3.2132 \(\int \frac{(a+b x) (d+e x)^{3/2}}{(a^2+2 a b x+b^2 x^2)^{3/2}} \, dx\)

Optimal. Leaf size=148 \[ -\frac{(d+e x)^{3/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e (a+b x) \sqrt{d+e x}}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e (a+b x) \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

[Out]

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

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Rubi [A]  time = 0.0921657, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {768, 646, 50, 63, 208} \[ -\frac{(d+e x)^{3/2}}{b \sqrt{a^2+2 a b x+b^2 x^2}}+\frac{3 e (a+b x) \sqrt{d+e x}}{b^2 \sqrt{a^2+2 a b x+b^2 x^2}}-\frac{3 e (a+b x) \sqrt{b d-a e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{d+e x}}{\sqrt{b d-a e}}\right )}{b^{5/2} \sqrt{a^2+2 a b x+b^2 x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 768

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Sim
p[(g*(d + e*x)^m*(a + b*x + c*x^2)^(p + 1))/(2*c*(p + 1)), x] - Dist[(e*g*m)/(2*c*(p + 1)), Int[(d + e*x)^(m -
1)*(a + b*x + c*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[2*c*f - b*g, 0] && LtQ[p, -1]
&& GtQ[m, 0]

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

Mathematica [C]  time = 0.0296636, size = 66, normalized size = 0.45 \[ \frac{2 e (a+b x) (d+e x)^{5/2} \, _2F_1\left (2,\frac{5}{2};\frac{7}{2};-\frac{b (d+e x)}{a e-b d}\right )}{5 \sqrt{(a+b x)^2} (a e-b d)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*e*(a + b*x)*(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*Sqrt[(a + b*x)^2])

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Maple [B]  time = 0.014, size = 222, normalized size = 1.5 \begin{align*}{\frac{ \left ( bx+a \right ) ^{2}}{{b}^{2}} \left ( -3\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) xab{e}^{2}+3\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) x{b}^{2}de+2\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}xbe-3\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ){a}^{2}{e}^{2}+3\,\arctan \left ({\frac{\sqrt{ex+d}b}{\sqrt{ \left ( ae-bd \right ) b}}} \right ) abde+3\,\sqrt{ \left ( ae-bd \right ) b}\sqrt{ex+d}ae-\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((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x)

[Out]

(-3*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*a*b*e^2+3*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*x*b^2*
d*e+2*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*x*b*e-3*arctan((e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a^2*e^2+3*arctan((
e*x+d)^(1/2)*b/((a*e-b*d)*b)^(1/2))*a*b*d*e+3*((a*e-b*d)*b)^(1/2)*(e*x+d)^(1/2)*a*e-((a*e-b*d)*b)^(1/2)*(e*x+d
)^(1/2)*b*d)*(b*x+a)^2/((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 (b x + a\right )}{\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((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*x + a)*(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 = 1.05173, size = 455, normalized size = 3.07 \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((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),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 [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right ) \left (d + e x\right )^{\frac{3}{2}}}{\left (\left (a + b x\right )^{2}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20585, size = 267, normalized size = 1.8 \begin{align*} \frac{3 \,{\left (b d e^{2} - a e^{3}\right )} \arctan \left (\frac{\sqrt{x e + d} b}{\sqrt{-b^{2} d + a b e}}\right ) e^{\left (-1\right )}}{\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{2 \, \sqrt{x e + d} e}{b^{2} \mathrm{sgn}\left ({\left (x e + d\right )} b e - b d e + a e^{2}\right )} - \frac{{\left (\sqrt{x e + d} b d e^{2} - \sqrt{x e + d} a e^{3}\right )} e^{\left (-1\right )}}{{\left ({\left (x e + d\right )} b - b d + a e\right )} 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((b*x+a)*(e*x+d)^(3/2)/(b^2*x^2+2*a*b*x+a^2)^(3/2),x, algorithm="giac")

[Out]

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