\(\int \frac {(c+d x^2)^3}{(a+b x^2)^{7/4}} \, dx\) [440]

Optimal result

Integrand size = 21, antiderivative size = 183 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {2 (b c-a d)^3 x}{3 a b^3 \left (a+b x^2\right )^{3/4}}+\frac {2 d^2 (21 b c-13 a d) x \sqrt [4]{a+b x^2}}{21 b^3}+\frac {2 d^3 x^3 \sqrt [4]{a+b x^2}}{7 b^2}+\frac {2 \left (7 b^3 c^3+42 a b^2 c^2 d-84 a^2 b c d^2+40 a^3 d^3\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{21 \sqrt {a} b^{7/2} \left (a+b x^2\right )^{3/4}} \] Output:

2/3*(-a*d+b*c)^3*x/a/b^3/(b*x^2+a)^(3/4)+2/21*d^2*(-13*a*d+21*b*c)*x*(b*x^ 
2+a)^(1/4)/b^3+2/7*d^3*x^3*(b*x^2+a)^(1/4)/b^2+2/21*(40*a^3*d^3-84*a^2*b*c 
*d^2+42*a*b^2*c^2*d+7*b^3*c^3)*(1+b*x^2/a)^(3/4)*InverseJacobiAM(1/2*arcta 
n(b^(1/2)*x/a^(1/2)),2^(1/2))/a^(1/2)/b^(7/2)/(b*x^2+a)^(3/4)
 

Mathematica [C] (verified)

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

Time = 15.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.85 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{7/4}} \, dx=\frac {x \left (14 b^3 c^3-40 a^3 d^3+4 a^2 b d^2 \left (21 c-5 d x^2\right )+6 a b^2 d \left (-7 c^2+7 c d x^2+d^2 x^4\right )+\left (7 b^3 c^3+42 a b^2 c^2 d-84 a^2 b c d^2+40 a^3 d^3\right ) \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {3}{2},-\frac {b x^2}{a}\right )\right )}{21 a b^3 \left (a+b x^2\right )^{3/4}} \] Input:

Integrate[(c + d*x^2)^3/(a + b*x^2)^(7/4),x]
 

Output:

(x*(14*b^3*c^3 - 40*a^3*d^3 + 4*a^2*b*d^2*(21*c - 5*d*x^2) + 6*a*b^2*d*(-7 
*c^2 + 7*c*d*x^2 + d^2*x^4) + (7*b^3*c^3 + 42*a*b^2*c^2*d - 84*a^2*b*c*d^2 
 + 40*a^3*d^3)*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/2, 3/4, 3/2, -((b 
*x^2)/a)]))/(21*a*b^3*(a + b*x^2)^(3/4))
 

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.25, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {315, 27, 403, 27, 299, 231, 229}

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.

\(\displaystyle \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{7/4}} \, dx\)

\(\Big \downarrow \) 315

\(\displaystyle \frac {2 \int \frac {\left (d x^2+c\right ) \left (c (b c+2 a d)-d (7 b c-10 a d) x^2\right )}{2 \left (b x^2+a\right )^{3/4}}dx}{3 a b}+\frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\left (d x^2+c\right ) \left (c (b c+2 a d)-d (7 b c-10 a d) x^2\right )}{\left (b x^2+a\right )^{3/4}}dx}{3 a b}+\frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 403

\(\displaystyle \frac {\frac {2 \int \frac {c \left (7 b^2 c^2+28 a b d c-20 a^2 d^2\right )-3 d \left (7 b^2 c^2-32 a b d c+20 a^2 d^2\right ) x^2}{2 \left (b x^2+a\right )^{3/4}}dx}{7 b}-\frac {2 d x \sqrt [4]{a+b x^2} \left (c+d x^2\right ) (7 b c-10 a d)}{7 b}}{3 a b}+\frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {c \left (7 b^2 c^2+28 a b d c-20 a^2 d^2\right )-3 d \left (7 b^2 c^2-32 a b d c+20 a^2 d^2\right ) x^2}{\left (b x^2+a\right )^{3/4}}dx}{7 b}-\frac {2 d x \sqrt [4]{a+b x^2} \left (c+d x^2\right ) (7 b c-10 a d)}{7 b}}{3 a b}+\frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 299

\(\displaystyle \frac {\frac {\frac {\left (40 a^3 d^3-84 a^2 b c d^2+42 a b^2 c^2 d+7 b^3 c^3\right ) \int \frac {1}{\left (b x^2+a\right )^{3/4}}dx}{b}-\frac {2 d x \sqrt [4]{a+b x^2} \left (20 a^2 d^2-32 a b c d+7 b^2 c^2\right )}{b}}{7 b}-\frac {2 d x \sqrt [4]{a+b x^2} \left (c+d x^2\right ) (7 b c-10 a d)}{7 b}}{3 a b}+\frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 231

