3.16.45 \(\int \frac {A+B x}{(d+e x)^3 \sqrt {a^2+2 a b x+b^2 x^2}} \, dx\)

Optimal. Leaf size=212 \[ \frac {(a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}-\frac {(a+b x) (B d-A e)}{2 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}+\frac {b (a+b x) (A b-a B) \log (a+b x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3}-\frac {b (a+b x) (A b-a B) \log (d+e x)}{\sqrt {a^2+2 a b x+b^2 x^2} (b d-a e)^3} \]

________________________________________________________________________________________

Rubi [A]  time = 0.15, antiderivative size = 212, 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, 77} \begin {gather*} \frac {(a+b x) (A b-a B)}{\sqrt {a^2+2 a b x+b^2 x^2} (d+e x) (b d-a e)^2}-\frac {(a+b x) (B d-A e)}{2 e \sqrt {a^2+2 a b x+b^2 x^2} (d+e x)^2 (b d-a e)}+\frac {b (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)^3}-\frac {b (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)^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

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)^3 \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)^3} \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=\frac {\left (a b+b^2 x\right ) \int \left (\frac {b (A b-a B)}{(b d-a e)^3 (a+b x)}+\frac {B d-A e}{b (b d-a e) (d+e x)^3}+\frac {(-A b+a B) e}{b (b d-a e)^2 (d+e x)^2}+\frac {(-A b+a B) e}{(b d-a e)^3 (d+e x)}\right ) \, dx}{\sqrt {a^2+2 a b x+b^2 x^2}}\\ &=-\frac {(B d-A e) (a+b x)}{2 e (b d-a e) (d+e x)^2 \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {(A b-a B) (a+b x)}{(b d-a e)^2 (d+e x) \sqrt {a^2+2 a b x+b^2 x^2}}+\frac {b (A b-a B) (a+b x) \log (a+b x)}{(b d-a e)^3 \sqrt {a^2+2 a b x+b^2 x^2}}-\frac {b (A b-a B) (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 = 0.15, size = 131, normalized size = 0.62 \begin {gather*} \frac {(a+b x) \left (\frac {2 (A b-a B)}{(d+e x) (b d-a e)^2}+\frac {B d-A e}{e (d+e x)^2 (a e-b d)}+\frac {2 b (A b-a B) \log (a+b x)}{(b d-a e)^3}-\frac {2 b (A b-a B) \log (d+e x)}{(b d-a e)^3}\right )}{2 \sqrt {(a+b x)^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

________________________________________________________________________________________

IntegrateAlgebraic [B]  time = 85.09, size = 8976, normalized size = 42.34 \begin {gather*} \text {Result too large to show} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

Result too large to show

________________________________________________________________________________________

fricas [B]  time = 0.44, size = 343, normalized size = 1.62 \begin {gather*} -\frac {B b^{2} d^{3} - 3 \, A b^{2} d^{2} e - A a^{2} e^{3} - {\left (B a^{2} - 4 \, A a b\right )} d e^{2} + 2 \, {\left ({\left (B a b - A b^{2}\right )} d e^{2} - {\left (B a^{2} - A a b\right )} e^{3}\right )} x + 2 \, {\left ({\left (B a b - A b^{2}\right )} e^{3} x^{2} + 2 \, {\left (B a b - A b^{2}\right )} d e^{2} x + {\left (B a b - A b^{2}\right )} d^{2} e\right )} \log \left (b x + a\right ) - 2 \, {\left ({\left (B a b - A b^{2}\right )} e^{3} x^{2} + 2 \, {\left (B a b - A b^{2}\right )} d e^{2} x + {\left (B a b - A b^{2}\right )} d^{2} e\right )} \log \left (e x + d\right )}{2 \, {\left (b^{3} d^{5} e - 3 \, a b^{2} d^{4} e^{2} + 3 \, a^{2} b d^{3} e^{3} - a^{3} d^{2} e^{4} + {\left (b^{3} d^{3} e^{3} - 3 \, a b^{2} d^{2} e^{4} + 3 \, a^{2} b d e^{5} - a^{3} e^{6}\right )} x^{2} + 2 \, {\left (b^{3} d^{4} e^{2} - 3 \, a b^{2} d^{3} e^{3} + 3 \, a^{2} b d^{2} e^{4} - a^{3} d e^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="fricas")

