3.845 \(\int \frac{15 d^2+20 d e x+8 e^2 x^2}{\sqrt{a+b x} \sqrt{d+e x}} \, dx\)

Optimal. Leaf size=122 \[ \frac{2 \left (3 a^2 e^2-8 a b d e+8 b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{5/2} \sqrt{e}}+\frac{4 e (a+b x)^{3/2} \sqrt{d+e x}}{b^2}+\frac{2 \sqrt{a+b x} \sqrt{d+e x} (7 b d-5 a e)}{b^2} \]

[Out]

(2*(7*b*d - 5*a*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/b^2 + (4*e*(a + b*x)^(3/2)*Sqrt[d + e*x])/b^2 + (2*(8*b^2*d^2
- 8*a*b*d*e + 3*a^2*e^2)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(b^(5/2)*Sqrt[e])

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Rubi [A]  time = 0.123517, antiderivative size = 122, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.132, Rules used = {951, 80, 63, 217, 206} \[ \frac{2 \left (3 a^2 e^2-8 a b d e+8 b^2 d^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{5/2} \sqrt{e}}+\frac{4 e (a+b x)^{3/2} \sqrt{d+e x}}{b^2}+\frac{2 \sqrt{a+b x} \sqrt{d+e x} (7 b d-5 a e)}{b^2} \]

Antiderivative was successfully verified.

[In]

Int[(15*d^2 + 20*d*e*x + 8*e^2*x^2)/(Sqrt[a + b*x]*Sqrt[d + e*x]),x]

[Out]

(2*(7*b*d - 5*a*e)*Sqrt[a + b*x]*Sqrt[d + e*x])/b^2 + (4*e*(a + b*x)^(3/2)*Sqrt[d + e*x])/b^2 + (2*(8*b^2*d^2
- 8*a*b*d*e + 3*a^2*e^2)*ArcTanh[(Sqrt[e]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt[d + e*x])])/(b^(5/2)*Sqrt[e])

Rule 951

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[(c^p*(d + e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{15 d^2+20 d e x+8 e^2 x^2}{\sqrt{a+b x} \sqrt{d+e x}} \, dx &=\frac{4 e (a+b x)^{3/2} \sqrt{d+e x}}{b^2}+\frac{\int \frac{2 e \left (15 b^2 d^2-6 a b d e-2 a^2 e^2\right )+4 b e^2 (7 b d-5 a e) x}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{2 b^2 e}\\ &=\frac{2 (7 b d-5 a e) \sqrt{a+b x} \sqrt{d+e x}}{b^2}+\frac{4 e (a+b x)^{3/2} \sqrt{d+e x}}{b^2}+\frac{\left (8 b^2 d^2-8 a b d e+3 a^2 e^2\right ) \int \frac{1}{\sqrt{a+b x} \sqrt{d+e x}} \, dx}{b^2}\\ &=\frac{2 (7 b d-5 a e) \sqrt{a+b x} \sqrt{d+e x}}{b^2}+\frac{4 e (a+b x)^{3/2} \sqrt{d+e x}}{b^2}+\frac{\left (2 \left (8 b^2 d^2-8 a b d e+3 a^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{d-\frac{a e}{b}+\frac{e x^2}{b}}} \, dx,x,\sqrt{a+b x}\right )}{b^3}\\ &=\frac{2 (7 b d-5 a e) \sqrt{a+b x} \sqrt{d+e x}}{b^2}+\frac{4 e (a+b x)^{3/2} \sqrt{d+e x}}{b^2}+\frac{\left (2 \left (8 b^2 d^2-8 a b d e+3 a^2 e^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{1-\frac{e x^2}{b}} \, dx,x,\frac{\sqrt{a+b x}}{\sqrt{d+e x}}\right )}{b^3}\\ &=\frac{2 (7 b d-5 a e) \sqrt{a+b x} \sqrt{d+e x}}{b^2}+\frac{4 e (a+b x)^{3/2} \sqrt{d+e x}}{b^2}+\frac{2 \left (8 b^2 d^2-8 a b d e+3 a^2 e^2\right ) \tanh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b} \sqrt{d+e x}}\right )}{b^{5/2} \sqrt{e}}\\ \end{align*}

