\(\int (c \sqrt {a+b x^2})^{3/2} \, dx\) [256]

Optimal result

Integrand size = 17, antiderivative size = 119 \[ \int \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {2}{5} x \left (c \sqrt {a+b x^2}\right )^{3/2}+\frac {6 a x \left (c \sqrt {a+b x^2}\right )^{3/2}}{5 \left (a+b x^2\right )}-\frac {6 \sqrt {a} \left (c \sqrt {a+b x^2}\right )^{3/2} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 \sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/4}} \]

[Out]

2/5*x*(c*(b*x^2+a)^(1/2))^(3/2)+6/5*a*x*(c*(b*x^2+a)^(1/2))^(3/2)/(b*x^2+a)-6/5*(cos(1/2*arctan(x*b^(1/2)/a^(1
/2)))^2)^(1/2)/cos(1/2*arctan(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arctan(x*b^(1/2)/a^(1/2))),2^(1/2))*a^(1/2
)*(c*(b*x^2+a)^(1/2))^(3/2)/(1+b*x^2/a)^(3/4)/b^(1/2)

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {1973, 201, 233, 202} \[ \int \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=-\frac {6 \sqrt {a} \left (c \sqrt {a+b x^2}\right )^{3/2} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 \sqrt {b} \left (\frac {b x^2}{a}+1\right )^{3/4}}+\frac {2}{5} x \left (c \sqrt {a+b x^2}\right )^{3/2}+\frac {6 a x \left (c \sqrt {a+b x^2}\right )^{3/2}}{5 \left (a+b x^2\right )} \]

[In]

Int[(c*Sqrt[a + b*x^2])^(3/2),x]

[Out]

(2*x*(c*Sqrt[a + b*x^2])^(3/2))/5 + (6*a*x*(c*Sqrt[a + b*x^2])^(3/2))/(5*(a + b*x^2)) - (6*Sqrt[a]*(c*Sqrt[a +
b*x^2])^(3/2)*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*Sqrt[b]*(1 + (b*x^2)/a)^(3/4))

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 202

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]))*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]
*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 233

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

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*} \text {integral}& = \frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \int \left (1+\frac {b x^2}{a}\right )^{3/4} \, dx}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \\ & = \frac {2}{5} x \left (c \sqrt {a+b x^2}\right )^{3/2}+\frac {\left (3 \left (c \sqrt {a+b x^2}\right )^{3/2}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{5 \left (1+\frac {b x^2}{a}\right )^{3/4}} \\ & = \frac {2}{5} x \left (c \sqrt {a+b x^2}\right )^{3/2}+\frac {6 a x \left (c \sqrt {a+b x^2}\right )^{3/2}}{5 \left (a+b x^2\right )}-\frac {\left (3 \left (c \sqrt {a+b x^2}\right )^{3/2}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{5 \left (1+\frac {b x^2}{a}\right )^{3/4}} \\ & = \frac {2}{5} x \left (c \sqrt {a+b x^2}\right )^{3/2}+\frac {6 a x \left (c \sqrt {a+b x^2}\right )^{3/2}}{5 \left (a+b x^2\right )}-\frac {6 \sqrt {a} \left (c \sqrt {a+b x^2}\right )^{3/2} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 \sqrt {b} \left (1+\frac {b x^2}{a}\right )^{3/4}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 0.01 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.44 \[ \int \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\frac {x \left (c \sqrt {a+b x^2}\right )^{3/2} \operatorname {Hypergeometric2F1}\left (-\frac {3}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )}{\left (1+\frac {b x^2}{a}\right )^{3/4}} \]

[In]

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

[Out]

(x*(c*Sqrt[a + b*x^2])^(3/2)*Hypergeometric2F1[-3/4, 1/2, 3/2, -((b*x^2)/a)])/(1 + (b*x^2)/a)^(3/4)

Maple [F]

\[\int \left (c \sqrt {b \,x^{2}+a}\right )^{\frac {3}{2}}d x\]

[In]

int((c*(b*x^2+a)^(1/2))^(3/2),x)

[Out]

int((c*(b*x^2+a)^(1/2))^(3/2),x)

Fricas [F]

\[ \int \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int { \left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^2 + a)*sqrt(sqrt(b*x^2 + a)*c)*c, x)

Sympy [F]

\[ \int \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int \left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}\, dx \]

[In]

integrate((c*(b*x**2+a)**(1/2))**(3/2),x)

[Out]

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

Maxima [F]

\[ \int \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int { \left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

\[ \int \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int { \left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}} \,d x } \]

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \left (c \sqrt {a+b x^2}\right )^{3/2} \, dx=\int {\left (c\,\sqrt {b\,x^2+a}\right )}^{3/2} \,d x \]

[In]

int((c*(a + b*x^2)^(1/2))^(3/2),x)

[Out]

int((c*(a + b*x^2)^(1/2))^(3/2), x)