[Out]

-1/2*(B*b^2*d^3 - 3*A*b^2*d^2*e - A*a^2*e^3 - (B*a^2 - 4*A*a*b)*d*e^2 + 2*((B*a*b - A*b^2)*d*e^2 - (B*a^2 - A*
a*b)*e^3)*x + 2*((B*a*b - A*b^2)*e^3*x^2 + 2*(B*a*b - A*b^2)*d*e^2*x + (B*a*b - A*b^2)*d^2*e)*log(b*x + a) - 2
*((B*a*b - A*b^2)*e^3*x^2 + 2*(B*a*b - A*b^2)*d*e^2*x + (B*a*b - A*b^2)*d^2*e)*log(e*x + d))/(b^3*d^5*e - 3*a*
b^2*d^4*e^2 + 3*a^2*b*d^3*e^3 - a^3*d^2*e^4 + (b^3*d^3*e^3 - 3*a*b^2*d^2*e^4 + 3*a^2*b*d*e^5 - a^3*e^6)*x^2 +
2*(b^3*d^4*e^2 - 3*a*b^2*d^3*e^3 + 3*a^2*b*d^2*e^4 - a^3*d*e^5)*x)

________________________________________________________________________________________

giac [A]  time = 0.18, size = 306, normalized size = 1.44 \begin {gather*} -\frac {{\left (B a b^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{3} \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | b x + a \right |}\right )}{b^{4} d^{3} - 3 \, a b^{3} d^{2} e + 3 \, a^{2} b^{2} d e^{2} - a^{3} b e^{3}} + \frac {{\left (B a b e \mathrm {sgn}\left (b x + a\right ) - A b^{2} e \mathrm {sgn}\left (b x + a\right )\right )} \log \left ({\left | x e + d \right |}\right )}{b^{3} d^{3} e - 3 \, a b^{2} d^{2} e^{2} + 3 \, a^{2} b d e^{3} - a^{3} e^{4}} - \frac {{\left (B b^{2} d^{3} \mathrm {sgn}\left (b x + a\right ) - 3 \, A b^{2} d^{2} e \mathrm {sgn}\left (b x + a\right ) - B a^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) + 4 \, A a b d e^{2} \mathrm {sgn}\left (b x + a\right ) - A a^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + 2 \, {\left (B a b d e^{2} \mathrm {sgn}\left (b x + a\right ) - A b^{2} d e^{2} \mathrm {sgn}\left (b x + a\right ) - B a^{2} e^{3} \mathrm {sgn}\left (b x + a\right ) + A a b e^{3} \mathrm {sgn}\left (b x + a\right )\right )} x\right )} e^{\left (-1\right )}}{2 \, {\left (b d - a e\right )}^{3} {\left (x e + d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)^3/((b*x+a)^2)^(1/2),x, algorithm="giac")

[Out]

-(B*a*b^2*sgn(b*x + a) - A*b^3*sgn(b*x + a))*log(abs(b*x + a))/(b^4*d^3 - 3*a*b^3*d^2*e + 3*a^2*b^2*d*e^2 - a^
3*b*e^3) + (B*a*b*e*sgn(b*x + a) - A*b^2*e*sgn(b*x + a))*log(abs(x*e + d))/(b^3*d^3*e - 3*a*b^2*d^2*e^2 + 3*a^
2*b*d*e^3 - a^3*e^4) - 1/2*(B*b^2*d^3*sgn(b*x + a) - 3*A*b^2*d^2*e*sgn(b*x + a) - B*a^2*d*e^2*sgn(b*x + a) + 4
*A*a*b*d*e^2*sgn(b*x + a) - A*a^2*e^3*sgn(b*x + a) + 2*(B*a*b*d*e^2*sgn(b*x + a) - A*b^2*d*e^2*sgn(b*x + a) -
B*a^2*e^3*sgn(b*x + a) + A*a*b*e^3*sgn(b*x + a))*x)*e^(-1)/((b*d - a*e)^3*(x*e + d)^2)

