x)/(-(b*d) + a*e) + (2*S
qrt[a^2 + 2*a*b*x + b^2*x^2])/(-(b*d) + a*e) - (2*b*x*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])
/a])/(-(b*d) + a*e) - (2*b^2*x^2*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])/(a*(-(b*d) + a*e
)) + (2*Sqrt[b^2]*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*x) + Sqrt[a^2 + 2*a*b*x + b^2*x^2])/a])
/(a*(-(b*d) + a*e)) + (2*b*x*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(-(b
*d) + a*e) + (2*b^2*x^2*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2])/(2*b*d - a*e)])/(a*(-(b*d
) + a*e)) - (2*Sqrt[b^2]*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*ArcTanh[(-(Sqrt[b^2]*e*x) + e*Sqrt[a^2 + 2*a*b*x + b^
2*x^2])/(2*b*d - a*e)])/(a*(-(b*d) + a*e)) + (Sqrt[b^2]*x*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]
])/(-(b*d) + a*e) + ((b^2)^(3/2)*x^2*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*b*(-(b*d) + a*e
)) - (b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(-(b*d) + a*
e)) + (Sqrt[b^2]*x*Log[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(-(b*d) + a*e) + ((b^2)^(3/2)*x^2*Log
[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*b*(-(b*d) + a*e)) - (b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*L
og[a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(-(b*d) + a*e)) - (Sqrt[b^2]*x*Log[2*b*d - a*e + Sqrt[
b^2]*e*x - e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(-(b*d) + a*e) - ((b^2)^(3/2)*x^2*Log[2*b*d - a*e + Sqrt[b^2]*e*x
- e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*b*(-(b*d) + a*e)) + (b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[2*b*d - a*e
+ Sqrt[b^2]*e*x - e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(-(b*d) + a*e)) - (Sqrt[b^2]*x*Log[2*b*d - a*e - Sqrt[
b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(-(b*d) + a*e) - ((b^2)^(3/2)*x^2*Log[2*b*d - a*e - Sqrt[b^2]*e*x
+ e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*b*(-(b*d) + a*e)) + (b*x*Sqrt[a^2 + 2*a*b*x + b^2*x^2]*Log[2*b*d - a*e
- Sqrt[b^2]*e*x + e*Sqrt[a^2 + 2*a*b*x + b^2*x^2]])/(a*(-(b*d) + a*e)))/((-a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b
*x + b^2*x^2])*(a - Sqrt[b^2]*x + Sqrt[a^2 + 2*a*b*x + b^2*x^2]))

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fricas [A]  time = 0.44, size = 93, normalized size = 0.78 \begin {gather*} -\frac {b d - a e + {\left (b e x + a e\right )} \log \left (b x + a\right ) - {\left (b e x + a e\right )} \log \left (e x + d\right )}{a b^{2} d^{2} - 2 \, a^{2} b d e + a^{3} e^{2} + {\left (b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.24, size = 203, normalized size = 1.69 \begin {gather*} -\frac {a e \log \left ({\left | b + \frac {a}{x} \right |}\right )}{a b^{2} d^{2} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) - 2 \, a^{2} b d e \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) + a^{3} e^{2} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right )} + \frac {d e \log \left ({\left | \frac {d}{x} + e \right |}\right )}{b^{2} d^{3} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) - 2 \, a b d^{2} e \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right ) + a^{2} d e^{2} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right )} + \frac {b^{2} d - a b e}{{\left (b d - a e\right )}^{2} a {\left (b + \frac {a}{x}\right )} \mathrm {sgn}\left (\frac {b}{x} + \frac {a}{x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 77, normalized size = 0.64 \begin {gather*} -\frac {\left (b e x \ln \left (b x +a \right )-b e x \ln \left (e x +d \right )+a e \ln \left (b x +a \right )-a e \ln \left (e x +d \right )-a e +b d \right ) \left (b x +a \right )^{2}}{\left (a e -b d \right )^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)/(e*x+d)/(b^2*x^2+2*a*b*x+a^2)^(3/2),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(((2*a*b)/e>0)', see `assume?`
for more details)Is ((2*a*b)/e    -(2*b^2*d)/e^2)    ^2    -(4*b^2       *((-(2*a*b*d)/e)        +(b^2*d^2)/e^
2+a^2))     /e^2 zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,x}{\left (d+e\,x\right )\,{\left (a^2+2\,a\,b\,x+b^2\,x^2\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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