\(\int \frac {(c+d x^2)^3}{(a+b x^2)^{5/4}} \, dx\) [435]

Optimal result

Integrand size = 21, antiderivative size = 194 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {2 d \left (45 b^2 c^2-63 a b c d+25 a^2 d^2\right ) x}{15 b^3 \sqrt [4]{a+b x^2}}+\frac {2 d^2 (9 b c-5 a d) x \left (a+b x^2\right )^{3/4}}{15 b^3}+\frac {2 d^3 x^3 \left (a+b x^2\right )^{3/4}}{9 b^2}+\frac {2 \left (15 b^3 c^3-90 a b^2 c^2 d+108 a^2 b c d^2-40 a^3 d^3\right ) \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 \sqrt {a} b^{7/2} \sqrt [4]{a+b x^2}} \] Output:

2/15*d*(25*a^2*d^2-63*a*b*c*d+45*b^2*c^2)*x/b^3/(b*x^2+a)^(1/4)+2/15*d^2*( 
-5*a*d+9*b*c)*x*(b*x^2+a)^(3/4)/b^3+2/9*d^3*x^3*(b*x^2+a)^(3/4)/b^2+2/15*( 
-40*a^3*d^3+108*a^2*b*c*d^2-90*a*b^2*c^2*d+15*b^3*c^3)*(1+b*x^2/a)^(1/4)*E 
llipticE(sin(1/2*arctan(b^(1/2)*x/a^(1/2))),2^(1/2))/a^(1/2)/b^(7/2)/(b*x^ 
2+a)^(1/4)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 15.10 (sec) , antiderivative size = 158, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{5/4}} \, dx=\frac {x \left (90 b^3 c^3-120 a^3 d^3+4 a^2 b d^2 \left (81 c-5 d x^2\right )+2 a b^2 d \left (-135 c^2+27 c d x^2+5 d^2 x^4\right )+3 \left (-15 b^3 c^3+90 a b^2 c^2 d-108 a^2 b c d^2+40 a^3 d^3\right ) \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{45 a b^3 \sqrt [4]{a+b x^2}} \] Input:

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

Output:

(x*(90*b^3*c^3 - 120*a^3*d^3 + 4*a^2*b*d^2*(81*c - 5*d*x^2) + 2*a*b^2*d*(- 
135*c^2 + 27*c*d*x^2 + 5*d^2*x^4) + 3*(-15*b^3*c^3 + 90*a*b^2*c^2*d - 108* 
a^2*b*c*d^2 + 40*a^3*d^3)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 1/2 
, 3/2, -((b*x^2)/a)]))/(45*a*b^3*(a + b*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 252, normalized size of antiderivative = 1.30, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {315, 27, 403, 27, 299, 227, 225, 212}

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.

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

\(\Big \downarrow \) 315

\(\displaystyle \frac {2 \int -\frac {\left (d x^2+c\right ) \left (d (9 b c-10 a d) x^2+c (b c-2 a d)\right )}{2 \sqrt [4]{b x^2+a}}dx}{a b}+\frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt [4]{a+b x^2}}-\frac {\int \frac {\left (d x^2+c\right ) \left (d (9 b c-10 a d) x^2+c (b c-2 a d)\right )}{\sqrt [4]{b x^2+a}}dx}{a b}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {2 \int \frac {d \left (45 b^2 c^2-112 a b d c+60 a^2 d^2\right ) x^2+c (3 b c-10 a d) (3 b c-2 a d)}{2 \sqrt [4]{b x^2+a}}dx}{9 b}+\frac {2 d x \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) (9 b c-10 a d)}{9 b}}{a b}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {\int \frac {d \left (45 b^2 c^2-112 a b d c+60 a^2 d^2\right ) x^2+c (3 b c-10 a d) (3 b c-2 a d)}{\sqrt [4]{b x^2+a}}dx}{9 b}+\frac {2 d x \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) (9 b c-10 a d)}{9 b}}{a b}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {\frac {3 \left (-40 a^3 d^3+108 a^2 b c d^2-90 a b^2 c^2 d+15 b^3 c^3\right ) \int \frac {1}{\sqrt [4]{b x^2+a}}dx}{5 b}+\frac {2 d x \left (a+b x^2\right )^{3/4} \left (60 a^2 d^2-112 a b c d+45 b^2 c^2\right )}{5 b}}{9 b}+\frac {2 d x \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) (9 b c-10 a d)}{9 b}}{a b}\)

\(\Big \downarrow \) 227

\(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {\frac {3 \sqrt [4]{\frac {b x^2}{a}+1} \left (-40 a^3 d^3+108 a^2 b c d^2-90 a b^2 c^2 d+15 b^3 c^3\right ) \int \frac {1}{\sqrt [4]{\frac {b x^2}{a}+1}}dx}{5 b \sqrt [4]{a+b x^2}}+\frac {2 d x \left (a+b x^2\right )^{3/4} \left (60 a^2 d^2-112 a b c d+45 b^2 c^2\right )}{5 b}}{9 b}+\frac {2 d x \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) (9 b c-10 a d)}{9 b}}{a b}\)