________________________________________________________________________________________

maple [A]  time = 0.07, size = 321, normalized size = 1.51 \begin {gather*} -\frac {\left (b x +a \right ) \left (2 A \,b^{2} e^{3} x^{2} \ln \left (b x +a \right )-2 A \,b^{2} e^{3} x^{2} \ln \left (e x +d \right )-2 B a b \,e^{3} x^{2} \ln \left (b x +a \right )+2 B a b \,e^{3} x^{2} \ln \left (e x +d \right )+4 A \,b^{2} d \,e^{2} x \ln \left (b x +a \right )-4 A \,b^{2} d \,e^{2} x \ln \left (e x +d \right )-4 B a b d \,e^{2} x \ln \left (b x +a \right )+4 B a b d \,e^{2} x \ln \left (e x +d \right )-2 A a b \,e^{3} x +2 A \,b^{2} d^{2} e \ln \left (b x +a \right )-2 A \,b^{2} d^{2} e \ln \left (e x +d \right )+2 A \,b^{2} d \,e^{2} x +2 B \,a^{2} e^{3} x -2 B a b \,d^{2} e \ln \left (b x +a \right )+2 B a b \,d^{2} e \ln \left (e x +d \right )-2 B a b d \,e^{2} x +A \,a^{2} e^{3}-4 A a b d \,e^{2}+3 A \,b^{2} d^{2} e +B \,a^{2} d \,e^{2}-B \,b^{2} d^{3}\right )}{2 \sqrt {\left (b x +a \right )^{2}}\, \left (a e -b d \right )^{3} \left (e x +d \right )^{2} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)/(e*x+d)^3/((b*x+a)^2)^(1/2),x)

[Out]

-1/2*(b*x+a)*(2*A*ln(b*x+a)*x^2*b^2*e^3-2*A*ln(e*x+d)*x^2*b^2*e^3-2*B*ln(b*x+a)*x^2*a*b*e^3+2*B*ln(e*x+d)*x^2*
a*b*e^3+4*A*ln(b*x+a)*x*b^2*d*e^2-4*A*ln(e*x+d)*x*b^2*d*e^2-4*B*ln(b*x+a)*x*a*b*d*e^2+4*B*ln(e*x+d)*x*a*b*d*e^
2+2*A*ln(b*x+a)*b^2*d^2*e-2*A*ln(e*x+d)*b^2*d^2*e-2*A*x*a*b*e^3+2*A*x*b^2*d*e^2-2*B*ln(b*x+a)*a*b*d^2*e+2*B*ln
(e*x+d)*a*b*d^2*e+2*B*x*a^2*e^3-2*B*x*a*b*d*e^2+A*a^2*e^3-4*A*a*b*d*e^2+3*A*b^2*d^2*e+B*d*a^2*e^2-B*b^2*d^3)/(
(b*x+a)^2)^(1/2)/(a*e-b*d)^3/e/(e*x+d)^2

________________________________________________________________________________________

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)^3/((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?

