3.920 \(\int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx\)

Optimal. Leaf size=118 \[ \frac {a^{3/2} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}+\frac {a c \sqrt {c x} \sqrt [4]{a+b x^2}}{6 b} \]

[Out]

1/3*(c*x)^(5/2)*(b*x^2+a)^(1/4)/c+1/6*a^(3/2)*(1+a/b/x^2)^(3/4)*(c*x)^(3/2)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2))
)^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticF(sin(1/2*arccot(x*b^(1/2)/a^(1/2))),2^(1/2))/(b*x^2+a)^
(3/4)/b^(1/2)+1/6*a*c*(b*x^2+a)^(1/4)*(c*x)^(1/2)/b

________________________________________________________________________________________

Rubi [A]  time = 0.08, antiderivative size = 118, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {279, 321, 329, 237, 335, 275, 231} \[ \frac {a^{3/2} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 \sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}+\frac {a c \sqrt {c x} \sqrt [4]{a+b x^2}}{6 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(3/2)*(a + b*x^2)^(1/4),x]

[Out]

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

Rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2*EllipticF[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(3/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 237

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[(x^3*(1 + a/(b*x^4))^(3/4))/(a + b*x^4)^(3/4), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
 + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
 c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
 && FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx &=\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}+\frac {1}{6} a \int \frac {(c x)^{3/2}}{\left (a+b x^2\right )^{3/4}} \, dx\\ &=\frac {a c \sqrt {c x} \sqrt [4]{a+b x^2}}{6 b}+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}-\frac {\left (a^2 c^2\right ) \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx}{12 b}\\ &=\frac {a c \sqrt {c x} \sqrt [4]{a+b x^2}}{6 b}+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}-\frac {\left (a^2 c\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+\frac {b x^4}{c^2}\right )^{3/4}} \, dx,x,\sqrt {c x}\right )}{6 b}\\ &=\frac {a c \sqrt {c x} \sqrt [4]{a+b x^2}}{6 b}+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}-\frac {\left (a^2 c \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a c^2}{b x^4}\right )^{3/4} x^3} \, dx,x,\sqrt {c x}\right )}{6 b \left (a+b x^2\right )^{3/4}}\\ &=\frac {a c \sqrt {c x} \sqrt [4]{a+b x^2}}{6 b}+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}+\frac {\left (a^2 c \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {x}{\left (1+\frac {a c^2 x^4}{b}\right )^{3/4}} \, dx,x,\frac {1}{\sqrt {c x}}\right )}{6 b \left (a+b x^2\right )^{3/4}}\\ &=\frac {a c \sqrt {c x} \sqrt [4]{a+b x^2}}{6 b}+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}+\frac {\left (a^2 c \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a c^2 x^2}{b}\right )^{3/4}} \, dx,x,\frac {1}{c x}\right )}{12 b \left (a+b x^2\right )^{3/4}}\\ &=\frac {a c \sqrt {c x} \sqrt [4]{a+b x^2}}{6 b}+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}+\frac {a^{3/2} \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} F\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{6 \sqrt {b} \left (a+b x^2\right )^{3/4}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 0.04, size = 85, normalized size = 0.72 \[ \frac {c \sqrt {c x} \sqrt [4]{a+b x^2} \left (\left (a+b x^2\right ) \sqrt [4]{\frac {b x^2}{a}+1}-a \, _2F_1\left (-\frac {1}{4},\frac {1}{4};\frac {5}{4};-\frac {b x^2}{a}\right )\right )}{3 b \sqrt [4]{\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c*Sqrt[c*x]*(a + b*x^2)^(1/4)*((a + b*x^2)*(1 + (b*x^2)/a)^(1/4) - a*Hypergeometric2F1[-1/4, 1/4, 5/4, -((b*x
^2)/a)]))/(3*b*(1 + (b*x^2)/a)^(1/4))

________________________________________________________________________________________

fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x} c x, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(3/2)*(b*x^2+a)^(1/4),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(1/4)*sqrt(c*x)*c*x, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(3/2)*(b*x^2+a)^(1/4),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(1/4)*(c*x)^(3/2), x)

________________________________________________________________________________________

maple [F]  time = 0.06, size = 0, normalized size = 0.00 \[ \int \left (c x \right )^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(3/2)*(b*x^2+a)^(1/4),x)

[Out]

int((c*x)^(3/2)*(b*x^2+a)^(1/4),x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {3}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(3/2)*(b*x^2+a)^(1/4),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(1/4)*(c*x)^(3/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (c\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^{1/4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(3/2)*(a + b*x^2)^(1/4),x)

[Out]

int((c*x)^(3/2)*(a + b*x^2)^(1/4), x)

________________________________________________________________________________________

sympy [C]  time = 2.71, size = 46, normalized size = 0.39 \[ \frac {\sqrt [4]{a} c^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(3/2)*(b*x**2+a)**(1/4),x)

[Out]

a**(1/4)*c**(3/2)*x**(5/2)*gamma(5/4)*hyper((-1/4, 5/4), (9/4,), b*x**2*exp_polar(I*pi)/a)/(2*gamma(9/4))

________________________________________________________________________________________