Mathematica [A]  time = 0.421291, size = 135, normalized size = 1.11 \[ \frac{2 \left (\frac{\sqrt{b d-a e} \left (3 a^2 e^2-8 a b d e+8 b^2 d^2\right ) \sqrt{\frac{b (d+e x)}{b d-a e}} \sinh ^{-1}\left (\frac{\sqrt{e} \sqrt{a+b x}}{\sqrt{b d-a e}}\right )}{\sqrt{e}}+b \sqrt{a+b x} (d+e x) (-3 a e+7 b d+2 b e x)\right )}{b^3 \sqrt{d+e x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(15*d^2 + 20*d*e*x + 8*e^2*x^2)/(Sqrt[a + b*x]*Sqrt[d + e*x]),x]

[Out]

(2*(b*Sqrt[a + b*x]*(d + e*x)*(7*b*d - 3*a*e + 2*b*e*x) + (Sqrt[b*d - a*e]*(8*b^2*d^2 - 8*a*b*d*e + 3*a^2*e^2)
*Sqrt[(b*(d + e*x))/(b*d - a*e)]*ArcSinh[(Sqrt[e]*Sqrt[a + b*x])/Sqrt[b*d - a*e]])/Sqrt[e]))/(b^3*Sqrt[d + e*x
])

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Maple [B]  time = 0.318, size = 247, normalized size = 2. \begin{align*}{\frac{1}{{b}^{2}} \left ( 3\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){a}^{2}{e}^{2}-8\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ) abde+8\,\ln \left ( 1/2\,{\frac{2\,bxe+2\,\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }\sqrt{be}+ae+bd}{\sqrt{be}}} \right ){b}^{2}{d}^{2}+4\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }xbe-6\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }ae+14\,\sqrt{be}\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }bd \right ) \sqrt{ex+d}\sqrt{bx+a}{\frac{1}{\sqrt{ \left ( bx+a \right ) \left ( ex+d \right ) }}}{\frac{1}{\sqrt{be}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(1/2)/(b*x+a)^(1/2),x)

[Out]

