3.1764 \(\int \frac {A+B x}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=148 \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (B d-A e)}{(d+e x) (b d-a e)^2}+\frac {(a+b x) (A b-a B) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {(a+b x) (A b-a B) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 148, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {769, 646, 36, 31} \[ \frac {\sqrt {a^2+2 a b x+b^2 x^2} (B d-A e)}{(d+e x) (b d-a e)^2}+\frac {(a+b x) (A b-a B) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2}-\frac {(a+b x) (A b-a B) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 646

Int[((d_.) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[(a + b*x + c*x^2)^Fra
cPart[p]/(c^IntPart[p]*(b/2 + c*x)^(2*FracPart[p])), Int[(d + e*x)^m*(b/2 + c*x)^(2*p), x], x] /; FreeQ[{a, b,
c, d, e, m, p}, x] && EqQ[b^2 - 4*a*c, 0] &&  !IntegerQ[p] && NeQ[2*c*d - b*e, 0]

Rule 769

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp
[(-2*c*(e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/((p + 1)*(2*c*d - b*e)^2), x] + Dist[(2*c*f -
b*g)/(2*c*d - b*e), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x]
&& EqQ[b^2 - 4*a*c, 0] && EqQ[m + 2*p + 3, 0] && NeQ[2*c*f - b*g, 0] && NeQ[2*c*d - b*e, 0]

Rubi steps

\begin {align*} \int \frac {A+B x}{(d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx &=\frac {(B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{(b d-a e)^2 (d+e x)}+\frac {(A b-a B) \int \frac {1}{(d+e x) \sqrt {a^2+2 a b x+b^2 x^2}} \, dx}{b d-a e}\\ &=\frac {(B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{(b d-a e)^2 (d+e x)}+\frac {\left ((A b-a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{\left (a b+b^2 x\right ) (d+e x)} \, dx}{(b d-a e) \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{(b d-a e)^2 (d+e x)}+\frac {\left (b (A b-a B) \left (a b+b^2 x\right )\right ) \int \frac {1}{a b+b^2 x} \, dx}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {\left ((A b-a B) e \left (a b+b^2 x\right )\right ) \int \frac {1}{d+e x} \, dx}{b (b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {(B d-A e) \sqrt {a^2+2 a b x+b^2 x^2}}{(b d-a e)^2 (d+e x)}+\frac {(A b-a B) (a+b x) \log (a+b x)}{(b d-a e)^2 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {(A b-a B) (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 = 0.12, size = 97, normalized size = 0.66 \[ \frac {(a+b x) \left (\frac {B d-A e}{e (d+e x) (a e-b d)}+\frac {(A b-a B) \log (a+b x)}{(b d-a e)^2}+\frac {(a B-A b) \log (d+e x)}{(b d-a e)^2}\right )}{\sqrt {(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.51, size = 148, normalized size = 1.00 \[ -\frac {B b d^{2} + A a e^{2} - {\left (B a + A b\right )} d e + {\left ({\left (B a - A b\right )} e^{2} x + {\left (B a - A b\right )} d e\right )} \log \left (b x + a\right ) - {\left ({\left (B a - A b\right )} e^{2} x + {\left (B a - A b\right )} d e\right )} \log \left (e x + d\right )}{b^{2} d^{3} e - 2 \, a b d^{2} e^{2} + a^{2} d e^{3} + {\left (b^{2} d^{2} e^{2} - 2 \, a b d e^{3} + a^{2} e^{4}\right )} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^2/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.17, size = 190, normalized size = 1.28 \[ -\frac {{\left (B a b \mathrm {sgn}\left (b x + a\right ) - A b^{2} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{3} d^{2} - 2 \, a b^{2} d e + a^{2} b e^{2}} + \frac {{\left (B a e \mathrm {sgn}\left (b x + a\right ) - A b e \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x e + d \right |}\right )}{b^{2} d^{2} e - 2 \, a b d e^{2} + a^{2} e^{3}} - \frac {{\left (B b d^{2} \mathrm {sgn}\left (b x + a\right ) - B a d e \mathrm {sgn}\left (b x + a\right ) - A b d e \mathrm {sgn}\left (b x + a\right ) + A a e^{2} \mathrm {sgn}\left (b x + a\right )\right )} e^{\left (-1\right )}}{{\left (b d - a e\right )}^{2} {\left (x e + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^2/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.06, size = 161, normalized size = 1.09 \[ \frac {\left (b x +a \right ) \left (A b \,e^{2} x \ln \left (b x +a \right )-A b \,e^{2} x \ln \left (e x +d \right )-B a \,e^{2} x \ln \left (b x +a \right )+B a \,e^{2} x \ln \left (e x +d \right )+A b d e \ln \left (b x +a \right )-A b d e \ln \left (e x +d \right )-B a d e \ln \left (b x +a \right )+B a d e \ln \left (e x +d \right )-A a \,e^{2}+A b d e +B a d e -B b \,d^{2}\right )}{\sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{2} \left (e x +d \right ) e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(b*x+a)*(A*ln(b*x+a)*x*b*e^2-A*ln(e*x+d)*x*b*e^2-B*ln(b*x+a)*x*a*e^2+B*ln(e*x+d)*x*a*e^2+A*ln(b*x+a)*b*d*e-A*l
n(e*x+d)*b*d*e-B*ln(b*x+a)*a*d*e+B*ln(e*x+d)*a*d*e-A*a*e^2+A*b*d*e+B*a*d*e-B*b*d^2)/((b*x+a)^2)^(1/2)/(a*e-b*d
)^2/e/(e*x+d)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^2/((b*x+a)^2)^(1/2),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 \[ \int \frac {A+B\,x}{\sqrt {{\left (a+b\,x\right )}^2}\,{\left (d+e\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 1.22, size = 355, normalized size = 2.40 \[ \frac {\left (- A b + B a\right ) \log {\left (x + \frac {- A a b e - A b^{2} d + B a^{2} e + B a b d - \frac {a^{3} e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} + \frac {3 a^{2} b d e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} - \frac {3 a b^{2} d^{2} e \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} + \frac {b^{3} d^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}}}{- 2 A b^{2} e + 2 B a b e} \right )}}{\left (a e - b d\right )^{2}} - \frac {\left (- A b + B a\right ) \log {\left (x + \frac {- A a b e - A b^{2} d + B a^{2} e + B a b d + \frac {a^{3} e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} - \frac {3 a^{2} b d e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} + \frac {3 a b^{2} d^{2} e \left (- A b + B a\right )}{\left (a e - b d\right )^{2}} - \frac {b^{3} d^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{2}}}{- 2 A b^{2} e + 2 B a b e} \right )}}{\left (a e - b d\right )^{2}} + \frac {- A e + B d}{a d e^{2} - b d^{2} e + x \left (a e^{3} - b d e^{2}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**2/((b*x+a)**2)**(1/2),x)

[Out]

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

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