\(\displaystyle \frac {\frac {\frac {\left (\frac {b x^2}{a}+1\right )^{3/4} \left (40 a^3 d^3-84 a^2 b c d^2+42 a b^2 c^2 d+7 b^3 c^3\right ) \int \frac {1}{\left (\frac {b x^2}{a}+1\right )^{3/4}}dx}{b \left (a+b x^2\right )^{3/4}}-\frac {2 d x \sqrt [4]{a+b x^2} \left (20 a^2 d^2-32 a b c d+7 b^2 c^2\right )}{b}}{7 b}-\frac {2 d x \sqrt [4]{a+b x^2} \left (c+d x^2\right ) (7 b c-10 a d)}{7 b}}{3 a b}+\frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/4}}\)

\(\Big \downarrow \) 229

\(\displaystyle \frac {\frac {\frac {2 \sqrt {a} \left (\frac {b x^2}{a}+1\right )^{3/4} \left (40 a^3 d^3-84 a^2 b c d^2+42 a b^2 c^2 d+7 b^3 c^3\right ) \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{b^{3/2} \left (a+b x^2\right )^{3/4}}-\frac {2 d x \sqrt [4]{a+b x^2} \left (20 a^2 d^2-32 a b c d+7 b^2 c^2\right )}{b}}{7 b}-\frac {2 d x \sqrt [4]{a+b x^2} \left (c+d x^2\right ) (7 b c-10 a d)}{7 b}}{3 a b}+\frac {2 x \left (c+d x^2\right )^2 (b c-a d)}{3 a b \left (a+b x^2\right )^{3/4}}\)

Input:

Int[(c + d*x^2)^3/(a + b*x^2)^(7/4),x]
 

Output:

(2*(b*c - a*d)*x*(c + d*x^2)^2)/(3*a*b*(a + b*x^2)^(3/4)) + ((-2*d*(7*b*c 
- 10*a*d)*x*(a + b*x^2)^(1/4)*(c + d*x^2))/(7*b) + ((-2*d*(7*b^2*c^2 - 32* 
a*b*c*d + 20*a^2*d^2)*x*(a + b*x^2)^(1/4))/b + (2*Sqrt[a]*(7*b^3*c^3 + 42* 
a*b^2*c^2*d - 84*a^2*b*c*d^2 + 40*a^3*d^3)*(1 + (b*x^2)/a)^(3/4)*EllipticF 
[ArcTan[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(b^(3/2)*(a + b*x^2)^(3/4)))/(7*b))/(3 
*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 229

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) 
)*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 231

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(1 + b*(x^2/a))^(3/4)/( 
a + b*x^2)^(3/4)   Int[1/(1 + b*(x^2/a))^(3/4), 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]

\[\int \frac {\left (x^{2} d +c \right )^{3}}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}d x\]

Input:

int((d*x^2+c)^3/(b*x^2+a)^(7/4),x)
 

Output:

int((d*x^2+c)^3/(b*x^2+a)^(7/4),x)
 

Fricas [F]

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

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

Sympy [F]

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

integrate((d*x**2+c)**3/(b*x**2+a)**(7/4),x)
 

Output:

Integral((c + d*x**2)**3/(a + b*x**2)**(7/4), x)
 

Maxima [F]

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

integrate((d*x^2+c)^3/(b*x^2+a)^(7/4),x, algorithm="maxima")
 

Output:

integrate((d*x^2 + c)^3/(b*x^2 + a)^(7/4), x)
 

Giac [F]

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

integrate((d*x^2+c)^3/(b*x^2+a)^(7/4),x, algorithm="giac")
 

Output:

integrate((d*x^2 + c)^3/(b*x^2 + a)^(7/4), x)
 

Mupad [F(-1)]

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

int((c + d*x^2)^3/(a + b*x^2)^(7/4),x)
 

Output:

int((c + d*x^2)^3/(a + b*x^2)^(7/4), x)
 

Reduce [F]

\[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{7/4}} \, dx=\left (\int \frac {x^{6}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a +\left (b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{2}}d x \right ) d^{3}+3 \left (\int \frac {x^{4}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a +\left (b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{2}}d x \right ) c \,d^{2}+3 \left (\int \frac {x^{2}}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a +\left (b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{2}}d x \right ) c^{2} d +\left (\int \frac {1}{\left (b \,x^{2}+a \right )^{\frac {3}{4}} a +\left (b \,x^{2}+a \right )^{\frac {3}{4}} b \,x^{2}}d x \right ) c^{3} \] Input:

int((d*x^2+c)^3/(b*x^2+a)^(7/4),x)
                                                                                    
                                                                                    
 

Output:

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