Integrand size = 23, antiderivative size = 83 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2}} \, dx=\frac {x \left (a+b x^2\right )^{3/4} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\left (1+\frac {b x^2}{a}\right )^{3/4} \sqrt [4]{c+d x^2}} \] Output:
x*(b*x^2+a)^(3/4)*(1+d*x^2/c)^(1/4)*AppellF1(1/2,-3/4,1/4,3/2,-b*x^2/a,-d* x^2/c)/(1+b*x^2/a)^(3/4)/(d*x^2+c)^(1/4)
Leaf count is larger than twice the leaf count of optimal. \(171\) vs. \(2(83)=166\).
Time = 2.45 (sec) , antiderivative size = 171, normalized size of antiderivative = 2.06 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2}} \, dx=\frac {6 a c x \left (a+b x^2\right )^{3/4} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\sqrt [4]{c+d x^2} \left (6 a c \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (-a d \operatorname {AppellF1}\left (\frac {3}{2},-\frac {3}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+3 b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )} \] Input:
Integrate[(a + b*x^2)^(3/4)/(c + d*x^2)^(1/4),x]
Output:
(6*a*c*x*(a + b*x^2)^(3/4)*AppellF1[1/2, -3/4, 1/4, 3/2, -((b*x^2)/a), -(( d*x^2)/c)])/((c + d*x^2)^(1/4)*(6*a*c*AppellF1[1/2, -3/4, 1/4, 3/2, -((b*x ^2)/a), -((d*x^2)/c)] + x^2*(-(a*d*AppellF1[3/2, -3/4, 5/4, 5/2, -((b*x^2) /a), -((d*x^2)/c)]) + 3*b*c*AppellF1[3/2, 1/4, 1/4, 5/2, -((b*x^2)/a), -(( d*x^2)/c)])))
Time = 0.20 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {334, 334, 333}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2}} \, dx\) |
\(\Big \downarrow \) 334 |
\(\displaystyle \frac {\left (a+b x^2\right )^{3/4} \int \frac {\left (\frac {b x^2}{a}+1\right )^{3/4}}{\sqrt [4]{d x^2+c}}dx}{\left (\frac {b x^2}{a}+1\right )^{3/4}}\) |
\(\Big \downarrow \) 334 |
\(\displaystyle \frac {\left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {d x^2}{c}+1} \int \frac {\left (\frac {b x^2}{a}+1\right )^{3/4}}{\sqrt [4]{\frac {d x^2}{c}+1}}dx}{\left (\frac {b x^2}{a}+1\right )^{3/4} \sqrt [4]{c+d x^2}}\) |
\(\Big \downarrow \) 333 |
\(\displaystyle \frac {x \left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {d x^2}{c}+1} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\left (\frac {b x^2}{a}+1\right )^{3/4} \sqrt [4]{c+d x^2}}\) |
Input:
Int[(a + b*x^2)^(3/4)/(c + d*x^2)^(1/4),x]
Output:
(x*(a + b*x^2)^(3/4)*(1 + (d*x^2)/c)^(1/4)*AppellF1[1/2, -3/4, 1/4, 3/2, - ((b*x^2)/a), -((d*x^2)/c)])/((1 + (b*x^2)/a)^(3/4)*(c + d*x^2)^(1/4))
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^p*c^q*x*AppellF1[1/2, -p, -q, 3/2, (-b)*(x^2/a), (-d)*(x^2/c)], x] /; F reeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[a^IntPart[p]*((a + b*x^2)^FracPart[p]/(1 + b*(x^2/a))^FracPart[p]) Int[ (1 + b*(x^2/a))^p*(c + d*x^2)^q, x], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && !(IntegerQ[p] || GtQ[a, 0])
\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{\left (x^{2} d +c \right )^{\frac {1}{4}}}d x\]
Input:
int((b*x^2+a)^(3/4)/(d*x^2+c)^(1/4),x)
Output:
int((b*x^2+a)^(3/4)/(d*x^2+c)^(1/4),x)
Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2}} \, dx=\text {Timed out} \] Input:
integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(1/4),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {3}{4}}}{\sqrt [4]{c + d x^{2}}}\, dx \] Input:
integrate((b*x**2+a)**(3/4)/(d*x**2+c)**(1/4),x)
Output:
Integral((a + b*x**2)**(3/4)/(c + d*x**2)**(1/4), x)
\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{{\left (d x^{2} + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(1/4),x, algorithm="maxima")
Output:
integrate((b*x^2 + a)^(3/4)/(d*x^2 + c)^(1/4), x)
\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{{\left (d x^{2} + c\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(1/4),x, algorithm="giac")
Output:
integrate((b*x^2 + a)^(3/4)/(d*x^2 + c)^(1/4), x)
Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/4}}{{\left (d\,x^2+c\right )}^{1/4}} \,d x \] Input:
int((a + b*x^2)^(3/4)/(c + d*x^2)^(1/4),x)
Output:
int((a + b*x^2)^(3/4)/(c + d*x^2)^(1/4), x)
\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\sqrt [4]{c+d x^2}} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{\left (d \,x^{2}+c \right )^{\frac {1}{4}}}d x \] Input:
int((b*x^2+a)^(3/4)/(d*x^2+c)^(1/4),x)
Output:
int((a + b*x**2)**(3/4)/(c + d*x**2)**(1/4),x)