3.10.44 \(\int \frac {(c x)^{3/2}}{\sqrt [4]{a+b x^2}} \, dx\) [944]

Optimal. Leaf size=117 \[ \frac {c \sqrt {c x} \left (a+b x^2\right )^{3/4}}{2 b}-\frac {a c^{3/2} \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{5/4}}-\frac {a c^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{5/4}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.05, antiderivative size = 117, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {327, 335, 246, 218, 214, 211} \begin {gather*} -\frac {a c^{3/2} \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{5/4}}-\frac {a c^{3/2} \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{4 b^{5/4}}+\frac {c \sqrt {c x} \left (a+b x^2\right )^{3/4}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x)^(3/2)/(a + b*x^2)^(1/4),x]

[Out]

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

Rule 211

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

Rule 214

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

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 246

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

Rule 327

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 335

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]

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]
time = 0.27, size = 97, normalized size = 0.83 \begin {gather*} \frac {(c x)^{3/2} \left (2 \sqrt [4]{b} \sqrt {x} \left (a+b x^2\right )^{3/4}-a \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )-a \tanh ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{4 b^{5/4} x^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (c x \right )^{\frac {3}{2}}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx\]

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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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((c*x)^(3/2)/(b*x^2 + a)^(1/4), x)

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (83) = 166\).
time = 1.61, size = 314, normalized size = 2.68 \begin {gather*} \frac {4 \, {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} c + 4 \, \left (\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} b \arctan \left (-\frac {\left (\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {3}{4}} {\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a b^{4} c - {\left (b^{5} x^{2} + a b^{4}\right )} \left (\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {3}{4}} \sqrt {\frac {\sqrt {b x^{2} + a} a^{2} c^{3} x + \sqrt {\frac {a^{4} c^{6}}{b^{5}}} {\left (b^{3} x^{2} + a b^{2}\right )}}{b x^{2} + a}}}{a^{4} b c^{6} x^{2} + a^{5} c^{6}}\right ) - \left (\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c + \left (\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} {\left (b^{2} x^{2} + a b\right )}}{b x^{2} + a}\right ) + \left (\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} b \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} a c - \left (\frac {a^{4} c^{6}}{b^{5}}\right )^{\frac {1}{4}} {\left (b^{2} x^{2} + a b\right )}}{b x^{2} + a}\right )}{8 \, b} \end {gather*}

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]

1/8*(4*(b*x^2 + a)^(3/4)*sqrt(c*x)*c + 4*(a^4*c^6/b^5)^(1/4)*b*arctan(-((a^4*c^6/b^5)^(3/4)*(b*x^2 + a)^(3/4)*
sqrt(c*x)*a*b^4*c - (b^5*x^2 + a*b^4)*(a^4*c^6/b^5)^(3/4)*sqrt((sqrt(b*x^2 + a)*a^2*c^3*x + sqrt(a^4*c^6/b^5)*
(b^3*x^2 + a*b^2))/(b*x^2 + a)))/(a^4*b*c^6*x^2 + a^5*c^6)) - (a^4*c^6/b^5)^(1/4)*b*log(((b*x^2 + a)^(3/4)*sqr
t(c*x)*a*c + (a^4*c^6/b^5)^(1/4)*(b^2*x^2 + a*b))/(b*x^2 + a)) + (a^4*c^6/b^5)^(1/4)*b*log(((b*x^2 + a)^(3/4)*
sqrt(c*x)*a*c - (a^4*c^6/b^5)^(1/4)*(b^2*x^2 + a*b))/(b*x^2 + a)))/b

________________________________________________________________________________________

Sympy [C] Result contains complex when optimal does not.
time = 1.29, size = 44, normalized size = 0.38 \begin {gather*} \frac {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 \sqrt [4]{a} \Gamma \left (\frac {9}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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*a**(1/4)*gamma(9/4))

________________________________________________________________________________________

Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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((c*x)^(3/2)/(b*x^2 + a)^(1/4), x)

________________________________________________________________________________________

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

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)

________________________________________________________________________________________