3.1.65 \(\int (a+b x^2)^{5/2} \, dx\) [65]

Optimal. Leaf size=84 \[ \frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}} \]

[Out]

5/24*a*x*(b*x^2+a)^(3/2)+1/6*x*(b*x^2+a)^(5/2)+5/16*a^3*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/b^(1/2)+5/16*a^2*x*
(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {201, 223, 212} \begin {gather*} \frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}+\frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(5/2),x]

[Out]

(5*a^2*x*Sqrt[a + b*x^2])/16 + (5*a*x*(a + b*x^2)^(3/2))/24 + (x*(a + b*x^2)^(5/2))/6 + (5*a^3*ArcTanh[(Sqrt[b
]*x)/Sqrt[a + b*x^2]])/(16*Sqrt[b])

Rule 201

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 )^{5/2} \, dx &=\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {1}{6} (5 a) \int \left (a+b x^2\right )^{3/2} \, dx\\ &=\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {1}{8} \left (5 a^2\right ) \int \sqrt {a+b x^2} \, dx\\ &=\frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {1}{16} \left (5 a^3\right ) \int \frac {1}{\sqrt {a+b x^2}} \, dx\\ &=\frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {1}{16} \left (5 a^3\right ) \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {x}{\sqrt {a+b x^2}}\right )\\ &=\frac {5}{16} a^2 x \sqrt {a+b x^2}+\frac {5}{24} a x \left (a+b x^2\right )^{3/2}+\frac {1}{6} x \left (a+b x^2\right )^{5/2}+\frac {5 a^3 \tanh ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{16 \sqrt {b}}\\ \end {align*}

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Mathematica [A]
time = 0.02, size = 71, normalized size = 0.85 \begin {gather*} \frac {1}{48} \sqrt {a+b x^2} \left (33 a^2 x+26 a b x^3+8 b^2 x^5\right )-\frac {5 a^3 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{16 \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(5/2),x]

[Out]

(Sqrt[a + b*x^2]*(33*a^2*x + 26*a*b*x^3 + 8*b^2*x^5))/48 - (5*a^3*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(16*Sqr
t[b])

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Maple [A]
time = 0.05, size = 68, normalized size = 0.81

method	result	size
risch	\(\frac {x \left (8 b^{2} x^{4}+26 a b \,x^{2}+33 a^{2}\right ) \sqrt {b \,x^{2}+a}}{48}+\frac {5 a^{3} \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{16 \sqrt {b}}\)	\(59\)
default	\(\frac {x \left (b \,x^{2}+a \right )^{\frac {5}{2}}}{6}+\frac {5 a \left (\frac {x \left (b \,x^{2}+a \right )^{\frac {3}{2}}}{4}+\frac {3 a \left (\frac {x \sqrt {b \,x^{2}+a}}{2}+\frac {a \ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{2 \sqrt {b}}\right )}{4}\right )}{6}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/6*x*(b*x^2+a)^(5/2)+5/6*a*(1/4*x*(b*x^2+a)^(3/2)+3/4*a*(1/2*x*(b*x^2+a)^(1/2)+1/2*a/b^(1/2)*ln(x*b^(1/2)+(b*
x^2+a)^(1/2))))

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Maxima [A]
time = 0.31, size = 58, normalized size = 0.69 \begin {gather*} \frac {1}{6} \, {\left (b x^{2} + a\right )}^{\frac {5}{2}} x + \frac {5}{24} \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a x + \frac {5}{16} \, \sqrt {b x^{2} + a} a^{2} x + \frac {5 \, a^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{16 \, \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

1/6*(b*x^2 + a)^(5/2)*x + 5/24*(b*x^2 + a)^(3/2)*a*x + 5/16*sqrt(b*x^2 + a)*a^2*x + 5/16*a^3*arcsinh(b*x/sqrt(
a*b))/sqrt(b)

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Fricas [A]
time = 0.50, size = 146, normalized size = 1.74 \begin {gather*} \left [\frac {15 \, a^{3} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (8 \, b^{3} x^{5} + 26 \, a b^{2} x^{3} + 33 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{96 \, b}, -\frac {15 \, a^{3} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (8 \, b^{3} x^{5} + 26 \, a b^{2} x^{3} + 33 \, a^{2} b x\right )} \sqrt {b x^{2} + a}}{48 \, b}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/96*(15*a^3*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(8*b^3*x^5 + 26*a*b^2*x^3 + 33*a^2*b
*x)*sqrt(b*x^2 + a))/b, -1/48*(15*a^3*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (8*b^3*x^5 + 26*a*b^2*x^3
+ 33*a^2*b*x)*sqrt(b*x^2 + a))/b]

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Sympy [A]
time = 2.78, size = 97, normalized size = 1.15 \begin {gather*} \frac {11 a^{\frac {5}{2}} x \sqrt {1 + \frac {b x^{2}}{a}}}{16} + \frac {13 a^{\frac {3}{2}} b x^{3} \sqrt {1 + \frac {b x^{2}}{a}}}{24} + \frac {\sqrt {a} b^{2} x^{5} \sqrt {1 + \frac {b x^{2}}{a}}}{6} + \frac {5 a^{3} \operatorname {asinh}{\left (\frac {\sqrt {b} x}{\sqrt {a}} \right )}}{16 \sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(5/2),x)

[Out]

11*a**(5/2)*x*sqrt(1 + b*x**2/a)/16 + 13*a**(3/2)*b*x**3*sqrt(1 + b*x**2/a)/24 + sqrt(a)*b**2*x**5*sqrt(1 + b*
x**2/a)/6 + 5*a**3*asinh(sqrt(b)*x/sqrt(a))/(16*sqrt(b))

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Giac [A]
time = 0.58, size = 63, normalized size = 0.75 \begin {gather*} -\frac {5 \, a^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{16 \, \sqrt {b}} + \frac {1}{48} \, {\left (2 \, {\left (4 \, b^{2} x^{2} + 13 \, a b\right )} x^{2} + 33 \, a^{2}\right )} \sqrt {b x^{2} + a} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(5/2),x, algorithm="giac")

[Out]

-5/16*a^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/sqrt(b) + 1/48*(2*(4*b^2*x^2 + 13*a*b)*x^2 + 33*a^2)*sqrt(b*x
^2 + a)*x

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Mupad [B]
time = 4.69, size = 37, normalized size = 0.44 \begin {gather*} \frac {x\,{\left (b\,x^2+a\right )}^{5/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,x^2}{a}\right )}{{\left (\frac {b\,x^2}{a}+1\right )}^{5/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(5/2),x)

[Out]

(x*(a + b*x^2)^(5/2)*hypergeom([-5/2, 1/2], 3/2, -(b*x^2)/a))/((b*x^2)/a + 1)^(5/2)

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