3.3.35 \(\int \frac {(c (a+b x^2)^2)^{3/2}}{x^2} \, dx\) [235]

Optimal. Leaf size=134 \[ -\frac {a^3 c \sqrt {c \left (a+b x^2\right )^2}}{x \left (a+b x^2\right )}+\frac {3 a^2 b c x \sqrt {c \left (a+b x^2\right )^2}}{a+b x^2}+\frac {a b^2 c x^3 \sqrt {c \left (a+b x^2\right )^2}}{a+b x^2}+\frac {b^3 c x^5 \sqrt {c \left (a+b x^2\right )^2}}{5 \left (a+b x^2\right )} \]

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {1973, 276} \begin {gather*} -\frac {a^3 c \sqrt {c \left (a+b x^2\right )^2}}{x \left (a+b x^2\right )}+\frac {3 a^2 b c x \sqrt {c \left (a+b x^2\right )^2}}{a+b x^2}+\frac {b^3 c x^5 \sqrt {c \left (a+b x^2\right )^2}}{5 \left (a+b x^2\right )}+\frac {a b^2 c x^3 \sqrt {c \left (a+b x^2\right )^2}}{a+b x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*(a + b*x^2)^2)^(3/2)/x^2,x]

[Out]

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

Rule 276

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]

Rule 1973

Int[(u_.)*((c_.)*((a_) + (b_.)*(x_)^(n_.))^(q_))^(p_), x_Symbol] :> Dist[Simp[(c*(a + b*x^n)^q)^p/(1 + b*(x^n/
a))^(p*q)], Int[u*(1 + b*(x^n/a))^(p*q), x], x] /; FreeQ[{a, b, c, n, p, q}, x] &&  !GeQ[a, 0]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 62, normalized size = 0.46 \begin {gather*} \frac {\left (c \left (a+b x^2\right )^2\right )^{3/2} \left (-5 a^3+15 a^2 b x^2+5 a b^2 x^4+b^3 x^6\right )}{5 x \left (a+b x^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((c*(a + b*x^2)^2)^(3/2)*(-5*a^3 + 15*a^2*b*x^2 + 5*a*b^2*x^4 + b^3*x^6))/(5*x*(a + b*x^2)^3)

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Maple [A]
time = 0.04, size = 60, normalized size = 0.45

method	result	size
gosper	\(-\frac {\left (-b^{3} x^{6}-5 a \,b^{2} x^{4}-15 a^{2} b \,x^{2}+5 a^{3}\right ) \left (c \left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{5 x \left (b \,x^{2}+a \right )^{3}}\)	\(60\)
default	\(-\frac {\left (-b^{3} x^{6}-5 a \,b^{2} x^{4}-15 a^{2} b \,x^{2}+5 a^{3}\right ) \left (c \left (b \,x^{2}+a \right )^{2}\right )^{\frac {3}{2}}}{5 x \left (b \,x^{2}+a \right )^{3}}\)	\(60\)
risch	\(\frac {c \sqrt {c \left (b \,x^{2}+a \right )^{2}}\, b \left (\frac {1}{5} b^{2} x^{5}+a b \,x^{3}+3 a^{2} x \right )}{b \,x^{2}+a}-\frac {a^{3} c \sqrt {c \left (b \,x^{2}+a \right )^{2}}}{x \left (b \,x^{2}+a \right )}\)	\(79\)
trager	\(\frac {c \left (b^{3} x^{5}+b^{3} x^{4}+5 a \,b^{2} x^{3}+b^{3} x^{3}+5 a \,b^{2} x^{2}+b^{3} x^{2}+15 a^{2} b x +5 a \,b^{2} x +b^{3} x +5 a^{3}\right ) \left (-1+x \right ) \sqrt {b^{2} c \,x^{4}+2 a b c \,x^{2}+a^{2} c}}{5 x \left (b \,x^{2}+a \right )}\)	\(114\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(b*x^2+a)^2)^(3/2)/x^2,x,method=_RETURNVERBOSE)

[Out]

-1/5*(-b^3*x^6-5*a*b^2*x^4-15*a^2*b*x^2+5*a^3)*(c*(b*x^2+a)^2)^(3/2)/x/(b*x^2+a)^3

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Maxima [A]
time = 0.30, size = 48, normalized size = 0.36 \begin {gather*} \frac {b^{3} c^{\frac {3}{2}} x^{6} + 5 \, a b^{2} c^{\frac {3}{2}} x^{4} + 15 \, a^{2} b c^{\frac {3}{2}} x^{2} - 5 \, a^{3} c^{\frac {3}{2}}}{5 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b^3*c^(3/2)*x^6 + 5*a*b^2*c^(3/2)*x^4 + 15*a^2*b*c^(3/2)*x^2 - 5*a^3*c^(3/2))/x

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Fricas [A]
time = 0.33, size = 72, normalized size = 0.54 \begin {gather*} \frac {{\left (b^{3} c x^{6} + 5 \, a b^{2} c x^{4} + 15 \, a^{2} b c x^{2} - 5 \, a^{3} c\right )} \sqrt {b^{2} c x^{4} + 2 \, a b c x^{2} + a^{2} c}}{5 \, {\left (b x^{3} + a x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b^3*c*x^6 + 5*a*b^2*c*x^4 + 15*a^2*b*c*x^2 - 5*a^3*c)*sqrt(b^2*c*x^4 + 2*a*b*c*x^2 + a^2*c)/(b*x^3 + a*x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (c \left (a + b x^{2}\right )^{2}\right )^{\frac {3}{2}}}{x^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 3.57, size = 69, normalized size = 0.51 \begin {gather*} \frac {1}{5} \, {\left (b^{3} x^{5} \mathrm {sgn}\left (b x^{2} + a\right ) + 5 \, a b^{2} x^{3} \mathrm {sgn}\left (b x^{2} + a\right ) + 15 \, a^{2} b x \mathrm {sgn}\left (b x^{2} + a\right ) - \frac {5 \, a^{3} \mathrm {sgn}\left (b x^{2} + a\right )}{x}\right )} c^{\frac {3}{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(b^3*x^5*sgn(b*x^2 + a) + 5*a*b^2*x^3*sgn(b*x^2 + a) + 15*a^2*b*x*sgn(b*x^2 + a) - 5*a^3*sgn(b*x^2 + a)/x)
*c^(3/2)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (c\,{\left (b\,x^2+a\right )}^2\right )}^{3/2}}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(a + b*x^2)^2)^(3/2)/x^2,x)

[Out]

int((c*(a + b*x^2)^2)^(3/2)/x^2, x)

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