\(\int \sqrt [4]{a+b x^2} (c+d x^2)^{5/4} \, dx\) [495]

Optimal result

Integrand size = 23, antiderivative size = 84 \[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^{5/4} \, dx=\frac {c x \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2} \operatorname {AppellF1}\left (\frac {1}{2},-\frac {1}{4},-\frac {5}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(220\) vs. \(2(84)=168\).

Time = 2.90 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.62 \[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^{5/4} \, dx=\frac {\left (5 b^2 c^2+10 a b c d-3 a^2 d^2\right ) x^3 \left (1+\frac {b x^2}{a}\right )^{3/4} \left (1+\frac {d x^2}{c}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},\frac {3}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+6 x \left (c+d x^2\right ) \left (\left (a+b x^2\right ) \left (9 b c+a d+4 b d x^2\right )+a (7 b c-a d) \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )\right )}{96 b \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \] Input:

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

Output:

((5*b^2*c^2 + 10*a*b*c*d - 3*a^2*d^2)*x^3*(1 + (b*x^2)/a)^(3/4)*(1 + (d*x^ 
2)/c)^(3/4)*AppellF1[3/2, 3/4, 3/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + 6*x 
*(c + d*x^2)*((a + b*x^2)*(9*b*c + a*d + 4*b*d*x^2) + a*(7*b*c - a*d)*((c* 
(a + b*x^2))/(a*(c + d*x^2)))^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, ((-(b 
*c) + a*d)*x^2)/(a*(c + d*x^2))]))/(96*b*(a + b*x^2)^(3/4)*(c + d*x^2)^(3/ 
4))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 84, 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.

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 334

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

\(\Big \downarrow \) 333

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

Input:

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

Output:

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

\[\int \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (x^{2} d +c \right )^{\frac {5}{4}}d x\]

Input:

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

Output:

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

Fricas [F(-1)]

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

integrate((b*x^2+a)^(1/4)*(d*x^2+c)^(5/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

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

integrate((b*x**2+a)**(1/4)*(d*x**2+c)**(5/4),x)
 

Output:

Integral((a + b*x**2)**(1/4)*(c + d*x**2)**(5/4), x)
 

Maxima [F]

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

integrate((b*x^2+a)^(1/4)*(d*x^2+c)^(5/4),x, algorithm="maxima")
 

Output:

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

Giac [F]

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

integrate((b*x^2+a)^(1/4)*(d*x^2+c)^(5/4),x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \sqrt [4]{a+b x^2} \left (c+d x^2\right )^{5/4} \, dx=\frac {2 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} a d x +18 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b c x +8 \left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} b d \,x^{3}-3 \left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} x^{2}}{b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}d x \right ) a^{2} d^{2}+10 \left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} x^{2}}{b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}d x \right ) a b c d +5 \left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}} x^{2}}{b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}d x \right ) b^{2} c^{2}-2 \left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}d x \right ) a^{2} c d +14 \left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {1}{4}} \left (b \,x^{2}+a \right )^{\frac {1}{4}}}{b d \,x^{4}+a d \,x^{2}+b c \,x^{2}+a c}d x \right ) a b \,c^{2}}{32 b} \] Input:

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

Output:

(2*(c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*a*d*x + 18*(c + d*x**2)**(1/4)* 
(a + b*x**2)**(1/4)*b*c*x + 8*(c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*b*d* 
x**3 - 3*int(((c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*x**2)/(a*c + a*d*x** 
2 + b*c*x**2 + b*d*x**4),x)*a**2*d**2 + 10*int(((c + d*x**2)**(1/4)*(a + b 
*x**2)**(1/4)*x**2)/(a*c + a*d*x**2 + b*c*x**2 + b*d*x**4),x)*a*b*c*d + 5* 
int(((c + d*x**2)**(1/4)*(a + b*x**2)**(1/4)*x**2)/(a*c + a*d*x**2 + b*c*x 
**2 + b*d*x**4),x)*b**2*c**2 - 2*int(((c + d*x**2)**(1/4)*(a + b*x**2)**(1 
/4))/(a*c + a*d*x**2 + b*c*x**2 + b*d*x**4),x)*a**2*c*d + 14*int(((c + d*x 
**2)**(1/4)*(a + b*x**2)**(1/4))/(a*c + a*d*x**2 + b*c*x**2 + b*d*x**4),x) 
*a*b*c**2)/(32*b)