\(\Big \downarrow \) 225

\(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {\frac {3 \sqrt [4]{\frac {b x^2}{a}+1} \left (-40 a^3 d^3+108 a^2 b c d^2-90 a b^2 c^2 d+15 b^3 c^3\right ) \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{5/4}}dx\right )}{5 b \sqrt [4]{a+b x^2}}+\frac {2 d x \left (a+b x^2\right )^{3/4} \left (60 a^2 d^2-112 a b c d+45 b^2 c^2\right )}{5 b}}{9 b}+\frac {2 d x \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) (9 b c-10 a d)}{9 b}}{a b}\)

\(\Big \downarrow \) 212

\(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt [4]{a+b x^2}}-\frac {\frac {\frac {2 d x \left (a+b x^2\right )^{3/4} \left (60 a^2 d^2-112 a b c d+45 b^2 c^2\right )}{5 b}+\frac {3 \sqrt [4]{\frac {b x^2}{a}+1} \left (-40 a^3 d^3+108 a^2 b c d^2-90 a b^2 c^2 d+15 b^3 c^3\right ) \left (\frac {2 x}{\sqrt [4]{\frac {b x^2}{a}+1}}-\frac {2 \sqrt {a} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{\sqrt {b}}\right )}{5 b \sqrt [4]{a+b x^2}}}{9 b}+\frac {2 d x \left (a+b x^2\right )^{3/4} \left (c+d x^2\right ) (9 b c-10 a d)}{9 b}}{a b}\)

Input:

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

Output:

(2*(b*c - a*d)*x*(c + d*x^2)^2)/(a*b*(a + b*x^2)^(1/4)) - ((2*d*(9*b*c - 1 
0*a*d)*x*(a + b*x^2)^(3/4)*(c + d*x^2))/(9*b) + ((2*d*(45*b^2*c^2 - 112*a* 
b*c*d + 60*a^2*d^2)*x*(a + b*x^2)^(3/4))/(5*b) + (3*(15*b^3*c^3 - 90*a*b^2 
*c^2*d + 108*a^2*b*c*d^2 - 40*a^3*d^3)*(1 + (b*x^2)/a)^(1/4)*((2*x)/(1 + ( 
b*x^2)/a)^(1/4) - (2*Sqrt[a]*EllipticE[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/ 
Sqrt[b]))/(5*b*(a + b*x^2)^(1/4)))/(9*b))/(a*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 225

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[2*(x/(a + b*x^2)^(1/4)) 
, x] - Simp[a   Int[1/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b}, x] && GtQ[ 
a, 0] && PosQ[b/a]
 

rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(1/4)/( 
a + b*x^2)^(1/4)   Int[1/(1 + b*(x^2/a))^(1/4), x], x] /; FreeQ[{a, b}, x] 
&& PosQ[a]
 

rule 299

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[d*x 
*((a + b*x^2)^(p + 1)/(b*(2*p + 3))), x] - Simp[(a*d - b*c*(2*p + 3))/(b*(2 
*p + 3))   Int[(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - 
 a*d, 0] && NeQ[2*p + 3, 0]
 

rule 315

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim 
p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), 
x] - Simp[1/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S 
imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) 
*x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 
1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
 

rule 403

Int[((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*( 
x_)^2), x_Symbol] :> Simp[f*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^q/(b*(2*(p + 
 q + 1) + 1))), x] + Simp[1/(b*(2*(p + q + 1) + 1))   Int[(a + b*x^2)^p*(c 
+ d*x^2)^(q - 1)*Simp[c*(b*e - a*f + b*e*2*(p + q + 1)) + (d*(b*e - a*f) + 
f*2*q*(b*c - a*d) + b*d*e*2*(p + q + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, 
 d, e, f, p}, x] && GtQ[q, 0] && NeQ[2*(p + q + 1) + 1, 0]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral((d^3*x^6 + 3*c*d^2*x^4 + 3*c^2*d*x^2 + c^3)*(b*x^2 + a)^(3/4)/(b^ 
2*x^4 + 2*a*b*x^2 + a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{5/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d^{3}+3 \left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c \,d^{2}+3 \left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c^{2} d +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {1}{4}} a +\left (b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c^{3} \] Input:

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

Output:

int(x**6/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2),x)*d**3 + 3* 
int(x**4/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2),x)*c*d**2 + 
3*int(x**2/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2),x)*c**2*d 
+ int(1/((a + b*x**2)**(1/4)*a + (a + b*x**2)**(1/4)*b*x**2),x)*c**3