\(\int \frac {(c+d x^2)^{7/4}}{(a+b x^2)^{5/4}} \, dx\) [527]

Optimal result

Integrand size = 23, antiderivative size = 87 \[ \int \frac {\left (c+d x^2\right )^{7/4}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {c x \sqrt [4]{1+\frac {b x^2}{a}} \left (c+d x^2\right )^{3/4} \operatorname {AppellF1}\left (\frac {1}{2},\frac {5}{4},-\frac {7}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{a \sqrt [4]{a+b x^2} \left (1+\frac {d x^2}{c}\right )^{3/4}} \] Output:

c*x*(1+b*x^2/a)^(1/4)*(d*x^2+c)^(3/4)*AppellF1(1/2,5/4,-7/4,3/2,-b*x^2/a,- 
d*x^2/c)/a/(b*x^2+a)^(1/4)/(1+d*x^2/c)^(3/4)
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(366\) vs. \(2(87)=174\).

Time = 5.14 (sec) , antiderivative size = 366, normalized size of antiderivative = 4.21 \[ \int \frac {\left (c+d x^2\right )^{7/4}}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {x \left (d (-4 b c+5 a d) x^2 \sqrt [4]{1+\frac {b x^2}{a}} \sqrt [4]{1+\frac {d x^2}{c}} \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+\frac {6 \left (3 a c \left (-2 a d^2 x^2+b c \left (c+2 d x^2\right )\right ) \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+(-b c+a d) x^2 \left (c+d x^2\right ) \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )\right )}{6 a c \operatorname {AppellF1}\left (\frac {1}{2},\frac {1}{4},\frac {1}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )-x^2 \left (a d \operatorname {AppellF1}\left (\frac {3}{2},\frac {1}{4},\frac {5}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+b c \operatorname {AppellF1}\left (\frac {3}{2},\frac {5}{4},\frac {1}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )\right )}\right )}{3 a b \sqrt [4]{a+b x^2} \sqrt [4]{c+d x^2}} \] Input:

Integrate[(c + d*x^2)^(7/4)/(a + b*x^2)^(5/4),x]
 

Output:

(x*(d*(-4*b*c + 5*a*d)*x^2*(1 + (b*x^2)/a)^(1/4)*(1 + (d*x^2)/c)^(1/4)*App 
ellF1[3/2, 1/4, 1/4, 5/2, -((b*x^2)/a), -((d*x^2)/c)] + (6*(3*a*c*(-2*a*d^ 
2*x^2 + b*c*(c + 2*d*x^2))*AppellF1[1/2, 1/4, 1/4, 3/2, -((b*x^2)/a), -((d 
*x^2)/c)] + (-(b*c) + a*d)*x^2*(c + d*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)])))/(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*(a + b*x^2)^(1/4)*(c + d*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 87, 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.

\(\displaystyle \int \frac {\left (c+d x^2\right )^{7/4}}{\left (a+b x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\sqrt [4]{\frac {b x^2}{a}+1} \int \frac {\left (d x^2+c\right )^{7/4}}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx}{a \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {c \sqrt [4]{\frac {b x^2}{a}+1} \left (c+d x^2\right )^{3/4} \int \frac {\left (\frac {d x^2}{c}+1\right )^{7/4}}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx}{a \sqrt [4]{a+b x^2} \left (\frac {d x^2}{c}+1\right )^{3/4}}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {c x \sqrt [4]{\frac {b x^2}{a}+1} \left (c+d x^2\right )^{3/4} \operatorname {AppellF1}\left (\frac {1}{2},\frac {5}{4},-\frac {7}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{a \sqrt [4]{a+b x^2} \left (\frac {d x^2}{c}+1\right )^{3/4}}\)

Input:

Int[(c + d*x^2)^(7/4)/(a + b*x^2)^(5/4),x]
 

Output:

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

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]

\[\int \frac {\left (x^{2} d +c \right )^{\frac {7}{4}}}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

Input:

int((d*x^2+c)^(7/4)/(b*x^2+a)^(5/4),x)
 

Output:

int((d*x^2+c)^(7/4)/(b*x^2+a)^(5/4),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right )^{7/4}}{\left (a+b x^2\right )^{5/4}} \, dx=\text {Timed out} \] Input:

integrate((d*x^2+c)^(7/4)/(b*x^2+a)^(5/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (c+d x^2\right )^{7/4}}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {\left (c + d x^{2}\right )^{\frac {7}{4}}}{\left (a + b x^{2}\right )^{\frac {5}{4}}}\, dx \] Input:

integrate((d*x**2+c)**(7/4)/(b*x**2+a)**(5/4),x)
 

Output:

Integral((c + d*x**2)**(7/4)/(a + b*x**2)**(5/4), x)
 

Maxima [F]

\[ \int \frac {\left (c+d x^2\right )^{7/4}}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {7}{4}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((d*x^2+c)^(7/4)/(b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^(7/4)/(b*x^2 + a)^(5/4), x)
 

Giac [F]

\[ \int \frac {\left (c+d x^2\right )^{7/4}}{\left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {7}{4}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] Input:

integrate((d*x^2+c)^(7/4)/(b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)^(7/4)/(b*x^2 + a)^(5/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right )^{7/4}}{\left (a+b x^2\right )^{5/4}} \, dx=\int \frac {{\left (d\,x^2+c\right )}^{7/4}}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:

int((c + d*x^2)^(7/4)/(a + b*x^2)^(5/4),x)
 

Output:

int((c + d*x^2)^(7/4)/(a + b*x^2)^(5/4), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^2\right )^{7/4}}{\left (a+b x^2\right )^{5/4}} \, dx=\left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {3}{4}}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c +\left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {3}{4}} x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d \] Input:

int((d*x^2+c)^(7/4)/(b*x^2+a)^(5/4),x)
 

Output:

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