Integrand size = 19, antiderivative size = 1011 \[ \int \frac {(c x)^{9/4}}{\sqrt {a+b x^2}} \, dx =\text {Too large to display} \] Output:
4/9*c*(c*x)^(5/4)*(b*x^2+a)^(1/2)/b+10/9*(2+2^(1/2))^(1/2)*a^(3/2)*c^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))/b^(5/4)/(b* x^2+a)^(1/2)/(a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))-10/9*(2+2^(1/2))^(1/2)* a^(3/2)*c^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)*El lipticE(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))/b^(5/4)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))-10/9*a^( 3/2)*c^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)*Ellipt icF(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)/b^(5/4)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^ (1/2))+10/9*a^(3/2)*c^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...
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.03 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.07 \[ \int \frac {(c x)^{9/4}}{\sqrt {a+b x^2}} \, dx=\frac {4 c (c x)^{5/4} \left (a+b x^2-a \sqrt {1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},-\frac {b x^2}{a}\right )\right )}{9 b \sqrt {a+b x^2}} \] Input:
Integrate[(c*x)^(9/4)/Sqrt[a + b*x^2],x]
Output:
(4*c*(c*x)^(5/4)*(a + b*x^2 - a*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/2, 5/8, 13/8, -((b*x^2)/a)]))/(9*b*Sqrt[a + b*x^2])
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.22 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {262, 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.
\(\displaystyle \int \frac {(c x)^{9/4}}{\sqrt {a+b x^2}} \, dx\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {4 c (c x)^{5/4} \sqrt {a+b x^2}}{9 b}-\frac {5 a c^2 \int \frac {\sqrt [4]{c x}}{\sqrt {b x^2+a}}dx}{9 b}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {4 c (c x)^{5/4} \sqrt {a+b x^2}}{9 b}-\frac {20 a c \int \frac {c x}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{9 b}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {4 c (c x)^{5/4} \sqrt {a+b x^2}}{9 b}-\frac {20 a c \sqrt {\frac {b x^2}{a}+1} \int \frac {c x}{\sqrt {\frac {b x^2}{a}+1}}d\sqrt [4]{c x}}{9 b \sqrt {a+b x^2}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {4 c (c x)^{5/4} \sqrt {a+b x^2}}{9 b}-\frac {4 a c (c x)^{5/4} \sqrt {\frac {b x^2}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},-\frac {b x^2}{a}\right )}{9 b \sqrt {a+b x^2}}\) |
Input:
Int[(c*x)^(9/4)/Sqrt[a + b*x^2],x]
Output:
(4*c*(c*x)^(5/4)*Sqrt[a + b*x^2])/(9*b) - (4*a*c*(c*x)^(5/4)*Sqrt[1 + (b*x ^2)/a]*Hypergeometric2F1[1/2, 5/8, 13/8, -((b*x^2)/a)])/(9*b*Sqrt[a + b*x^ 2])
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[((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])
\[\int \frac {\left (c x \right )^{\frac {9}{4}}}{\sqrt {b \,x^{2}+a}}d x\]
Input:
int((c*x)^(9/4)/(b*x^2+a)^(1/2),x)
Output:
int((c*x)^(9/4)/(b*x^2+a)^(1/2),x)
\[ \int \frac {(c x)^{9/4}}{\sqrt {a+b x^2}} \, dx=\int { \frac {\left (c x\right )^{\frac {9}{4}}}{\sqrt {b x^{2} + a}} \,d x } \] Input:
integrate((c*x)^(9/4)/(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
integral((c*x)^(1/4)*c^2*x^2/sqrt(b*x^2 + a), x)
Result contains complex when optimal does not.
Time = 49.75 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.04 \[ \int \frac {(c x)^{9/4}}{\sqrt {a+b x^2}} \, dx=\frac {c^{\frac {9}{4}} x^{\frac {13}{4}} \Gamma \left (\frac {13}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {13}{8} \\ \frac {21}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt {a} \Gamma \left (\frac {21}{8}\right )} \] Input:
integrate((c*x)**(9/4)/(b*x**2+a)**(1/2),x)
Output:
c**(9/4)*x**(13/4)*gamma(13/8)*hyper((1/2, 13/8), (21/8,), b*x**2*exp_pola r(I*pi)/a)/(2*sqrt(a)*gamma(21/8))
\[ \int \frac {(c x)^{9/4}}{\sqrt {a+b x^2}} \, dx=\int { \frac {\left (c x\right )^{\frac {9}{4}}}{\sqrt {b x^{2} + a}} \,d x } \] Input:
integrate((c*x)^(9/4)/(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
integrate((c*x)^(9/4)/sqrt(b*x^2 + a), x)
\[ \int \frac {(c x)^{9/4}}{\sqrt {a+b x^2}} \, dx=\int { \frac {\left (c x\right )^{\frac {9}{4}}}{\sqrt {b x^{2} + a}} \,d x } \] Input:
integrate((c*x)^(9/4)/(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
integrate((c*x)^(9/4)/sqrt(b*x^2 + a), x)
Timed out. \[ \int \frac {(c x)^{9/4}}{\sqrt {a+b x^2}} \, dx=\int \frac {{\left (c\,x\right )}^{9/4}}{\sqrt {b\,x^2+a}} \,d x \] Input:
int((c*x)^(9/4)/(a + b*x^2)^(1/2),x)
Output:
int((c*x)^(9/4)/(a + b*x^2)^(1/2), x)
\[ \int \frac {(c x)^{9/4}}{\sqrt {a+b x^2}} \, dx=\frac {c^{\frac {9}{4}} \left (4 x^{\frac {5}{4}} \sqrt {b \,x^{2}+a}-5 \left (\int \frac {x^{\frac {1}{4}} \sqrt {b \,x^{2}+a}}{b \,x^{2}+a}d x \right ) a \right )}{9 b} \] Input:
int((c*x)^(9/4)/(b*x^2+a)^(1/2),x)
Output:
(c**(1/4)*c**2*(4*x**(1/4)*sqrt(a + b*x**2)*x - 5*int((x**(1/4)*sqrt(a + b *x**2))/(a + b*x**2),x)*a))/(9*b)