3.2078 \(\int \frac{(a+b x) \sqrt{d+e x}}{(a^2+2 a b x+b^2 x^2)^2} \, dx\)
Optimal. Leaf size=110 \[ \frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{3/2}}-\frac{e \sqrt{d+e x}}{4 b (a+b x) (b d-a e)}-\frac{\sqrt{d+e x}}{2 b (a+b x)^2} \]
[Out]
-Sqrt[d + e*x]/(2*b*(a + b*x)^2) - (e*Sqrt[d + e*x])/(4*b*(b*d - a*e)*(a + b*x)) + (e^2*ArcTanh[(Sqrt[b]*Sqrt[
d + e*x])/Sqrt[b*d - a*e]])/(4*b^(3/2)*(b*d - a*e)^(3/2))
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Rubi [A] time = 0.0550826, antiderivative size = 110, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used =
{27, 47, 51, 63, 208} \[ \frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{3/2}}-\frac{e \sqrt{d+e x}}{4 b (a+b x) (b d-a e)}-\frac{\sqrt{d+e x}}{2 b (a+b x)^2} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*x)*Sqrt[d + e*x])/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
[Out]
-Sqrt[d + e*x]/(2*b*(a + b*x)^2) - (e*Sqrt[d + e*x])/(4*b*(b*d - a*e)*(a + b*x)) + (e^2*ArcTanh[(Sqrt[b]*Sqrt[
d + e*x])/Sqrt[b*d - a*e]])/(4*b^(3/2)*(b*d - a*e)^(3/2))
Rule 27
Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]
Rule 47
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]
Rule 51
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]
Rule 63
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{(a+b x) \sqrt{d+e x}}{\left (a^2+2 a b x+b^2 x^2\right )^2} \, dx &=\int \frac{\sqrt{d+e x}}{(a+b x)^3} \, dx\\ &=-\frac{\sqrt{d+e x}}{2 b (a+b x)^2}+\frac{e \int \frac{1}{(a+b x)^2 \sqrt{d+e x}} \, dx}{4 b}\\ &=-\frac{\sqrt{d+e x}}{2 b (a+b x)^2}-\frac{e \sqrt{d+e x}}{4 b (b d-a e) (a+b x)}-\frac{e^2 \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{8 b (b d-a e)}\\ &=-\frac{\sqrt{d+e x}}{2 b (a+b x)^2}-\frac{e \sqrt{d+e x}}{4 b (b d-a e) (a+b x)}-\frac{e \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 b (b d-a e)}\\ &=-\frac{\sqrt{d+e x}}{2 b (a+b x)^2}-\frac{e \sqrt{d+e x}}{4 b (b d-a e) (a+b x)}+\frac{e^2 \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{3/2} (b d-a e)^{3/2}}\\ \end{align*}
Mathematica [C] time = 0.0129316, size = 52, normalized size = 0.47 \[ \frac{2 e^2 (d+e x)^{3/2} \, _2F_1\left (\frac{3}{2},3;\frac{5}{2};-\frac{b (d+e x)}{a e-b d}\right )}{3 (a e-b d)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*x)*Sqrt[d + e*x])/(a^2 + 2*a*b*x + b^2*x^2)^2,x]
[Out]
(2*e^2*(d + e*x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(3*(-(b*d) + a*e)^3)
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Maple [A] time = 0.013, size = 111, normalized size = 1. \begin{align*}{\frac{{e}^{2}}{4\, \left ( bex+ae \right ) ^{2} \left ( ae-bd \right ) } \left ( ex+d \right ) ^{{\frac{3}{2}}}}-{\frac{{e}^{2}}{4\, \left ( bex+ae \right ) ^{2}b}\sqrt{ex+d}}+{\frac{{e}^{2}}{ \left ( 4\,ae-4\,bd \right ) b}\arctan \left ({b\sqrt{ex+d}{\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x)
[Out]
