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

Optimal. Leaf size=65 \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}+\frac {3}{8} a x \sqrt {a+b x^2}+\frac {1}{4} x \left (a+b x^2\right )^{3/2} \]

[Out]

1/4*x*(b*x^2+a)^(3/2)+3/8*a^2*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(1/2)+3/8*a*x*(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.01, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {195, 217, 206} \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 \sqrt {b}}+\frac {3}{8} a x \sqrt {a+b x^2}+\frac {1}{4} x \left (a+b x^2\right )^{3/2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2),x]

[Out]

(3*a*x*Sqrt[a + b*x^2])/8 + (x*(a + b*x^2)^(3/2))/4 + (3*a^2*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(8*Sqrt[b])

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.09, size = 65, normalized size = 1.00 \[ \frac {1}{8} \sqrt {a+b x^2} \left (\frac {3 a^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {\frac {b x^2}{a}+1}}+5 a x+2 b x^3\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2),x]

[Out]

(Sqrt[a + b*x^2]*(5*a*x + 2*b*x^3 + (3*a^(3/2)*ArcSinh[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[b]*Sqrt[1 + (b*x^2)/a])))/8

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fricas [A]  time = 1.06, size = 124, normalized size = 1.91 \[ \left [\frac {3 \, a^{2} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, b^{2} x^{3} + 5 \, a b x\right )} \sqrt {b x^{2} + a}}{16 \, b}, -\frac {3 \, a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, b^{2} x^{3} + 5 \, a b x\right )} \sqrt {b x^{2} + a}}{8 \, b}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/16*(3*a^2*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*b^2*x^3 + 5*a*b*x)*sqrt(b*x^2 + a)
)/b, -1/8*(3*a^2*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*b^2*x^3 + 5*a*b*x)*sqrt(b*x^2 + a))/b]

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giac [A]  time = 0.63, size = 49, normalized size = 0.75 \[ \frac {1}{8} \, {\left (2 \, b x^{2} + 5 \, a\right )} \sqrt {b x^{2} + a} x - \frac {3 \, a^{2} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 51, normalized size = 0.78 \[ \frac {3 a^{2} \ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{8 \sqrt {b}}+\frac {3 \sqrt {b \,x^{2}+a}\, a x}{8}+\frac {\left (b \,x^{2}+a \right )^{\frac {3}{2}} x}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*x*(b*x^2+a)^(3/2)+3/8*a*x*(b*x^2+a)^(1/2)+3/8*a^2/b^(1/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))

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maxima [A]  time = 1.31, size = 43, normalized size = 0.66 \[ \frac {1}{4} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} x + \frac {3}{8} \, \sqrt {b x^{2} + a} a x + \frac {3 \, a^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(b*x^2 + a)^(3/2)*x + 3/8*sqrt(b*x^2 + a)*a*x + 3/8*a^2*arcsinh(b*x/sqrt(a*b))/sqrt(b)

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mupad [B]  time = 4.61, size = 37, normalized size = 0.57 \[ \frac {x\,{\left (b\,x^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2),x)

[Out]

(x*(a + b*x^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(3/2)

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sympy [A]  time = 2.90, size = 70, normalized size = 1.08 \[ \frac {5 a^{\frac {3}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{8} + \frac {\sqrt {a} b x^{3} \sqrt {1 + \frac {b x^{2}}{a}}}{4} + \frac {3 a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{8 \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

5*a**(3/2)*x*sqrt(1 + b*x**2/a)/8 + sqrt(a)*b*x**3*sqrt(1 + b*x**2/a)/4 + 3*a**2*asinh(sqrt(b)*x/sqrt(a))/(8*s
qrt(b))

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