Integrand size = 19, antiderivative size = 118 \[ \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx=\frac {a c \sqrt {c x} \sqrt [4]{a+b x^2}}{6 b}+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}+\frac {a^{3/2} \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{6 \sqrt {b} \left (a+b x^2\right )^{3/4}} \] Output:
1/6*a*c*(c*x)^(1/2)*(b*x^2+a)^(1/4)/b+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)*InverseJacobiAM(1/2*arccot(b^(1/2) *x/a^(1/2)),2^(1/2))/b^(1/2)/(b*x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.72 \[ \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx=\frac {c \sqrt {c x} \sqrt [4]{a+b x^2} \left (\left (a+b x^2\right ) \sqrt [4]{1+\frac {b x^2}{a}}-a \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {5}{4},-\frac {b x^2}{a}\right )\right )}{3 b \sqrt [4]{1+\frac {b x^2}{a}}} \] Input:
Integrate[(c*x)^(3/2)*(a + b*x^2)^(1/4),x]
Output:
(c*Sqrt[c*x]*(a + b*x^2)^(1/4)*((a + b*x^2)*(1 + (b*x^2)/a)^(1/4) - a*Hype rgeometric2F1[-1/4, 1/4, 5/4, -((b*x^2)/a)]))/(3*b*(1 + (b*x^2)/a)^(1/4))
Time = 0.28 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.02, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {248, 262, 266, 768, 858, 807, 229}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{6} a \int \frac {(c x)^{3/2}}{\left (b x^2+a\right )^{3/4}}dx+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{6} a \left (\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}-\frac {a c^2 \int \frac {1}{\sqrt {c x} \left (b x^2+a\right )^{3/4}}dx}{2 b}\right )+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {1}{6} a \left (\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}-\frac {a c \int \frac {1}{\left (b x^2+a\right )^{3/4}}d\sqrt {c x}}{b}\right )+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {1}{6} a \left (\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}-\frac {a c (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4} (c x)^{3/2}}d\sqrt {c x}}{b \left (a+b x^2\right )^{3/4}}\right )+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {1}{6} a \left (\frac {a c (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\sqrt {c x} \left (\frac {a x^2 c^4}{b}+1\right )^{3/4}}d\frac {1}{\sqrt {c x}}}{b \left (a+b x^2\right )^{3/4}}+\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}\right )+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{6} a \left (\frac {a c (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a x c^3}{b}+1\right )^{3/4}}d(c x)}{2 b \left (a+b x^2\right )^{3/4}}+\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}\right )+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle \frac {1}{6} a \left (\frac {\sqrt {a} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} c^2 x}{\sqrt {b}}\right ),2\right )}{\sqrt {b} \left (a+b x^2\right )^{3/4}}+\frac {c \sqrt {c x} \sqrt [4]{a+b x^2}}{b}\right )+\frac {(c x)^{5/2} \sqrt [4]{a+b x^2}}{3 c}\) |
Input:
Int[(c*x)^(3/2)*(a + b*x^2)^(1/4),x]
Output:
((c*x)^(5/2)*(a + b*x^2)^(1/4))/(3*c) + (a*((c*Sqrt[c*x]*(a + b*x^2)^(1/4) )/b + (Sqrt[a]*(1 + a/(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcTan[(Sqrt[a] *c^2*x)/Sqrt[b]]/2, 2])/(Sqrt[b]*(a + b*x^2)^(3/4))))/6
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[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]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[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]
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] && Int egerQ[m]
\[\int \left (c x \right )^{\frac {3}{2}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}d x\]
Input:
int((c*x)^(3/2)*(b*x^2+a)^(1/4),x)
Output:
int((c*x)^(3/2)*(b*x^2+a)^(1/4),x)
\[ \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((c*x)^(3/2)*(b*x^2+a)^(1/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/4)*sqrt(c*x)*c*x, x)
Result contains complex when optimal does not.
Time = 1.21 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.39 \[ \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx=\frac {\sqrt [4]{a} c^{\frac {3}{2}} x^{\frac {5}{2}} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{4} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate((c*x)**(3/2)*(b*x**2+a)**(1/4),x)
Output:
a**(1/4)*c**(3/2)*x**(5/2)*gamma(5/4)*hyper((-1/4, 5/4), (9/4,), b*x**2*ex p_polar(I*pi)/a)/(2*gamma(9/4))
\[ \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((c*x)^(3/2)*(b*x^2+a)^(1/4),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(1/4)*(c*x)^(3/2), x)
\[ \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx=\int { {\left (b x^{2} + a\right )}^{\frac {1}{4}} \left (c x\right )^{\frac {3}{2}} \,d x } \] Input:
integrate((c*x)^(3/2)*(b*x^2+a)^(1/4),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(1/4)*(c*x)^(3/2), x)
Timed out. \[ \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx=\int {\left (c\,x\right )}^{3/2}\,{\left (b\,x^2+a\right )}^{1/4} \,d x \] Input:
int((c*x)^(3/2)*(a + b*x^2)^(1/4),x)
Output:
int((c*x)^(3/2)*(a + b*x^2)^(1/4), x)
\[ \int (c x)^{3/2} \sqrt [4]{a+b x^2} \, dx=\frac {\sqrt {c}\, c \left (2 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} a +4 \sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}-\left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b \,x^{3}+a x}d x \right ) a^{2}\right )}{12 b} \] Input:
int((c*x)^(3/2)*(b*x^2+a)^(1/4),x)
Output:
(sqrt(c)*c*(2*sqrt(x)*(a + b*x**2)**(1/4)*a + 4*sqrt(x)*(a + b*x**2)**(1/4 )*b*x**2 - int((sqrt(x)*(a + b*x**2)**(1/4))/(a*x + b*x**3),x)*a**2))/(12* b)