3.21.30 \(\int \frac {a+b x}{(d+e x) (a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\) [2030]

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

[Out]

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

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Rubi [A]
time = 0.06, 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 = {784, 21, 46} \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 46

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] && Lt
Q[m + n + 2, 0])

Rule 784

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 = 0.02, size = 57, normalized size = 0.48 \begin {gather*} \frac {-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}} \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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Maple [A]
time = 0.09, size = 77, normalized size = 0.64

method	result	size
default	\(-\frac {\left (\ln \left (b x +a \right ) b e x -\ln \left (e x +d \right ) b e x +\ln \left (b x +a \right ) a e -\ln \left (e x +d \right ) a e -a e +b d \right ) \left (b x +a \right )^{2}}{\left (a e -b d \right )^{2} \left (\left (b x +a \right )^{2}\right )^{\frac {3}{2}}}\)	\(77\)
risch	\(\frac {\sqrt {\left (b x +a \right )^{2}}}{\left (b x +a \right )^{2} \left (a e -b d \right )}-\frac {\sqrt {\left (b x +a \right )^{2}}\, e \ln \left (b x +a \right )}{\left (b x +a \right ) \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}+\frac {\sqrt {\left (b x +a \right )^{2}}\, e \ln \left (-e x -d \right )}{\left (b x +a \right ) \left (a^{2} e^{2}-2 a b d e +b^{2} d^{2}\right )}\)	\(127\)

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,method=_RETURNVERBOSE)

[Out]

-(ln(b*x+a)*b*e*x-ln(e*x+d)*b*e*x+ln(b*x+a)*a*e-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*%e^-1*a*b-2*%e^-2*b^2*d)^2>
0)', see `as

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Fricas [A]
time = 2.11, size = 91, normalized size = 0.76 \begin {gather*} -\frac {{\left (b x + a\right )} e \log \left (b x + a\right ) - {\left (b x + a\right )} e \log \left (x e + d\right ) + b d - a e}{b^{3} d^{2} x + a b^{2} d^{2} + {\left (a^{2} b x + a^{3}\right )} e^{2} - 2 \, {\left (a b^{2} d x + a^{2} b d\right )} e} \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*x + a)*e*log(b*x + a) - (b*x + a)*e*log(x*e + d) + b*d - a*e)/(b^3*d^2*x + a*b^2*d^2 + (a^2*b*x + a^3)*e^
2 - 2*(a*b^2*d*x + a^2*b*d)*e)

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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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Giac [A]
time = 1.43, size = 139, normalized size = 1.16 \begin {gather*} -\frac {b e \log \left ({\left | b x + a \right |}\right )}{b^{3} d^{2} \mathrm {sgn}\left (b x + a\right ) - 2 \, a b^{2} d e \mathrm {sgn}\left (b x + a\right ) + a^{2} b e^{2} \mathrm {sgn}\left (b x + a\right )} + \frac {e^{2} \log \left ({\left | x e + d \right |}\right )}{b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - 2 \, a b d e^{2} \mathrm {sgn}\left (b x + a\right ) + a^{2} e^{3} \mathrm {sgn}\left (b x + a\right )} - \frac {1}{{\left (b d - a e\right )} {\left (b x + a\right )} \mathrm {sgn}\left (b x + a\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]

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

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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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