\(\int \frac {(a+b x^2)^{3/4}}{(c+d x^2)^{5/4}} \, dx\) [504]

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{5/4}} \, dx=\frac {x \left (a+b x^2\right )^{3/4} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},\frac {5}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c \left (1+\frac {b x^2}{a}\right )^{3/4} \sqrt [4]{c+d x^2}} \] Output:

x*(b*x^2+a)^(3/4)*(1+d*x^2/c)^(1/4)*AppellF1(1/2,-3/4,5/4,3/2,-b*x^2/a,-d* 
x^2/c)/c/(1+b*x^2/a)^(3/4)/(d*x^2+c)^(1/4)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(255\) vs. \(2(86)=172\).

Time = 4.57 (sec) , antiderivative size = 255, normalized size of antiderivative = 2.97 \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{5/4}} \, dx=\frac {2 x \left (\frac {3 \left (a+b x^2\right )}{c}-\frac {2 b x^2 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c}+\frac {9 a^2 \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{-6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+x^2 \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}\right )}{3 \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:

Integrate[(a + b*x^2)^(3/4)/(c + d*x^2)^(5/4),x]
 

Output:

(2*x*((3*(a + b*x^2))/c - (2*b*x^2*(1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c)^( 
1/4)*AppellF1[3/2, 1/4, 1/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)])/c + (9*a^2* 
AppellF1[1/2, 1/4, 1/4, 3/2, -((b*x^2)/a), -((d*x^2)/c)])/(-6*a*c*AppellF1 
[1/2, 1/4, 1/4, 3/2, -((b*x^2)/a), -((d*x^2)/c)] + x^2*(a*d*AppellF1[3/2, 
1/4, 5/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + b*c*AppellF1[3/2, 5/4, 1/4, 5 
/2, -((b*x^2)/a), -((d*x^2)/c)]))))/(3*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4) 
)
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 86, 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.

\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\left (a+b x^2\right )^{3/4} \int \frac {\left (\frac {b x^2}{a}+1\right )^{3/4}}{\left (d x^2+c\right )^{5/4}}dx}{\left (\frac {b x^2}{a}+1\right )^{3/4}}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {d x^2}{c}+1} \int \frac {\left (\frac {b x^2}{a}+1\right )^{3/4}}{\left (\frac {d x^2}{c}+1\right )^{5/4}}dx}{c \left (\frac {b x^2}{a}+1\right )^{3/4} \sqrt [4]{c+d x^2}}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {x \left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {d x^2}{c}+1} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {3}{4},\frac {5}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{c \left (\frac {b x^2}{a}+1\right )^{3/4} \sqrt [4]{c+d x^2}}\)

Input:

Int[(a + b*x^2)^(3/4)/(c + d*x^2)^(5/4),x]
 

Output:

(x*(a + b*x^2)^(3/4)*(1 + (d*x^2)/c)^(1/4)*AppellF1[1/2, -3/4, 5/4, 3/2, - 
((b*x^2)/a), -((d*x^2)/c)])/(c*(1 + (b*x^2)/a)^(3/4)*(c + d*x^2)^(1/4))
 

Defintions of rubi rules used

rule 333

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])
 

rule 334

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])
 

Maple [F]

\[\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{\left (x^{2} d +c \right )^{\frac {5}{4}}}d x\]

Input:

int((b*x^2+a)^(3/4)/(d*x^2+c)^(5/4),x)
 

Output:

int((b*x^2+a)^(3/4)/(d*x^2+c)^(5/4),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{5/4}} \, dx=\text {Timed out} \] Input:

integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(5/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{5/4}} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {3}{4}}}{\left (c + d x^{2}\right )^{\frac {5}{4}}}\, dx \] Input:

integrate((b*x**2+a)**(3/4)/(d*x**2+c)**(5/4),x)
 

Output:

Integral((a + b*x**2)**(3/4)/(c + d*x**2)**(5/4), x)
 

Maxima [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{{\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(5/4),x, algorithm="maxima")
 

Output:

integrate((b*x^2 + a)^(3/4)/(d*x^2 + c)^(5/4), x)
 

Giac [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{5/4}} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}}}{{\left (d x^{2} + c\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((b*x^2+a)^(3/4)/(d*x^2+c)^(5/4),x, algorithm="giac")
 

Output:

integrate((b*x^2 + a)^(3/4)/(d*x^2 + c)^(5/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{5/4}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/4}}{{\left (d\,x^2+c\right )}^{5/4}} \,d x \] Input:

int((a + b*x^2)^(3/4)/(c + d*x^2)^(5/4),x)
 

Output:

int((a + b*x^2)^(3/4)/(c + d*x^2)^(5/4), x)
 

Reduce [F]

\[ \int \frac {\left (a+b x^2\right )^{3/4}}{\left (c+d x^2\right )^{5/4}} \, dx=\int \frac {\left (b \,x^{2}+a \right )^{\frac {3}{4}}}{\left (d \,x^{2}+c \right )^{\frac {1}{4}} c +\left (d \,x^{2}+c \right )^{\frac {1}{4}} d \,x^{2}}d x \] Input:

int((b*x^2+a)^(3/4)/(d*x^2+c)^(5/4),x)
 

Output:

int((a + b*x**2)**(3/4)/((c + d*x**2)**(1/4)*c + (c + d*x**2)**(1/4)*d*x** 
2),x)