________________________________________________________________________________________

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

[Out]

int((A + B*x)/(((a + b*x)^2)^(1/2)*(d + e*x)^3), x)

________________________________________________________________________________________

sympy [B]  time = 1.99, size = 558, normalized size = 2.63 \begin {gather*} - \frac {b \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{2} e - A b^{3} d + B a^{2} b e + B a b^{2} d - \frac {a^{4} b e^{4} \left (- A b + B a\right )}{\left (a e - b d\right )^{3}} + \frac {4 a^{3} b^{2} d e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{3}} - \frac {6 a^{2} b^{3} d^{2} e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{3}} + \frac {4 a b^{4} d^{3} e \left (- A b + B a\right )}{\left (a e - b d\right )^{3}} - \frac {b^{5} d^{4} \left (- A b + B a\right )}{\left (a e - b d\right )^{3}}}{- 2 A b^{3} e + 2 B a b^{2} e} \right )}}{\left (a e - b d\right )^{3}} + \frac {b \left (- A b + B a\right ) \log {\left (x + \frac {- A a b^{2} e - A b^{3} d + B a^{2} b e + B a b^{2} d + \frac {a^{4} b e^{4} \left (- A b + B a\right )}{\left (a e - b d\right )^{3}} - \frac {4 a^{3} b^{2} d e^{3} \left (- A b + B a\right )}{\left (a e - b d\right )^{3}} + \frac {6 a^{2} b^{3} d^{2} e^{2} \left (- A b + B a\right )}{\left (a e - b d\right )^{3}} - \frac {4 a b^{4} d^{3} e \left (- A b + B a\right )}{\left (a e - b d\right )^{3}} + \frac {b^{5} d^{4} \left (- A b + B a\right )}{\left (a e - b d\right )^{3}}}{- 2 A b^{3} e + 2 B a b^{2} e} \right )}}{\left (a e - b d\right )^{3}} + \frac {- A a e^{2} + 3 A b d e - B a d e - B b d^{2} + x \left (2 A b e^{2} - 2 B a e^{2}\right )}{2 a^{2} d^{2} e^{3} - 4 a b d^{3} e^{2} + 2 b^{2} d^{4} e + x^{2} \left (2 a^{2} e^{5} - 4 a b d e^{4} + 2 b^{2} d^{2} e^{3}\right ) + x \left (4 a^{2} d e^{4} - 8 a b d^{2} e^{3} + 4 b^{2} d^{3} e^{2}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)/(e*x+d)**3/((b*x+a)**2)**(1/2),x)

[Out]

-b*(-A*b + B*a)*log(x + (-A*a*b**2*e - A*b**3*d + B*a**2*b*e + B*a*b**2*d - a**4*b*e**4*(-A*b + B*a)/(a*e - b*
d)**3 + 4*a**3*b**2*d*e**3*(-A*b + B*a)/(a*e - b*d)**3 - 6*a**2*b**3*d**2*e**2*(-A*b + B*a)/(a*e - b*d)**3 + 4
*a*b**4*d**3*e*(-A*b + B*a)/(a*e - b*d)**3 - b**5*d**4*(-A*b + B*a)/(a*e - b*d)**3)/(-2*A*b**3*e + 2*B*a*b**2*
e))/(a*e - b*d)**3 + b*(-A*b + B*a)*log(x + (-A*a*b**2*e - A*b**3*d + B*a**2*b*e + B*a*b**2*d + a**4*b*e**4*(-
A*b + B*a)/(a*e - b*d)**3 - 4*a**3*b**2*d*e**3*(-A*b + B*a)/(a*e - b*d)**3 + 6*a**2*b**3*d**2*e**2*(-A*b + B*a
)/(a*e - b*d)**3 - 4*a*b**4*d**3*e*(-A*b + B*a)/(a*e - b*d)**3 + b**5*d**4*(-A*b + B*a)/(a*e - b*d)**3)/(-2*A*
b**3*e + 2*B*a*b**2*e))/(a*e - b*d)**3 + (-A*a*e**2 + 3*A*b*d*e - B*a*d*e - B*b*d**2 + x*(2*A*b*e**2 - 2*B*a*e
**2))/(2*a**2*d**2*e**3 - 4*a*b*d**3*e**2 + 2*b**2*d**4*e + x**2*(2*a**2*e**5 - 4*a*b*d*e**4 + 2*b**2*d**2*e**
3) + x*(4*a**2*d*e**4 - 8*a*b*d**2*e**3 + 4*b**2*d**3*e**2))

________________________________________________________________________________________