3.1648 \(\int \frac{(d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx\)
Optimal. Leaf size=85 \[ -\frac{3 e \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{5/2}}-\frac{(d+e x)^{3/2}}{b (a+b x)}+\frac{3 e \sqrt{d+e x}}{b^2} \]
[Out]
(3*e*Sqrt[d + e*x])/b^2 - (d + e*x)^(3/2)/(b*(a + b*x)) - (3*e*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])
/Sqrt[b*d - a*e]])/b^(5/2)
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Rubi [A] time = 0.0400498, antiderivative size = 85, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used =
{27, 47, 50, 63, 208} \[ -\frac{3 e \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{5/2}}-\frac{(d+e x)^{3/2}}{b (a+b x)}+\frac{3 e \sqrt{d+e x}}{b^2} \]
Antiderivative was successfully verified.
[In]
Int[(d + e*x)^(3/2)/(a^2 + 2*a*b*x + b^2*x^2),x]
[Out]
(3*e*Sqrt[d + e*x])/b^2 - (d + e*x)^(3/2)/(b*(a + b*x)) - (3*e*Sqrt[b*d - a*e]*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])
/Sqrt[b*d - a*e]])/b^(5/2)
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 50
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && 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{(d+e x)^{3/2}}{a^2+2 a b x+b^2 x^2} \, dx &=\int \frac{(d+e x)^{3/2}}{(a+b x)^2} \, dx\\ &=-\frac{(d+e x)^{3/2}}{b (a+b x)}+\frac{(3 e) \int \frac{\sqrt{d+e x}}{a+b x} \, dx}{2 b}\\ &=\frac{3 e \sqrt{d+e x}}{b^2}-\frac{(d+e x)^{3/2}}{b (a+b x)}+\frac{(3 e (b d-a e)) \int \frac{1}{(a+b x) \sqrt{d+e x}} \, dx}{2 b^2}\\ &=\frac{3 e \sqrt{d+e x}}{b^2}-\frac{(d+e x)^{3/2}}{b (a+b x)}+\frac{(3 (b d-a e)) \operatorname{Subst}\left (\int \frac{1}{a-\frac{b d}{e}+\frac{b x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{b^2}\\ &=\frac{3 e \sqrt{d+e x}}{b^2}-\frac{(d+e x)^{3/2}}{b (a+b x)}-\frac{3 e \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{5/2}}\\ \end{align*}
Mathematica [C] time = 0.0140168, size = 50, normalized size = 0.59 \[ \frac{2 e (d+e x)^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};-\frac{b (d+e x)}{a e-b d}\right )}{5 (a e-b d)^2} \]
Antiderivative was successfully verified.
[In]
Integrate[(d + e*x)^(3/2)/(a^2 + 2*a*b*x + b^2*x^2),x]
[Out]
(2*e*(d + e*x)^(5/2)*Hypergeometric2F1[2, 5/2, 7/2, -((b*(d + e*x))/(-(b*d) + a*e))])/(5*(-(b*d) + a*e)^2)
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Maple [B] time = 0.2, size = 148, normalized size = 1.7 \begin{align*} 2\,{\frac{e\sqrt{ex+d}}{{b}^{2}}}+{\frac{a{e}^{2}}{{b}^{2} \left ( bxe+ae \right ) }\sqrt{ex+d}}-{\frac{de}{b \left ( bxe+ae \right ) }\sqrt{ex+d}}-3\,{\frac{a{e}^{2}}{{b}^{2}\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) }+3\,{\frac{de}{b\sqrt{ \left ( ae-bd \right ) b}}\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) } \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x)
[Out]
2*e*(e*x+d)^(1/2)/b^2+1/b^2*(e*x+d)^(1/2)/(b*e*x+a*e)*a*e^2-e/b*(e*x+d)^(1/2)/(b*e*x+a*e)*d-3/b^2/((a*e-b*d)*b
)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a*e-b*d)*b)^(1/2))*a*e^2+3*e/b/((a*e-b*d)*b)^(1/2)*arctan(b*(e*x+d)^(1/2)/((a
