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

Optimal. Leaf size=107 \[ \frac {(a+b x) (A b-a B) \log (a+b x)}{b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}+\frac {(a+b x) (B d-A e) \log (d+e x)}{e \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \]

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Rubi [A]  time = 0.08, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.061, Rules used = {770, 72} \begin {gather*} \frac {(a+b x) (A b-a B) \log (a+b x)}{b \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)}+\frac {(a+b x) (B d-A e) \log (d+e x)}{e \sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 72

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

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

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Mathematica [A]  time = 0.04, size = 66, normalized size = 0.62 \begin {gather*} \frac {(a+b x) (e (A b-a B) \log (a+b x)+b (B d-A e) \log (d+e x))}{b e \sqrt {(a+b x)^2} (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [B]  time = 0.81, size = 352, normalized size = 3.29 \begin {gather*} -\frac {\left (\sqrt {b^2}+b\right ) \left (A \sqrt {b^2}-a B\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}-a-\sqrt {b^2} x\right )}{2 b \sqrt {b^2} (b d-a e)}+\frac {\left (b-\sqrt {b^2}\right ) \left (a B+A \sqrt {b^2}\right ) \log \left (\sqrt {a^2+2 a b x+b^2 x^2}+a-\sqrt {b^2} x\right )}{2 b \sqrt {b^2} (b d-a e)}+\frac {\left (A \sqrt {b^2} e+A b e-\sqrt {b^2} B d-b B d\right ) \log \left (-e \sqrt {a^2+2 a b x+b^2 x^2}-a e+\sqrt {b^2} e x+2 b d\right )}{2 b e (b d-a e)}+\frac {\left (A \sqrt {b^2} e-A b e-\sqrt {b^2} B d+b B d\right ) \log \left (e \sqrt {a^2+2 a b x+b^2 x^2}-a e-\sqrt {b^2} e x+2 b d\right )}{2 b e (b d-a e)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.42, size = 53, normalized size = 0.50 \begin {gather*} -\frac {{\left (B a - A b\right )} e \log \left (b x + a\right ) - {\left (B b d - A b e\right )} \log \left (e x + d\right )}{b^{2} d e - a b e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 87, normalized size = 0.81 \begin {gather*} -\frac {{\left (B a \mathrm {sgn}\left (b x + a\right ) - A b \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{2} d - a b e} + \frac {{\left (B d \mathrm {sgn}\left (b x + a\right ) - A e \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x e + d \right |}\right )}{b d e - a e^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.06, size = 76, normalized size = 0.71 \begin {gather*} -\frac {\left (b x +a \right ) \left (A b e \ln \left (b x +a \right )-A b e \ln \left (e x +d \right )-B a e \ln \left (b x +a \right )+B b d \ln \left (e x +d \right )\right )}{\sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right ) b e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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*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(((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}{\sqrt {{\left (a+b\,x\right )}^2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [B]  time = 1.45, size = 226, normalized size = 2.11 \begin {gather*} - \frac {\left (- A e + B d\right ) \log {\left (x + \frac {- A a e - A b d + 2 B a d - \frac {a^{2} e \left (- A e + B d\right )}{a e - b d} + \frac {2 a b d \left (- A e + B d\right )}{a e - b d} - \frac {b^{2} d^{2} \left (- A e + B d\right )}{e \left (a e - b d\right )}}{- 2 A b e + B a e + B b d} \right )}}{e \left (a e - b d\right )} + \frac {\left (- A b + B a\right ) \log {\left (x + \frac {- A a e - A b d + 2 B a d + \frac {a^{2} e^{2} \left (- A b + B a\right )}{b \left (a e - b d\right )} - \frac {2 a d e \left (- A b + B a\right )}{a e - b d} + \frac {b d^{2} \left (- A b + B a\right )}{a e - b d}}{- 2 A b e + B a e + B b d} \right )}}{b \left (a e - b d\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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