3.233 \(\int (c (a+b x^2)^2)^{3/2} \, dx\)

Optimal. Leaf size=135 \[ \frac{a^2 b c x^3 \sqrt{c \left (a+b x^2\right )^2}}{a+b x^2}+\frac{a^3 c x \sqrt{c \left (a+b x^2\right )^2}}{a+b x^2}+\frac{b^3 c x^7 \sqrt{c \left (a+b x^2\right )^2}}{7 \left (a+b x^2\right )}+\frac{3 a b^2 c x^5 \sqrt{c \left (a+b x^2\right )^2}}{5 \left (a+b x^2\right )} \]

[Out]

(a^3*c*x*Sqrt[c*(a + b*x^2)^2])/(a + b*x^2) + (a^2*b*c*x^3*Sqrt[c*(a + b*x^2)^2])/(a + b*x^2) + (3*a*b^2*c*x^5
*Sqrt[c*(a + b*x^2)^2])/(5*(a + b*x^2)) + (b^3*c*x^7*Sqrt[c*(a + b*x^2)^2])/(7*(a + b*x^2))

________________________________________________________________________________________

Rubi [A]  time = 0.0530723, antiderivative size = 175, normalized size of antiderivative = 1.3, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {1988, 1088, 194} \[ \frac{b^3 x^7 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{7 \left (a+b x^2\right )^3}+\frac{3 a b^2 x^5 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{5 \left (a+b x^2\right )^3}+\frac{a^2 b x^3 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac{a^3 x \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{\left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x^2)^2)^(3/2),x]

[Out]

(a^3*x*(a^2*c + 2*a*b*c*x^2 + b^2*c*x^4)^(3/2))/(a + b*x^2)^3 + (a^2*b*x^3*(a^2*c + 2*a*b*c*x^2 + b^2*c*x^4)^(
3/2))/(a + b*x^2)^3 + (3*a*b^2*x^5*(a^2*c + 2*a*b*c*x^2 + b^2*c*x^4)^(3/2))/(5*(a + b*x^2)^3) + (b^3*x^7*(a^2*
c + 2*a*b*c*x^2 + b^2*c*x^4)^(3/2))/(7*(a + b*x^2)^3)

Rule 1988

Int[(u_)^(p_), x_Symbol] :> Int[ExpandToSum[u, x]^p, x] /; FreeQ[p, x] && TrinomialQ[u, x] &&  !TrinomialMatch
Q[u, x]

Rule 1088

Int[((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Dist[(a + b*x^2 + c*x^4)^p/(b + 2*c*x^2)^(2*p), In
t[(b + 2*c*x^2)^(2*p), x], x] /; FreeQ[{a, b, c, p}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p - 1/2]

Rule 194

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

Rubi steps

\begin{align*} \int \left (c \left (a+b x^2\right )^2\right )^{3/2} \, dx &=\int \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2} \, dx\\ &=\frac{\left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2} \int \left (2 a b c+2 b^2 c x^2\right )^3 \, dx}{\left (2 a b c+2 b^2 c x^2\right )^3}\\ &=\frac{\left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2} \int \left (8 a^3 b^3 c^3+24 a^2 b^4 c^3 x^2+24 a b^5 c^3 x^4+8 b^6 c^3 x^6\right ) \, dx}{\left (2 a b c+2 b^2 c x^2\right )^3}\\ &=\frac{a^3 x \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac{a^2 b x^3 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{\left (a+b x^2\right )^3}+\frac{3 a b^2 x^5 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{5 \left (a+b x^2\right )^3}+\frac{b^3 x^7 \left (a^2 c+2 a b c x^2+b^2 c x^4\right )^{3/2}}{7 \left (a+b x^2\right )^3}\\ \end{align*}

Mathematica [A]  time = 0.014702, size = 61, normalized size = 0.45 \[ \frac{\left (35 a^2 b x^3+35 a^3 x+21 a b^2 x^5+5 b^3 x^7\right ) \left (c \left (a+b x^2\right )^2\right )^{3/2}}{35 \left (a+b x^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*(a + b*x^2)^2)^(3/2),x]

[Out]

((c*(a + b*x^2)^2)^(3/2)*(35*a^3*x + 35*a^2*b*x^3 + 21*a*b^2*x^5 + 5*b^3*x^7))/(35*(a + b*x^2)^3)

________________________________________________________________________________________

Maple [A]  time = 0.003, size = 58, normalized size = 0.4 \begin{align*}{\frac{x \left ( 5\,{b}^{3}{x}^{6}+21\,a{b}^{2}{x}^{4}+35\,{a}^{2}b{x}^{2}+35\,{a}^{3} \right ) }{35\, \left ( b{x}^{2}+a \right ) ^{3}} \left ( c \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((c*(b*x^2+a)^2)^(3/2),x)

[Out]

1/35*x*(5*b^3*x^6+21*a*b^2*x^4+35*a^2*b*x^2+35*a^3)*(c*(b*x^2+a)^2)^(3/2)/(b*x^2+a)^3

________________________________________________________________________________________

Maxima [A]  time = 1.03473, size = 58, normalized size = 0.43 \begin{align*} \frac{1}{7} \, b^{3} c^{\frac{3}{2}} x^{7} + \frac{3}{5} \, a b^{2} c^{\frac{3}{2}} x^{5} + a^{2} b c^{\frac{3}{2}} x^{3} + a^{3} c^{\frac{3}{2}} x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/7*b^3*c^(3/2)*x^7 + 3/5*a*b^2*c^(3/2)*x^5 + a^2*b*c^(3/2)*x^3 + a^3*c^(3/2)*x

________________________________________________________________________________________

Fricas [A]  time = 1.43896, size = 158, normalized size = 1.17 \begin{align*} \frac{{\left (5 \, b^{3} c x^{7} + 21 \, a b^{2} c x^{5} + 35 \, a^{2} b c x^{3} + 35 \, a^{3} c x\right )} \sqrt{b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{35 \,{\left (b x^{2} + a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*b^3*c*x^7 + 21*a*b^2*c*x^5 + 35*a^2*b*c*x^3 + 35*a^3*c*x)*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)/(b*x^2
+ a)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \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((c*(b*x**2+a)**2)**(3/2),x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.18606, size = 62, normalized size = 0.46 \begin{align*} \frac{1}{35} \,{\left (5 \, b^{3} x^{7} + 21 \, a b^{2} x^{5} + 35 \, a^{2} b x^{3} + 35 \, a^{3} x\right )} c^{\frac{3}{2}} \mathrm{sgn}\left (b x^{2} + a\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*b^3*x^7 + 21*a*b^2*x^5 + 35*a^2*b*x^3 + 35*a^3*x)*c^(3/2)*sgn(b*x^2 + a)