*e-b*d)*b)^(1/2))*d
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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((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [A] time = 1.89191, size = 455, normalized size = 5.35 \begin{align*} \left [\frac{3 \,{\left (b e x + a e\right )} \sqrt{\frac{b d - a e}{b}} \log \left (\frac{b e x + 2 \, b d - a e - 2 \, \sqrt{e x + d} b \sqrt{\frac{b d - a e}{b}}}{b x + a}\right ) + 2 \,{\left (2 \, b e x - b d + 3 \, a e\right )} \sqrt{e x + d}}{2 \,{\left (b^{3} x + a b^{2}\right )}}, -\frac{3 \,{\left (b e x + a e\right )} \sqrt{-\frac{b d - a e}{b}} \arctan \left (-\frac{\sqrt{e x + d} b \sqrt{-\frac{b d - a e}{b}}}{b d - a e}\right ) -{\left (2 \, b e x - b d + 3 \, a e\right )} \sqrt{e x + d}}{b^{3} x + a b^{2}}\right ] \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="fricas")
[Out]
[1/2*(3*(b*e*x + a*e)*sqrt((b*d - a*e)/b)*log((b*e*x + 2*b*d - a*e - 2*sqrt(e*x + d)*b*sqrt((b*d - a*e)/b))/(b
*x + a)) + 2*(2*b*e*x - b*d + 3*a*e)*sqrt(e*x + d))/(b^3*x + a*b^2), -(3*(b*e*x + a*e)*sqrt(-(b*d - a*e)/b)*ar
ctan(-sqrt(e*x + d)*b*sqrt(-(b*d - a*e)/b)/(b*d - a*e)) - (2*b*e*x - b*d + 3*a*e)*sqrt(e*x + d))/(b^3*x + a*b^
2)]
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Sympy [B] time = 108.324, size = 923, normalized size = 10.86 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)**(3/2)/(b**2*x**2+2*a*b*x+a**2),x)
[Out]
2*a**2*e**3*sqrt(d + e*x)/(2*a**2*b**2*e**2 - 2*a*b**3*d*e + 2*a*b**3*e**2*x - 2*b**4*d*e*x) - a**2*e**3*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**2) + a**2*e**3*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**2) - 4*a*d*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) + a*d*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))/b - a*d*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))/b - 4*a*e**2*atan(sqrt(d + e*x)/sqrt(a*e/b - d))/(b**3*sqrt(a*e/b - d)
) - d**2*e*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 + d**2*e*sqrt(-1/(b*(a*e - b*d)**3))*lo
g(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 + 2*d**2*e*sqrt(d + e*x)/(2*a**2*e**2 - 2*a*b*d*e + 2*a*b*e**2*x - 2*b**2*d*e*x)
+ 4*d*e*atan(sqrt(d + e*x)/sqrt(a*e/b - d))/(b**2*sqrt(a*e/b - d)) + 2*e*sqrt(d + e*x)/b**2
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Giac [A] time = 1.16355, size = 165, normalized size = 1.94 \begin{align*} \frac{3 \,{\left (b d e - a e^{2}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right )}{\sqrt{-b^{2} d + a b e} b^{2}} + \frac{2 \, \sqrt{x e + d} e}{b^{2}} - \frac{\sqrt{x e + d} b d e - \sqrt{x e + d} a e^{2}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} b^{2}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2),x, algorithm="giac")
[Out]
3*(b*d*e - a*e^2)*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))/(sqrt(-b^2*d + a*b*e)*b^2) + 2*sqrt(x*e + d)*e/
b^2 - (sqrt(x*e + d)*b*d*e - sqrt(x*e + d)*a*e^2)/(((x*e + d)*b - b*d + a*e)*b^2)