1/4*e^2/(b*e*x+a*e)^2/(a*e-b*d)*(e*x+d)^(3/2)-1/4*e^2/(b*e*x+a*e)^2/b*(e*x+d)^(1/2)+1/4*e^2/(a*e-b*d)/b/((a*e-
b*d)*b)^(1/2)*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 1.07169, size = 949, normalized size = 8.63 \begin{align*} \left [-\frac{{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt{b^{2} d - a b e} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{b^{2} d - a b e} \sqrt{e x + d}}{b x + a}\right ) + 2 \,{\left (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} +{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} +{\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \,{\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}, -\frac{{\left (b^{2} e^{2} x^{2} + 2 \, a b e^{2} x + a^{2} e^{2}\right )} \sqrt{-b^{2} d + a b e} \arctan \left (\frac{\sqrt{-b^{2} d + a b e} \sqrt{e x + d}}{b e x + b d}\right ) +{\left (2 \, b^{3} d^{2} - 3 \, a b^{2} d e + a^{2} b e^{2} +{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (a^{2} b^{4} d^{2} - 2 \, a^{3} b^{3} d e + a^{4} b^{2} e^{2} +{\left (b^{6} d^{2} - 2 \, a b^{5} d e + a^{2} b^{4} e^{2}\right )} x^{2} + 2 \,{\left (a b^{5} d^{2} - 2 \, a^{2} b^{4} d e + a^{3} b^{3} e^{2}\right )} x\right )}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="fricas")
[Out]
[-1/8*((b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(b^2*d - a*b*e)*log((b*e*x + 2*b*d - a*e - 2*sqrt(b^2*d - a*b
*e)*sqrt(e*x + d))/(b*x + a)) + 2*(2*b^3*d^2 - 3*a*b^2*d*e + a^2*b*e^2 + (b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d
))/(a^2*b^4*d^2 - 2*a^3*b^3*d*e + a^4*b^2*e^2 + (b^6*d^2 - 2*a*b^5*d*e + a^2*b^4*e^2)*x^2 + 2*(a*b^5*d^2 - 2*a
^2*b^4*d*e + a^3*b^3*e^2)*x), -1/4*((b^2*e^2*x^2 + 2*a*b*e^2*x + a^2*e^2)*sqrt(-b^2*d + a*b*e)*arctan(sqrt(-b^
2*d + a*b*e)*sqrt(e*x + d)/(b*e*x + b*d)) + (2*b^3*d^2 - 3*a*b^2*d*e + a^2*b*e^2 + (b^3*d*e - a*b^2*e^2)*x)*sq
rt(e*x + d))/(a^2*b^4*d^2 - 2*a^3*b^3*d*e + a^4*b^2*e^2 + (b^6*d^2 - 2*a*b^5*d*e + a^2*b^4*e^2)*x^2 + 2*(a*b^5
*d^2 - 2*a^2*b^4*d*e + a^3*b^3*e^2)*x)]
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Sympy [B] time = 177.155, size = 1658, normalized size = 15.07 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)*(e*x+d)**(1/2)/(b**2*x**2+2*a*b*x+a**2)**2,x)
[Out]
-10*a**2*e**4*sqrt(d + e*x)/(8*a**4*b*e**4 - 16*a**3*b**2*d*e**3 + 16*a**3*b**2*e**4*x - 48*a**2*b**3*d*e**3*x
+ 8*a**2*b**3*e**2*(d + e*x)**2 + 16*a*b**4*d**3*e + 48*a*b**4*d**2*e**2*x - 16*a*b**4*d*e*(d + e*x)**2 - 8*b
**5*d**4 - 16*b**5*d**3*e*x + 8*b**5*d**2*(d + e*x)**2) + 20*a*d*e**3*sqrt(d + e*x)/(8*a**4*e**4 - 16*a**3*b*d
*e**3 + 16*a**3*b*e**4*x - 48*a**2*b**2*d*e**3*x + 8*a**2*b**2*e**2*(d + e*x)**2 + 16*a*b**3*d**3*e + 48*a*b**
3*d**2*e**2*x - 16*a*b**3*d*e*(d + e*x)**2 - 8*b**4*d**4 - 16*b**4*d**3*e*x + 8*b**4*d**2*(d + e*x)**2) - 6*a*
e**3*(d + e*x)**(3/2)/(8*a**4*e**4 - 16*a**3*b*d*e**3 + 16*a**3*b*e**4*x - 48*a**2*b**2*d*e**3*x + 8*a**2*b**2