(3*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a^2*e^2-8*ln(1/2*(2*b*x*e+2*((b
*x+a)*(e*x+d))^(1/2)*(b*e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*a*b*d*e+8*ln(1/2*(2*b*x*e+2*((b*x+a)*(e*x+d))^(1/2)*(b*
e)^(1/2)+a*e+b*d)/(b*e)^(1/2))*b^2*d^2+4*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*x*b*e-6*(b*e)^(1/2)*((b*x+a)*(e*x
+d))^(1/2)*a*e+14*(b*e)^(1/2)*((b*x+a)*(e*x+d))^(1/2)*b*d)*(e*x+d)^(1/2)*(b*x+a)^(1/2)/(b*e)^(1/2)/b^2/((b*x+a
)*(e*x+d))^(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((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(1/2)/(b*x+a)^(1/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.97896, size = 710, normalized size = 5.82 \begin{align*} \left [\frac{{\left (8 \, b^{2} d^{2} - 8 \, a b d e + 3 \, a^{2} e^{2}\right )} \sqrt{b e} \log \left (8 \, b^{2} e^{2} x^{2} + b^{2} d^{2} + 6 \, a b d e + a^{2} e^{2} + 4 \,{\left (2 \, b e x + b d + a e\right )} \sqrt{b e} \sqrt{b x + a} \sqrt{e x + d} + 8 \,{\left (b^{2} d e + a b e^{2}\right )} x\right ) + 4 \,{\left (2 \, b^{2} e^{2} x + 7 \, b^{2} d e - 3 \, a b e^{2}\right )} \sqrt{b x + a} \sqrt{e x + d}}{2 \, b^{3} e}, -\frac{{\left (8 \, b^{2} d^{2} - 8 \, a b d e + 3 \, a^{2} e^{2}\right )} \sqrt{-b e} \arctan \left (\frac{{\left (2 \, b e x + b d + a e\right )} \sqrt{-b e} \sqrt{b x + a} \sqrt{e x + d}}{2 \,{\left (b^{2} e^{2} x^{2} + a b d e +{\left (b^{2} d e + a b e^{2}\right )} x\right )}}\right ) - 2 \,{\left (2 \, b^{2} e^{2} x + 7 \, b^{2} d e - 3 \, a b e^{2}\right )} \sqrt{b x + a} \sqrt{e x + d}}{b^{3} e}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(1/2)/(b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[1/2*((8*b^2*d^2 - 8*a*b*d*e + 3*a^2*e^2)*sqrt(b*e)*log(8*b^2*e^2*x^2 + b^2*d^2 + 6*a*b*d*e + a^2*e^2 + 4*(2*b
*e*x + b*d + a*e)*sqrt(b*e)*sqrt(b*x + a)*sqrt(e*x + d) + 8*(b^2*d*e + a*b*e^2)*x) + 4*(2*b^2*e^2*x + 7*b^2*d*
e - 3*a*b*e^2)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e), -((8*b^2*d^2 - 8*a*b*d*e + 3*a^2*e^2)*sqrt(-b*e)*arctan(1
/2*(2*b*e*x + b*d + a*e)*sqrt(-b*e)*sqrt(b*x + a)*sqrt(e*x + d)/(b^2*e^2*x^2 + a*b*d*e + (b^2*d*e + a*b*e^2)*x
)) - 2*(2*b^2*e^2*x + 7*b^2*d*e - 3*a*b*e^2)*sqrt(b*x + a)*sqrt(e*x + d))/(b^3*e)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{15 d^{2} + 20 d e x + 8 e^{2} x^{2}}{\sqrt{a + b x} \sqrt{d + e x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e**2*x**2+20*d*e*x+15*d**2)/(e*x+d)**(1/2)/(b*x+a)**(1/2),x)

[Out]

Integral((15*d**2 + 20*d*e*x + 8*e**2*x**2)/(sqrt(a + b*x)*sqrt(d + e*x)), x)

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Giac [A]  time = 1.19239, size = 196, normalized size = 1.61 \begin{align*} \frac{2 \,{\left (\sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \sqrt{b x + a}{\left (\frac{2 \,{\left (b x + a\right )} e}{b^{3}} + \frac{{\left (7 \, b^{6} d e^{2} - 5 \, a b^{5} e^{3}\right )} e^{\left (-2\right )}}{b^{8}}\right )} - \frac{{\left (8 \, b^{2} d^{2} - 8 \, a b d e + 3 \, a^{2} e^{2}\right )} e^{\left (-\frac{1}{2}\right )} \log \left ({\left | -\sqrt{b x + a} \sqrt{b} e^{\frac{1}{2}} + \sqrt{b^{2} d +{\left (b x + a\right )} b e - a b e} \right |}\right )}{b^{\frac{5}{2}}}\right )} b}{{\left | b \right |}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((8*e^2*x^2+20*d*e*x+15*d^2)/(e*x+d)^(1/2)/(b*x+a)^(1/2),x, algorithm="giac")

[Out]

2*(sqrt(b^2*d + (b*x + a)*b*e - a*b*e)*sqrt(b*x + a)*(2*(b*x + a)*e/b^3 + (7*b^6*d*e^2 - 5*a*b^5*e^3)*e^(-2)/b
^8) - (8*b^2*d^2 - 8*a*b*d*e + 3*a^2*e^2)*e^(-1/2)*log(abs(-sqrt(b*x + a)*sqrt(b)*e^(1/2) + sqrt(b^2*d + (b*x
+ a)*b*e - a*b*e)))/b^(5/2))*b/abs(b)