Integrand size = 19, antiderivative size = 1018 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/4}} \, dx =\text {Too large to display} \] Output:
4/3*(b*x^2+a)^(1/2)/c/(c*x)^(1/4)+8/3*(2+2^(1/2))^(1/2)*a^(3/4)*b^(1/2)*(c *x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^(1/2)/x)^(1/2)*((a^(1/4)*c^(1/2)+b^(1/4)*( c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^(1/2)/(c*x)^(1/2))^(1/2)*EllipticE(1/2*(-a ^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b^(1/2)*x/a^(1/2)-2*b^(1/4)*(c*x)^(1/2)/a^ (1/4)/c^(1/2))/b^(1/4)/(c*x)^(1/2))^(1/2),(-2+2*2^(1/2))^(1/2))/c^(3/2)/(b *x^2+a)^(1/2)/(a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))+8/3*(2+2^(1/2))^(1/2)* a^(3/4)*b^(1/2)*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^(1/2)/x)^(1/2)*(-(a^(1/4 )*c^(1/2)-b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^(1/2)/(c*x)^(1/2))^(1/2 )*EllipticE(1/2*(a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b^(1/2)*x/a^(1/2)+2*b^(1 /4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^(1/2))^(1/2),(-2+2*2^(1/2)) ^(1/2))/c^(3/2)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))-8/3* a^(3/4)*b^(1/2)*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^(1/2)/x)^(1/2)*((a^(1/4) *c^(1/2)+b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^(1/2)/(c*x)^(1/2))^(1/2) *EllipticF(1/2*(-a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b^(1/2)*x/a^(1/2)-2*b^(1 /4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^(1/2))^(1/2),(-2+2*2^(1/2)) ^(1/2))/(2+2^(1/2))^(1/2)/c^(3/2)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2)+b^(1/4) *(c*x)^(1/2))-8/3*a^(3/4)*b^(1/2)*(c*x)^(3/4)*(-(b*x^2+a)/a^(1/2)/b^(1/2)/ x)^(1/2)*(-(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))^2/a^(1/4)/b^(1/4)/c^(1/2) /(c*x)^(1/2))^(1/2)*EllipticF(1/2*(a^(1/4)*c^(1/2)*(2^(1/2)+2^(1/2)*b^(1/2 )*x/a^(1/2)+2*b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))/b^(1/4)/(c*x)^(1/2))...
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.05 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/4}} \, dx=-\frac {4 x \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{8},\frac {7}{8},-\frac {b x^2}{a}\right )}{(c x)^{5/4} \sqrt {1+\frac {b x^2}{a}}} \] Input:
Integrate[Sqrt[a + b*x^2]/(c*x)^(5/4),x]
Output:
(-4*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-1/2, -1/8, 7/8, -((b*x^2)/a)])/(( c*x)^(5/4)*Sqrt[1 + (b*x^2)/a])
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.22 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {247, 266, 889, 888}
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.