*e**2*(d + e*x)**2 + 16*a*b**3*d**3*e + 48*a*b**3*d**2*e**2*x - 16*a*b**3*d*e*(d + e*x)**2 - 8*b**4*d**4 - 16*
b**4*d**3*e*x + 8*b**4*d**2*(d + e*x)**2) + 3*a*e**3*sqrt(-1/(b*(a*e - b*d)**5))*log(-a**3*e**3*sqrt(-1/(b*(a*
e - b*d)**5)) + 3*a**2*b*d*e**2*sqrt(-1/(b*(a*e - b*d)**5)) - 3*a*b**2*d**2*e*sqrt(-1/(b*(a*e - b*d)**5)) + b*
*3*d**3*sqrt(-1/(b*(a*e - b*d)**5)) + sqrt(d + e*x))/(8*b) - 3*a*e**3*sqrt(-1/(b*(a*e - b*d)**5))*log(a**3*e**
3*sqrt(-1/(b*(a*e - b*d)**5)) - 3*a**2*b*d*e**2*sqrt(-1/(b*(a*e - b*d)**5)) + 3*a*b**2*d**2*e*sqrt(-1/(b*(a*e
- b*d)**5)) - b**3*d**3*sqrt(-1/(b*(a*e - b*d)**5)) + sqrt(d + e*x))/(8*b) - 10*b*d**2*e**2*sqrt(d + e*x)/(8*a
**4*e**4 - 16*a**3*b*d*e**3 + 16*a**3*b*e**4*x - 48*a**2*b**2*d*e**3*x + 8*a**2*b**2*e**2*(d + e*x)**2 + 16*a*
b**3*d**3*e + 48*a*b**3*d**2*e**2*x - 16*a*b**3*d*e*(d + e*x)**2 - 8*b**4*d**4 - 16*b**4*d**3*e*x + 8*b**4*d**
2*(d + e*x)**2) + 6*b*d*e**2*(d + e*x)**(3/2)/(8*a**4*e**4 - 16*a**3*b*d*e**3 + 16*a**3*b*e**4*x - 48*a**2*b**
2*d*e**3*x + 8*a**2*b**2*e**2*(d + e*x)**2 + 16*a*b**3*d**3*e + 48*a*b**3*d**2*e**2*x - 16*a*b**3*d*e*(d + e*x
)**2 - 8*b**4*d**4 - 16*b**4*d**3*e*x + 8*b**4*d**2*(d + e*x)**2) - 3*d*e**2*sqrt(-1/(b*(a*e - b*d)**5))*log(-
a**3*e**3*sqrt(-1/(b*(a*e - b*d)**5)) + 3*a**2*b*d*e**2*sqrt(-1/(b*(a*e - b*d)**5)) - 3*a*b**2*d**2*e*sqrt(-1/
(b*(a*e - b*d)**5)) + b**3*d**3*sqrt(-1/(b*(a*e - b*d)**5)) + sqrt(d + e*x))/8 + 3*d*e**2*sqrt(-1/(b*(a*e - b*
d)**5))*log(a**3*e**3*sqrt(-1/(b*(a*e - b*d)**5)) - 3*a**2*b*d*e**2*sqrt(-1/(b*(a*e - b*d)**5)) + 3*a*b**2*d**
2*e*sqrt(-1/(b*(a*e - b*d)**5)) - b**3*d**3*sqrt(-1/(b*(a*e - b*d)**5)) + sqrt(d + e*x))/8 + 2*e**2*sqrt(d + e
*x)/(2*a**2*b*e**2 - 2*a*b**2*d*e + 2*a*b**2*e**2*x - 2*b**3*d*e*x) - e**2*sqrt(-1/(b*(a*e - b*d)**3))*log(-a*
*2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) + 2*a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) - b**2*d**2*sqrt(-1/(b*(a*e - b*d)
**3)) + sqrt(d + e*x))/(2*b) + e**2*sqrt(-1/(b*(a*e - b*d)**3))*log(a**2*e**2*sqrt(-1/(b*(a*e - b*d)**3)) - 2*
a*b*d*e*sqrt(-1/(b*(a*e - b*d)**3)) + b**2*d**2*sqrt(-1/(b*(a*e - b*d)**3)) + sqrt(d + e*x))/(2*b)
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Giac [A] time = 1.118, size = 178, normalized size = 1.62 \begin{align*} -\frac{\arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{4 \,{\left (b^{2} d - a b e\right )} \sqrt{-b^{2} d + a b e}} - \frac{{\left (x e + d\right )}^{\frac{3}{2}} b e^{2} + \sqrt{x e + d} b d e^{2} - \sqrt{x e + d} a e^{3}}{4 \,{\left (b^{2} d - a b e\right )}{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((b*x+a)*(e*x+d)^(1/2)/(b^2*x^2+2*a*b*x+a^2)^2,x, algorithm="giac")
[Out]
-1/4*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/((b^2*d - a*b*e)*sqrt(-b^2*d + a*b*e)) - 1/4*((x*e + d)^
(3/2)*b*e^2 + sqrt(x*e + d)*b*d*e^2 - sqrt(x*e + d)*a*e^3)/((b^2*d - a*b*e)*((x*e + d)*b - b*d + a*e)^2)