3.1714 \(\int \frac{(d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=158 \[ -\frac{3 e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{b d-a e}}-\frac{3 e \sqrt{d+e x}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

(-3*e*Sqrt[d + e*x])/(4*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (d + e*x)^(3/2)/(2*b*(a + b*x)*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) - (3*e^2*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*b^(5/2)*Sqrt[b*d - a*e]*S
qrt[a^2 + 2*a*b*x + b^2*x^2])

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Rubi [A]  time = 0.075453, antiderivative size = 158, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {646, 47, 63, 208} \[ -\frac{3 e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{5/2} \sqrt{a^2+2 a b x+b^2 x^2} \sqrt{b d-a e}}-\frac{3 e \sqrt{d+e x}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*e*Sqrt[d + e*x])/(4*b^2*Sqrt[a^2 + 2*a*b*x + b^2*x^2]) - (d + e*x)^(3/2)/(2*b*(a + b*x)*Sqrt[a^2 + 2*a*b*x
+ b^2*x^2]) - (3*e^2*(a + b*x)*ArcTanh[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[b*d - a*e]])/(4*b^(5/2)*Sqrt[b*d - a*e]*S
qrt[a^2 + 2*a*b*x + b^2*x^2])

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 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 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}}{\left (a^2+2 a b x+b^2 x^2\right )^{3/2}} \, dx &=\frac{\left (b^2 \left (a b+b^2 x\right )\right ) \int \frac{(d+e x)^{3/2}}{\left (a b+b^2 x\right )^3} \, dx}{\sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{(d+e x)^{3/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 e \left (a b+b^2 x\right )\right ) \int \frac{\sqrt{d+e x}}{\left (a b+b^2 x\right )^2} \, dx}{4 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{3 e \sqrt{d+e x}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 e^2 \left (a b+b^2 x\right )\right ) \int \frac{1}{\left (a b+b^2 x\right ) \sqrt{d+e x}} \, dx}{8 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{3 e \sqrt{d+e x}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{\left (3 e \left (a b+b^2 x\right )\right ) \operatorname{Subst}\left (\int \frac{1}{a b-\frac{b^2 d}{e}+\frac{b^2 x^2}{e}} \, dx,x,\sqrt{d+e x}\right )}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}\\ &=-\frac{3 e \sqrt{d+e x}}{4 b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{(d+e x)^{3/2}}{2 b (a+b x) \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e^2 (a+b x) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{4 b^{5/2} \sqrt{b d-a e} \sqrt{a^2+2 a b x+b^2 x^2}}\\ \end{align*}

Mathematica [A]  time = 0.108416, size = 110, normalized size = 0.7 \[ \frac{\frac{3 e^2 (a+b x)^2 \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{a e-b d}}\right )}{\sqrt{a e-b d}}-\sqrt{b} \sqrt{d+e x} (3 a e+2 b d+5 b e x)}{4 b^{5/2} (a+b x) \sqrt{(a+b x)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(3/2)/(a^2 + 2*a*b*x + b^2*x^2)^(3/2),x]

[Out]

(-(Sqrt[b]*Sqrt[d + e*x]*(2*b*d + 3*a*e + 5*b*e*x)) + (3*e^2*(a + b*x)^2*ArcTan[(Sqrt[b]*Sqrt[d + e*x])/Sqrt[-
(b*d) + a*e]])/Sqrt[-(b*d) + a*e])/(4*b^(5/2)*(a + b*x)*Sqrt[(a + b*x)^2])

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Maple [A]  time = 0.271, size = 194, normalized size = 1.2 \begin{align*} -{\frac{bx+a}{4\,{b}^{2}} \left ( -3\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){x}^{2}{b}^{2}{e}^{2}-6\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xab{e}^{2}+5\,\sqrt{ \left ( ae-bd \right ) b} \left ( ex+d \right ) ^{3/2}b-3\,\arctan \left ({\frac{b\sqrt{ex+d}}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}{e}^{2}+3\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}ae-3\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}bd \right ){\frac{1}{\sqrt{ \left ( ae-bd \right ) b}}} \left ( \left ( bx+a \right ) ^{2} \right ) ^{-{\frac{3}{2}}}} \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)^(3/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (e x + d\right )}^{\frac{3}{2}}}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2}\right )}^{\frac{3}{2}}}\,{d x} \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)^(3/2),x, algorithm="maxima")

[Out]

integrate((e*x + d)^(3/2)/(b^2*x^2 + 2*a*b*x + a^2)^(3/2), x)

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Fricas [A]  time = 1.6693, size = 795, normalized size = 5.03 \begin{align*} \left [\frac{3 \,{\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} + a b^{2} d e - 3 \, a^{2} b e^{2} + 5 \,{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{8 \,{\left (a^{2} b^{4} d - a^{3} b^{3} e +{\left (b^{6} d - a b^{5} e\right )} x^{2} + 2 \,{\left (a b^{5} d - a^{2} b^{4} e\right )} x\right )}}, \frac{3 \,{\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} + a b^{2} d e - 3 \, a^{2} b e^{2} + 5 \,{\left (b^{3} d e - a b^{2} e^{2}\right )} x\right )} \sqrt{e x + d}}{4 \,{\left (a^{2} b^{4} d - a^{3} b^{3} e +{\left (b^{6} d - a b^{5} e\right )} x^{2} + 2 \,{\left (a b^{5} d - a^{2} b^{4} e\right )} x\right )}}\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)^(3/2),x, algorithm="fricas")

[Out]

[1/8*(3*(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 + a*b^2*d*e - 3*a^2*b*e^2 + 5*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x
+ d))/(a^2*b^4*d - a^3*b^3*e + (b^6*d - a*b^5*e)*x^2 + 2*(a*b^5*d - a^2*b^4*e)*x), 1/4*(3*(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 +
a*b^2*d*e - 3*a^2*b*e^2 + 5*(b^3*d*e - a*b^2*e^2)*x)*sqrt(e*x + d))/(a^2*b^4*d - a^3*b^3*e + (b^6*d - a*b^5*e
)*x^2 + 2*(a*b^5*d - a^2*b^4*e)*x)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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)**(3/2),x)

[Out]

Timed out

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Giac [A]  time = 1.1912, size = 216, normalized size = 1.37 \begin{align*} \frac{3 \, \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{2}}{4 \, \sqrt{-b^{2} d + a b e} b^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{5 \,{\left (x e + d\right )}^{\frac{3}{2}} b e^{2} - 3 \, \sqrt{x e + d} b d e^{2} + 3 \, \sqrt{x e + d} a e^{3}}{4 \,{\left ({\left (x e + d\right )} b - b d + a e\right )}^{2} b^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{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)^(3/2),x, algorithm="giac")

[Out]

3/4*arctan(sqrt(x*e + d)*b/sqrt(-b^2*d + a*b*e))*e^2/(sqrt(-b^2*d + a*b*e)*b^2*sgn((x*e + d)*b*e - b*d*e + a*e
^2)) - 1/4*(5*(x*e + d)^(3/2)*b*e^2 - 3*sqrt(x*e + d)*b*d*e^2 + 3*sqrt(x*e + d)*a*e^3)/(((x*e + d)*b - b*d + a
*e)^2*b^2*sgn((x*e + d)*b*e - b*d*e + a*e^2))