\(\displaystyle \int \frac {\sqrt {a+b x^2}}{(c x)^{5/4}} \, dx\) |
\(\Big \downarrow \) 247 |
\(\displaystyle \frac {4 b \int \frac {(c x)^{3/4}}{\sqrt {b x^2+a}}dx}{c^2}-\frac {4 \sqrt {a+b x^2}}{c \sqrt [4]{c x}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {16 b \int \frac {(c x)^{3/2}}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{c^3}-\frac {4 \sqrt {a+b x^2}}{c \sqrt [4]{c x}}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {16 b \sqrt {\frac {b x^2}{a}+1} \int \frac {(c x)^{3/2}}{\sqrt {\frac {b x^2}{a}+1}}d\sqrt [4]{c x}}{c^3 \sqrt {a+b x^2}}-\frac {4 \sqrt {a+b x^2}}{c \sqrt [4]{c x}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {16 b (c x)^{7/4} \sqrt {\frac {b x^2}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {7}{8},\frac {15}{8},-\frac {b x^2}{a}\right )}{7 c^3 \sqrt {a+b x^2}}-\frac {4 \sqrt {a+b x^2}}{c \sqrt [4]{c x}}\) |
Input:
Int[Sqrt[a + b*x^2]/(c*x)^(5/4),x]
Output:
(-4*Sqrt[a + b*x^2])/(c*(c*x)^(1/4)) + (16*b*(c*x)^(7/4)*Sqrt[1 + (b*x^2)/ a]*Hypergeometric2F1[1/2, 7/8, 15/8, -((b*x^2)/a)])/(7*c^3*Sqrt[a + b*x^2] )
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1))) Int[ (c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + 2*p + 3)/2, 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[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(c*x) ^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0 ] && !(ILtQ[p, 0] || GtQ[a, 0])
\[\int \frac {\sqrt {b \,x^{2}+a}}{\left (c x \right )^{\frac {5}{4}}}d x\]
Input:
int((b*x^2+a)^(1/2)/(c*x)^(5/4),x)
Output:
int((b*x^2+a)^(1/2)/(c*x)^(5/4),x)
\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/4}} \, dx=\int { \frac {\sqrt {b x^{2} + a}}{\left (c x\right )^{\frac {5}{4}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)/(c*x)^(5/4),x, algorithm="fricas")
Output:
integral(sqrt(b*x^2 + a)*(c*x)^(3/4)/(c^2*x^2), x)
Result contains complex when optimal does not.
Time = 1.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.05 \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/4}} \, dx=\frac {\sqrt {a} \Gamma \left (- \frac {1}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{8} \\ \frac {7}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {5}{4}} \sqrt [4]{x} \Gamma \left (\frac {7}{8}\right )} \] Input:
integrate((b*x**2+a)**(1/2)/(c*x)**(5/4),x)
Output:
sqrt(a)*gamma(-1/8)*hyper((-1/2, -1/8), (7/8,), b*x**2*exp_polar(I*pi)/a)/ (2*c**(5/4)*x**(1/4)*gamma(7/8))
\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/4}} \, dx=\int { \frac {\sqrt {b x^{2} + a}}{\left (c x\right )^{\frac {5}{4}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)/(c*x)^(5/4),x, algorithm="maxima")
Output:
integrate(sqrt(b*x^2 + a)/(c*x)^(5/4), x)
\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/4}} \, dx=\int { \frac {\sqrt {b x^{2} + a}}{\left (c x\right )^{\frac {5}{4}}} \,d x } \] Input:
integrate((b*x^2+a)^(1/2)/(c*x)^(5/4),x, algorithm="giac")
Output:
integrate(sqrt(b*x^2 + a)/(c*x)^(5/4), x)
Timed out. \[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/4}} \, dx=\int \frac {\sqrt {b\,x^2+a}}{{\left (c\,x\right )}^{5/4}} \,d x \] Input:
int((a + b*x^2)^(1/2)/(c*x)^(5/4),x)
Output:
int((a + b*x^2)^(1/2)/(c*x)^(5/4), x)
\[ \int \frac {\sqrt {a+b x^2}}{(c x)^{5/4}} \, dx=\frac {\frac {4 \sqrt {b \,x^{2}+a}}{3}+\frac {4 x^{\frac {1}{4}} \left (\int \frac {\sqrt {b \,x^{2}+a}}{x^{\frac {5}{4}} a +x^{\frac {13}{4}} b}d x \right ) a}{3}}{x^{\frac {1}{4}} c^{\frac {5}{4}}} \] Input:
int((b*x^2+a)^(1/2)/(c*x)^(5/4),x)
Output:
(4*(sqrt(a + b*x**2) + x**(1/4)*int(sqrt(a + b*x**2)/(x**(1/4)*a*x + x**(1 /4)*b*x**3),x)*a))/(3*x**(1/4)*c**(1/4)*c)