Integrand size = 19, antiderivative size = 998 \[ \int \sqrt [4]{c x} \sqrt {a+b x^2} \, dx =\text {Too large to display} \] Output:
4/9*(c*x)^(5/4)*(b*x^2+a)^(1/2)/c-8/9*(2+2^(1/2))^(1/2)*a^(3/2)*(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^(1/4)/(b*x^2+a)^ (1/2)/(a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))+8/9*(2+2^(1/2))^(1/2)*a^(3/2)* (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^(1/4)/ (b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2)-b^(1/4)*(c*x)^(1/2))+8/9*a^(3/2)*(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 )/b^(1/4)/(b*x^2+a)^(1/2)/(a^(1/4)*c^(1/2)+b^(1/4)*(c*x)^(1/2))-8/9*a^(3/2 )*(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+...
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.06 \[ \int \sqrt [4]{c x} \sqrt {a+b x^2} \, dx=\frac {4 x \sqrt [4]{c x} \sqrt {a+b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{2},\frac {5}{8},\frac {13}{8},-\frac {b x^2}{a}\right )}{5 \sqrt {1+\frac {b x^2}{a}}} \] Input:
Integrate[(c*x)^(1/4)*Sqrt[a + b*x^2],x]
Output:
(4*x*(c*x)^(1/4)*Sqrt[a + b*x^2]*Hypergeometric2F1[-1/2, 5/8, 13/8, -((b*x ^2)/a)])/(5*Sqrt[1 + (b*x^2)/a])
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.21 (sec) , antiderivative size = 85, 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 = {248, 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.
\(\displaystyle \int \sqrt [4]{c x} \sqrt {a+b x^2} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {4}{9} a \int \frac {\sqrt [4]{c x}}{\sqrt {b x^2+a}}dx+\frac {4 (c x)^{5/4} \sqrt {a+b x^2}}{9 c}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {16 a \int \frac {c x}{\sqrt {b x^2+a}}d\sqrt [4]{c x}}{9 c}+\frac {4 (c x)^{5/4} \sqrt {a+b x^2}}{9 c}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {16 a \sqrt {\frac {b x^2}{a}+1} \int \frac {c x}{\sqrt {\frac {b x^2}{a}+1}}d\sqrt [4]{c x}}{9 c \sqrt {a+b x^2}}+\frac {4 (c x)^{5/4} \sqrt {a+b x^2}}{9 c}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {16 a (c x)^{5/4} \sqrt {\frac {b x^2}{a}+1} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {5}{8},\frac {13}{8},-\frac {b x^2}{a}\right )}{45 c \sqrt {a+b x^2}}+\frac {4 (c x)^{5/4} \sqrt {a+b x^2}}{9 c}\) |
Input:
Int[(c*x)^(1/4)*Sqrt[a + b*x^2],x]
Output:
(4*(c*x)^(5/4)*Sqrt[a + b*x^2])/(9*c) + (16*a*(c*x)^(5/4)*Sqrt[1 + (b*x^2) /a]*Hypergeometric2F1[1/2, 5/8, 13/8, -((b*x^2)/a)])/(45*c*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 + 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] :> 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])
\[\int \left (c x \right )^{\frac {1}{4}} \sqrt {b \,x^{2}+a}d x\]
Input:
int((c*x)^(1/4)*(b*x^2+a)^(1/2),x)
Output:
int((c*x)^(1/4)*(b*x^2+a)^(1/2),x)
\[ \int \sqrt [4]{c x} \sqrt {a+b x^2} \, dx=\int { \sqrt {b x^{2} + a} \left (c x\right )^{\frac {1}{4}} \,d x } \] Input:
integrate((c*x)^(1/4)*(b*x^2+a)^(1/2),x, algorithm="fricas")
Output:
integral(sqrt(b*x^2 + a)*(c*x)^(1/4), x)
Result contains complex when optimal does not.
Time = 0.79 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.05 \[ \int \sqrt [4]{c x} \sqrt {a+b x^2} \, dx=\frac {\sqrt {a} \sqrt [4]{c} x^{\frac {5}{4}} \Gamma \left (\frac {5}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, \frac {5}{8} \\ \frac {13}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \Gamma \left (\frac {13}{8}\right )} \] Input:
integrate((c*x)**(1/4)*(b*x**2+a)**(1/2),x)
Output:
sqrt(a)*c**(1/4)*x**(5/4)*gamma(5/8)*hyper((-1/2, 5/8), (13/8,), b*x**2*ex p_polar(I*pi)/a)/(2*gamma(13/8))
\[ \int \sqrt [4]{c x} \sqrt {a+b x^2} \, dx=\int { \sqrt {b x^{2} + a} \left (c x\right )^{\frac {1}{4}} \,d x } \] Input:
integrate((c*x)^(1/4)*(b*x^2+a)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*x^2 + a)*(c*x)^(1/4), x)
\[ \int \sqrt [4]{c x} \sqrt {a+b x^2} \, dx=\int { \sqrt {b x^{2} + a} \left (c x\right )^{\frac {1}{4}} \,d x } \] Input:
integrate((c*x)^(1/4)*(b*x^2+a)^(1/2),x, algorithm="giac")
Output:
integrate(sqrt(b*x^2 + a)*(c*x)^(1/4), x)
Timed out. \[ \int \sqrt [4]{c x} \sqrt {a+b x^2} \, dx=\int {\left (c\,x\right )}^{1/4}\,\sqrt {b\,x^2+a} \,d x \] Input:
int((c*x)^(1/4)*(a + b*x^2)^(1/2),x)
Output:
int((c*x)^(1/4)*(a + b*x^2)^(1/2), x)
\[ \int \sqrt [4]{c x} \sqrt {a+b x^2} \, dx=\frac {4 c^{\frac {1}{4}} \left (x^{\frac {5}{4}} \sqrt {b \,x^{2}+a}+\left (\int \frac {x^{\frac {1}{4}} \sqrt {b \,x^{2}+a}}{b \,x^{2}+a}d x \right ) a \right )}{9} \] Input:
int((c*x)^(1/4)*(b*x^2+a)^(1/2),x)
Output:
(4*c**(1/4)*(x**(1/4)*sqrt(a + b*x**2)*x + int((x**(1/4)*sqrt(a + b*x**2)) /(a + b*x**2),x)*a))/9