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

Optimal. Leaf size=195 \[ \frac{2 b^3 (d x)^{17/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{17 d^7 \left (a+b x^2\right )}+\frac{6 a b^2 (d x)^{13/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 d^5 \left (a+b x^2\right )}+\frac{2 a^2 b (d x)^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d^3 \left (a+b x^2\right )}+\frac{2 a^3 (d x)^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )} \]

[Out]

(2*a^3*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d*(a + b*x^2)) + (2*a^2*b*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4])/(3*d^3*(a + b*x^2)) + (6*a*b^2*(d*x)^(13/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*d^5*(a + b*x^
2)) + (2*b^3*(d*x)^(17/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(17*d^7*(a + b*x^2))

________________________________________________________________________________________

Rubi [A]  time = 0.0583482, antiderivative size = 195, 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} \[ \frac{2 b^3 (d x)^{17/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{17 d^7 \left (a+b x^2\right )}+\frac{6 a b^2 (d x)^{13/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 d^5 \left (a+b x^2\right )}+\frac{2 a^2 b (d x)^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d^3 \left (a+b x^2\right )}+\frac{2 a^3 (d x)^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^3*(d*x)^(5/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(5*d*(a + b*x^2)) + (2*a^2*b*(d*x)^(9/2)*Sqrt[a^2 + 2*a*b*
x^2 + b^2*x^4])/(3*d^3*(a + b*x^2)) + (6*a*b^2*(d*x)^(13/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(13*d^5*(a + b*x^
2)) + (2*b^3*(d*x)^(17/2)*Sqrt[a^2 + 2*a*b*x^2 + b^2*x^4])/(17*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 (d x)^{3/2} \left (a^2+2 a b x^2+b^2 x^4\right )^{3/2} \, dx &=\frac{\sqrt{a^2+2 a b x^2+b^2 x^4} \int (d x)^{3/2} \left (a b+b^2 x^2\right )^3 \, 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 (a^3 b^3 (d x)^{3/2}+\frac{3 a^2 b^4 (d x)^{7/2}}{d^2}+\frac{3 a b^5 (d x)^{11/2}}{d^4}+\frac{b^6 (d x)^{15/2}}{d^6}\right ) \, dx}{b^2 \left (a b+b^2 x^2\right )}\\ &=\frac{2 a^3 (d x)^{5/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{5 d \left (a+b x^2\right )}+\frac{2 a^2 b (d x)^{9/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{3 d^3 \left (a+b x^2\right )}+\frac{6 a b^2 (d x)^{13/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{13 d^5 \left (a+b x^2\right )}+\frac{2 b^3 (d x)^{17/2} \sqrt{a^2+2 a b x^2+b^2 x^4}}{17 d^7 \left (a+b x^2\right )}\\ \end{align*}

Mathematica [A]  time = 0.0237474, size = 66, normalized size = 0.34 \[ \frac{2 x (d x)^{3/2} \sqrt{\left (a+b x^2\right )^2} \left (1105 a^2 b x^2+663 a^3+765 a b^2 x^4+195 b^3 x^6\right )}{3315 \left (a+b x^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*x*(d*x)^(3/2)*Sqrt[(a + b*x^2)^2]*(663*a^3 + 1105*a^2*b*x^2 + 765*a*b^2*x^4 + 195*b^3*x^6))/(3315*(a + b*x^
2))

________________________________________________________________________________________

Maple [A]  time = 0.165, size = 61, normalized size = 0.3 \begin{align*}{\frac{2\,x \left ( 195\,{b}^{3}{x}^{6}+765\,a{x}^{4}{b}^{2}+1105\,{a}^{2}b{x}^{2}+663\,{a}^{3} \right ) }{3315\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( dx \right ) ^{{\frac{3}{2}}} \left ( \left ( b{x}^{2}+a \right ) ^{2} \right ) ^{{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3315*x*(195*b^3*x^6+765*a*b^2*x^4+1105*a^2*b*x^2+663*a^3)*(d*x)^(3/2)*((b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

________________________________________________________________________________________

Maxima [A]  time = 0.9979, size = 112, normalized size = 0.57 \begin{align*} \frac{2}{221} \,{\left (13 \, b^{3} d^{\frac{3}{2}} x^{3} + 17 \, a b^{2} d^{\frac{3}{2}} x\right )} x^{\frac{11}{2}} + \frac{4}{117} \,{\left (9 \, a b^{2} d^{\frac{3}{2}} x^{3} + 13 \, a^{2} b d^{\frac{3}{2}} x\right )} x^{\frac{7}{2}} + \frac{2}{45} \,{\left (5 \, a^{2} b d^{\frac{3}{2}} x^{3} + 9 \, a^{3} d^{\frac{3}{2}} x\right )} x^{\frac{3}{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/221*(13*b^3*d^(3/2)*x^3 + 17*a*b^2*d^(3/2)*x)*x^(11/2) + 4/117*(9*a*b^2*d^(3/2)*x^3 + 13*a^2*b*d^(3/2)*x)*x^
(7/2) + 2/45*(5*a^2*b*d^(3/2)*x^3 + 9*a^3*d^(3/2)*x)*x^(3/2)

________________________________________________________________________________________

Fricas [A]  time = 1.20209, size = 117, normalized size = 0.6 \begin{align*} \frac{2}{3315} \,{\left (195 \, b^{3} d x^{8} + 765 \, a b^{2} d x^{6} + 1105 \, a^{2} b d x^{4} + 663 \, a^{3} d x^{2}\right )} \sqrt{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3315*(195*b^3*d*x^8 + 765*a*b^2*d*x^6 + 1105*a^2*b*d*x^4 + 663*a^3*d*x^2)*sqrt(d*x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.20944, size = 123, normalized size = 0.63 \begin{align*} \frac{2}{17} \, \sqrt{d x} b^{3} d x^{8} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{6}{13} \, \sqrt{d x} a b^{2} d x^{6} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{2}{3} \, \sqrt{d x} a^{2} b d x^{4} \mathrm{sgn}\left (b x^{2} + a\right ) + \frac{2}{5} \, \sqrt{d x} a^{3} d x^{2} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/17*sqrt(d*x)*b^3*d*x^8*sgn(b*x^2 + a) + 6/13*sqrt(d*x)*a*b^2*d*x^6*sgn(b*x^2 + a) + 2/3*sqrt(d*x)*a^2*b*d*x^
4*sgn(b*x^2 + a) + 2/5*sqrt(d*x)*a^3*d*x^2*sgn(b*x^2 + a)