Integrand size = 19, antiderivative size = 83 \[ \int \frac {1}{\sqrt {c x} \sqrt [4]{a+b x^2}} \, dx=\frac {\arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{\sqrt [4]{b} \sqrt {c}}+\frac {\text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{\sqrt [4]{b} \sqrt {c}} \] Output:
arctan(b^(1/4)*(c*x)^(1/2)/c^(1/2)/(b*x^2+a)^(1/4))/b^(1/4)/c^(1/2)+arctan h(b^(1/4)*(c*x)^(1/2)/c^(1/2)/(b*x^2+a)^(1/4))/b^(1/4)/c^(1/2)
Time = 0.18 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.78 \[ \int \frac {1}{\sqrt {c x} \sqrt [4]{a+b x^2}} \, dx=\frac {\sqrt {x} \left (\arctan \left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )+\text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a+b x^2}}\right )\right )}{\sqrt [4]{b} \sqrt {c x}} \] Input:
Integrate[1/(Sqrt[c*x]*(a + b*x^2)^(1/4)),x]
Output:
(Sqrt[x]*(ArcTan[(b^(1/4)*Sqrt[x])/(a + b*x^2)^(1/4)] + ArcTanh[(b^(1/4)*S qrt[x])/(a + b*x^2)^(1/4)]))/(b^(1/4)*Sqrt[c*x])
Time = 0.21 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.13, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {266, 770, 756, 218, 221}
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.
\(\displaystyle \int \frac {1}{\sqrt {c x} \sqrt [4]{a+b x^2}} \, dx\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 \int \frac {1}{\sqrt [4]{b x^2+a}}d\sqrt {c x}}{c}\) |
\(\Big \downarrow \) 770 |
\(\displaystyle \frac {2 \int \frac {1}{1-b x^2}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}}{c}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {2 \left (\frac {1}{2} c \int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}+\frac {1}{2} c \int \frac {1}{\sqrt {b} x c+c}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}\right )}{c}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {2 \left (\frac {1}{2} c \int \frac {1}{c-\sqrt {b} c x}d\frac {\sqrt {c x}}{\sqrt [4]{b x^2+a}}+\frac {\sqrt {c} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{c}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \left (\frac {\sqrt {c} \arctan \left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}+\frac {\sqrt {c} \text {arctanh}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt {c} \sqrt [4]{a+b x^2}}\right )}{2 \sqrt [4]{b}}\right )}{c}\) |
Input:
Int[1/(Sqrt[c*x]*(a + b*x^2)^(1/4)),x]
Output:
(2*((Sqrt[c]*ArcTan[(b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4))])/(2*b ^(1/4)) + (Sqrt[c]*ArcTanh[(b^(1/4)*Sqrt[c*x])/(Sqrt[c]*(a + b*x^2)^(1/4)) ])/(2*b^(1/4))))/c
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
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]
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)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^(p + 1/n) Subst[In t[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]
\[\int \frac {1}{\sqrt {c x}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}}}d x\]
Input:
int(1/(c*x)^(1/2)/(b*x^2+a)^(1/4),x)
Output:
int(1/(c*x)^(1/2)/(b*x^2+a)^(1/4),x)
Result contains complex when optimal does not.
Time = 0.10 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.89 \[ \int \frac {1}{\sqrt {c x} \sqrt [4]{a+b x^2}} \, dx=\frac {1}{2} \, \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} + {\left (b c x^{2} + a c\right )} \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) - \frac {1}{2} \, \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} - {\left (b c x^{2} + a c\right )} \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) - \frac {1}{2} i \, \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} - {\left (i \, b c x^{2} + i \, a c\right )} \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) + \frac {1}{2} i \, \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}} \log \left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} - {\left (-i \, b c x^{2} - i \, a c\right )} \left (\frac {1}{b c^{2}}\right )^{\frac {1}{4}}}{b x^{2} + a}\right ) \] Input:
integrate(1/(c*x)^(1/2)/(b*x^2+a)^(1/4),x, algorithm="fricas")
Output:
1/2*(1/(b*c^2))^(1/4)*log(((b*x^2 + a)^(3/4)*sqrt(c*x) + (b*c*x^2 + a*c)*( 1/(b*c^2))^(1/4))/(b*x^2 + a)) - 1/2*(1/(b*c^2))^(1/4)*log(((b*x^2 + a)^(3 /4)*sqrt(c*x) - (b*c*x^2 + a*c)*(1/(b*c^2))^(1/4))/(b*x^2 + a)) - 1/2*I*(1 /(b*c^2))^(1/4)*log(((b*x^2 + a)^(3/4)*sqrt(c*x) - (I*b*c*x^2 + I*a*c)*(1/ (b*c^2))^(1/4))/(b*x^2 + a)) + 1/2*I*(1/(b*c^2))^(1/4)*log(((b*x^2 + a)^(3 /4)*sqrt(c*x) - (-I*b*c*x^2 - I*a*c)*(1/(b*c^2))^(1/4))/(b*x^2 + a))
Result contains complex when optimal does not.
Time = 0.69 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.53 \[ \int \frac {1}{\sqrt {c x} \sqrt [4]{a+b x^2}} \, dx=\frac {\sqrt {x} \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{4} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \sqrt {c} \Gamma \left (\frac {5}{4}\right )} \] Input:
integrate(1/(c*x)**(1/2)/(b*x**2+a)**(1/4),x)
Output:
sqrt(x)*gamma(1/4)*hyper((1/4, 1/4), (5/4,), b*x**2*exp_polar(I*pi)/a)/(2* a**(1/4)*sqrt(c)*gamma(5/4))
\[ \int \frac {1}{\sqrt {c x} \sqrt [4]{a+b x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x}} \,d x } \] Input:
integrate(1/(c*x)^(1/2)/(b*x^2+a)^(1/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(1/4)*sqrt(c*x)), x)
\[ \int \frac {1}{\sqrt {c x} \sqrt [4]{a+b x^2}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {1}{4}} \sqrt {c x}} \,d x } \] Input:
integrate(1/(c*x)^(1/2)/(b*x^2+a)^(1/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(1/4)*sqrt(c*x)), x)
Timed out. \[ \int \frac {1}{\sqrt {c x} \sqrt [4]{a+b x^2}} \, dx=\int \frac {1}{\sqrt {c\,x}\,{\left (b\,x^2+a\right )}^{1/4}} \,d x \] Input:
int(1/((c*x)^(1/2)*(a + b*x^2)^(1/4)),x)
Output:
int(1/((c*x)^(1/2)*(a + b*x^2)^(1/4)), x)
Time = 0.22 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.45 \[ \int \frac {1}{\sqrt {c x} \sqrt [4]{a+b x^2}} \, dx=\frac {\sqrt {x}\, \sqrt {c}\, \left (2 b \,x^{2}+2 a \right )}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} \sqrt {b \,x^{2}+a}\, c} \] Input:
int(1/(c*x)^(1/2)/(b*x^2+a)^(1/4),x)
Output:
(sqrt(x)*sqrt(c)*(a + b*x**2)**(1/4)*(a + b*x**2 + a + b*x**2))/(sqrt(a + b*x**2)*c*(a + b*x**2))