3.738 \(\int \frac{(a^2+2 a b x^2+b^2 x^4)^{3/2}}{\sqrt{d x}} \, dx\)

Optimal. Leaf size=193 \[ \frac{2 b^3 (d x)^{13/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 d^7 \left (a+b x^2\right )}+\frac{2 a b^2 (d x)^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac{6 a^2 b (d x)^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}+\frac{2 a^3 \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )} \]

[Out]

(2*a^3*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(d*(a + b*x^2)) + (6*a^2*b*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4])/(5*d^3*(a + b*x^2)) + (2*a*b^2*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d^5*(a + b*x^2)) +
(2*b^3*(d*x)^(13/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*d^7*(a + b*x^2))

________________________________________________________________________________________

Rubi [A]  time = 0.0544002, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.067, Rules used = {1112, 270} \[ \frac{2 b^3 (d x)^{13/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 d^7 \left (a+b x^2\right )}+\frac{2 a b^2 (d x)^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac{6 a^2 b (d x)^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}+\frac{2 a^3 \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/Sqrt[d*x],x]

[Out]

(2*a^3*Sqrt[d*x]*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(d*(a + b*x^2)) + (6*a^2*b*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2
+ b^2*x^4])/(5*d^3*(a + b*x^2)) + (2*a*b^2*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(3*d^5*(a + b*x^2)) +
(2*b^3*(d*x)^(13/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*d^7*(a + b*x^2))

Rule 1112

Int[((d_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^FracPa
rt[p]/(c^IntPart[p]*(b/2 + c*x^2)^(2*FracPart[p])), Int[(d*x)^m*(b/2 + c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c,
d, m, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\left (a^2+2 a b x^2+b^2 x^4\right )^{3/2}}{\sqrt{d x}} \, dx &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \frac{\left (a b+b^2 x^2\right )^3}{\sqrt{d x}} \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int \left (\frac{a^3 b^3}{\sqrt{d x}}+\frac{3 a^2 b^4 (d x)^{3/2}}{d^2}+\frac{3 a b^5 (d x)^{7/2}}{d^4}+\frac{b^6 (d x)^{11/2}}{d^6}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac{2 a^3 \sqrt{d x} \sqrt{a^2+2 a b x^2+b^2 x^4}}{d \left (a+b x^2\right )}+\frac{6 a^2 b (d x)^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d^3 \left (a+b x^2\right )}+\frac{2 a b^2 (d x)^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d^5 \left (a+b x^2\right )}+\frac{2 b^3 (d x)^{13/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 d^7 \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0216541, size = 66, normalized size = 0.34 \[ \frac{2 \sqrt{\left (a+b x^2\right )^2} \left (117 a^2 b x^3+195 a^3 x+65 a b^2 x^5+15 b^3 x^7\right )}{195 \sqrt{d x} \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a^2 + 2*a*b*x^2 + b^2*x^4)^(3/2)/Sqrt[d*x],x]

[Out]

(2*Sqrt[(a + b*x^2)^2]*(195*a^3*x + 117*a^2*b*x^3 + 65*a*b^2*x^5 + 15*b^3*x^7))/(195*Sqrt[d*x]*(a + b*x^2))

________________________________________________________________________________________

Maple [A]  time = 0.164, size = 61, normalized size = 0.3 \begin{align*}{\frac{2\, \left ( 15\,{b}^{3}{x}^{6}+65\,a{x}^{4}{b}^{2}+117\,{a}^{2}b{x}^{2}+195\,{a}^{3} \right ) x}{195\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{dx}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x)

[Out]

2/195*x*(15*b^3*x^6+65*a*b^2*x^4+117*a^2*b*x^2+195*a^3)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3/(d*x)^(1/2)

________________________________________________________________________________________

Maxima [A]  time = 1.00415, size = 117, normalized size = 0.61 \begin{align*} \frac{2 \,{\left (5 \,{\left (9 \, b^{3} \sqrt{d} x^{3} + 13 \, a b^{2} \sqrt{d} x\right )} x^{\frac{7}{2}} + 26 \,{\left (5 \, a b^{2} \sqrt{d} x^{3} + 9 \, a^{2} b \sqrt{d} x\right )} x^{\frac{3}{2}} + \frac{117 \,{\left (a^{2} b \sqrt{d} x^{3} + 5 \, a^{3} \sqrt{d} x\right )}}{\sqrt{x}}\right )}}{585 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x, algorithm="maxima")

[Out]

2/585*(5*(9*b^3*sqrt(d)*x^3 + 13*a*b^2*sqrt(d)*x)*x^(7/2) + 26*(5*a*b^2*sqrt(d)*x^3 + 9*a^2*b*sqrt(d)*x)*x^(3/
2) + 117*(a^2*b*sqrt(d)*x^3 + 5*a^3*sqrt(d)*x)/sqrt(x))/d

________________________________________________________________________________________

Fricas [A]  time = 1.30149, size = 99, normalized size = 0.51 \begin{align*} \frac{2 \,{\left (15 \, b^{3} x^{6} + 65 \, a b^{2} x^{4} + 117 \, a^{2} b x^{2} + 195 \, a^{3}\right )} \sqrt{d x}}{195 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x, algorithm="fricas")

[Out]

2/195*(15*b^3*x^6 + 65*a*b^2*x^4 + 117*a^2*b*x^2 + 195*a^3)*sqrt(d*x)/d

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b**2*x**4+2*a*b*x**2+a**2)**(3/2)/(d*x)**(1/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.23547, size = 120, normalized size = 0.62 \begin{align*} \frac{2 \,{\left (15 \, \sqrt{d x} b^{3} x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) + 65 \, \sqrt{d x} a b^{2} x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + 117 \, \sqrt{d x} a^{2} b x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) + 195 \, \sqrt{d x} a^{3} \mathrm{sgn}\left (b x^{2} + a\right )\right )}}{195 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b^2*x^4+2*a*b*x^2+a^2)^(3/2)/(d*x)^(1/2),x, algorithm="giac")

[Out]

2/195*(15*sqrt(d*x)*b^3*x^6*sgn(b*x^2 + a) + 65*sqrt(d*x)*a*b^2*x^4*sgn(b*x^2 + a) + 117*sqrt(d*x)*a^2*b*x^2*s
gn(b*x^2 + a) + 195*sqrt(d*x)*a^3*sgn(b*x^2 + a))/d