\(\int \frac {(c+d x^2)^{9/4}}{(a+b x^2)^{3/4}} \, dx\) [518]

Optimal result

Integrand size = 23, antiderivative size = 86 \[ \int \frac {\left (c+d x^2\right )^{9/4}}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {c^2 x \left (1+\frac {b x^2}{a}\right )^{3/4} \sqrt [4]{c+d x^2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},-\frac {9}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\left (a+b x^2\right )^{3/4} \sqrt [4]{1+\frac {d x^2}{c}}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(237\) vs. \(2(86)=172\).

Time = 5.49 (sec) , antiderivative size = 237, normalized size of antiderivative = 2.76 \[ \int \frac {\left (c+d x^2\right )^{9/4}}{\left (a+b x^2\right )^{3/4}} \, dx=\frac {d \left (15 b^2 c^2-18 a b c d+7 a^2 d^2\right ) x^3 \left (1+\frac {b x^2}{a}\right )^{3/4} \left (1+\frac {d x^2}{c}\right )^{3/4} \operatorname {AppellF1}\left (\frac {3}{2},\frac {3}{4},\frac {3}{4},\frac {5}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )+2 x \left (c+d x^2\right ) \left (-d \left (a+b x^2\right ) \left (-17 b c+7 a d-4 b d x^2\right )+\left (16 b^2 c^2-17 a b c d+7 a^2 d^2\right ) \left (\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},\frac {(-b c+a d) x^2}{a \left (c+d x^2\right )}\right )\right )}{32 b^2 \left (a+b x^2\right )^{3/4} \left (c+d x^2\right )^{3/4}} \] Input:

Integrate[(c + d*x^2)^(9/4)/(a + b*x^2)^(3/4),x]
 

Output:

(d*(15*b^2*c^2 - 18*a*b*c*d + 7*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)] + 
2*x*(c + d*x^2)*(-(d*(a + b*x^2)*(-17*b*c + 7*a*d - 4*b*d*x^2)) + (16*b^2* 
c^2 - 17*a*b*c*d + 7*a^2*d^2)*((c*(a + b*x^2))/(a*(c + d*x^2)))^(3/4)*Hype 
rgeometric2F1[1/2, 3/4, 3/2, ((-(b*c) + a*d)*x^2)/(a*(c + d*x^2))]))/(32*b 
^2*(a + b*x^2)^(3/4)*(c + d*x^2)^(3/4))
 

Rubi [A] (verified)

Time = 0.21 (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.

\(\displaystyle \int \frac {\left (c+d x^2\right )^{9/4}}{\left (a+b x^2\right )^{3/4}} \, dx\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \int \frac {\left (d x^2+c\right )^{9/4}}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{\left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 334

\(\displaystyle \frac {c^2 \left (\frac {b x^2}{a}+1\right )^{3/4} \sqrt [4]{c+d x^2} \int \frac {\left (\frac {d x^2}{c}+1\right )^{9/4}}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{\left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {d x^2}{c}+1}}\)

\(\Big \downarrow \) 333

\(\displaystyle \frac {c^2 x \left (\frac {b x^2}{a}+1\right )^{3/4} \sqrt [4]{c+d x^2} \operatorname {AppellF1}\left (\frac {1}{2},\frac {3}{4},-\frac {9}{4},\frac {3}{2},-\frac {b x^2}{a},-\frac {d x^2}{c}\right )}{\left (a+b x^2\right )^{3/4} \sqrt [4]{\frac {d x^2}{c}+1}}\)

Input:

Int[(c + d*x^2)^(9/4)/(a + b*x^2)^(3/4),x]
 

Output:

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

\[\int \frac {\left (x^{2} d +c \right )^{\frac {9}{4}}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]

Input:

int((d*x^2+c)^(9/4)/(b*x^2+a)^(3/4),x)
 

Output:

int((d*x^2+c)^(9/4)/(b*x^2+a)^(3/4),x)
 

Fricas [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right )^{9/4}}{\left (a+b x^2\right )^{3/4}} \, dx=\text {Timed out} \] Input:

integrate((d*x^2+c)^(9/4)/(b*x^2+a)^(3/4),x, algorithm="fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {\left (c+d x^2\right )^{9/4}}{\left (a+b x^2\right )^{3/4}} \, dx=\int \frac {\left (c + d x^{2}\right )^{\frac {9}{4}}}{\left (a + b x^{2}\right )^{\frac {3}{4}}}\, dx \] Input:

integrate((d*x**2+c)**(9/4)/(b*x**2+a)**(3/4),x)
 

Output:

Integral((c + d*x**2)**(9/4)/(a + b*x**2)**(3/4), x)
 

Maxima [F]

\[ \int \frac {\left (c+d x^2\right )^{9/4}}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {9}{4}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((d*x^2+c)^(9/4)/(b*x^2+a)^(3/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^(9/4)/(b*x^2 + a)^(3/4), x)
 

Giac [F]

\[ \int \frac {\left (c+d x^2\right )^{9/4}}{\left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{\frac {9}{4}}}{{\left (b x^{2} + a\right )}^{\frac {3}{4}}} \,d x } \] Input:

integrate((d*x^2+c)^(9/4)/(b*x^2+a)^(3/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)^(9/4)/(b*x^2 + a)^(3/4), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right )^{9/4}}{\left (a+b x^2\right )^{3/4}} \, dx=\int \frac {{\left (d\,x^2+c\right )}^{9/4}}{{\left (b\,x^2+a\right )}^{3/4}} \,d x \] Input:

int((c + d*x^2)^(9/4)/(a + b*x^2)^(3/4),x)
 

Output:

int((c + d*x^2)^(9/4)/(a + b*x^2)^(3/4), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^2\right )^{9/4}}{\left (a+b x^2\right )^{3/4}} \, dx=\left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {1}{4}}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) c^{2}+\left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {1}{4}} x^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) d^{2}+2 \left (\int \frac {\left (d \,x^{2}+c \right )^{\frac {1}{4}} x^{2}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x \right ) c d \] Input:

int((d*x^2+c)^(9/4)/(b*x^2+a)^(3/4),x)
 

Output:

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