Integrand size = 22, antiderivative size = 200 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt [4]{a-b x^2}} \, dx=-\frac {2 d \left (117 b^2 c^2+78 a b c d+20 a^2 d^2\right ) x \left (a-b x^2\right )^{3/4}}{195 b^3}-\frac {2 d^2 (39 b c+10 a d) x^3 \left (a-b x^2\right )^{3/4}}{117 b^2}-\frac {2 d^3 x^5 \left (a-b x^2\right )^{3/4}}{13 b}+\frac {2 \sqrt {a} \left (195 b^3 c^3+2 a d \left (117 b^2 c^2+78 a b c d+20 a^2 d^2\right )\right ) \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{195 b^{7/2} \sqrt [4]{a-b x^2}} \] Output:
-2/195*d*(20*a^2*d^2+78*a*b*c*d+117*b^2*c^2)*x*(-b*x^2+a)^(3/4)/b^3-2/117* d^2*(10*a*d+39*b*c)*x^3*(-b*x^2+a)^(3/4)/b^2-2/13*d^3*x^5*(-b*x^2+a)^(3/4) /b+2/195*a^(1/2)*(195*b^3*c^3+2*a*d*(20*a^2*d^2+78*a*b*c*d+117*b^2*c^2))*( 1-b*x^2/a)^(1/4)*EllipticE(sin(1/2*arcsin(b^(1/2)*x/a^(1/2))),2^(1/2))/b^( 7/2)/(-b*x^2+a)^(1/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 15.11 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.77 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt [4]{a-b x^2}} \, dx=\frac {x \left (-2 d \left (a-b x^2\right ) \left (60 a^2 d^2+2 a b d \left (117 c+25 d x^2\right )+3 b^2 \left (117 c^2+65 c d x^2+15 d^2 x^4\right )\right )+3 \left (195 b^3 c^3+234 a b^2 c^2 d+156 a^2 b c d^2+40 a^3 d^3\right ) \sqrt [4]{1-\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {b x^2}{a}\right )\right )}{585 b^3 \sqrt [4]{a-b x^2}} \] Input:
Integrate[(c + d*x^2)^3/(a - b*x^2)^(1/4),x]
Output:
(x*(-2*d*(a - b*x^2)*(60*a^2*d^2 + 2*a*b*d*(117*c + 25*d*x^2) + 3*b^2*(117 *c^2 + 65*c*d*x^2 + 15*d^2*x^4)) + 3*(195*b^3*c^3 + 234*a*b^2*c^2*d + 156* 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]))/(585*b^3*(a - b*x^2)^(1/4))
Time = 0.34 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.12, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {318, 27, 403, 27, 299, 227, 226}
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.
\(\displaystyle \int \frac {\left (c+d x^2\right )^3}{\sqrt [4]{a-b x^2}} \, dx\) |
\(\Big \downarrow \) 318 |
\(\displaystyle -\frac {2 \int -\frac {\left (d x^2+c\right ) \left (d (21 b c+10 a d) x^2+c (13 b c+2 a d)\right )}{2 \sqrt [4]{a-b x^2}}dx}{13 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )^2}{13 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\left (d x^2+c\right ) \left (d (21 b c+10 a d) x^2+c (13 b c+2 a d)\right )}{\sqrt [4]{a-b x^2}}dx}{13 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )^2}{13 b}\) |
\(\Big \downarrow \) 403 |
\(\displaystyle \frac {-\frac {2 \int -\frac {d \left (201 b^2 c^2+184 a b d c+60 a^2 d^2\right ) x^2+c \left (117 b^2 c^2+60 a b d c+20 a^2 d^2\right )}{2 \sqrt [4]{a-b x^2}}dx}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+21 b c)}{9 b}}{13 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )^2}{13 b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {d \left (201 b^2 c^2+184 a b d c+60 a^2 d^2\right ) x^2+c \left (117 b^2 c^2+60 a b d c+20 a^2 d^2\right )}{\sqrt [4]{a-b x^2}}dx}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+21 b c)}{9 b}}{13 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )^2}{13 b}\) |
\(\Big \downarrow \) 299 |
\(\displaystyle \frac {\frac {\frac {3 \left (40 a^3 d^3+156 a^2 b c d^2+234 a b^2 c^2 d+195 b^3 c^3\right ) \int \frac {1}{\sqrt [4]{a-b x^2}}dx}{5 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (60 a^2 d^2+184 a b c d+201 b^2 c^2\right )}{5 b}}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+21 b c)}{9 b}}{13 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )^2}{13 b}\) |
\(\Big \downarrow \) 227 |
\(\displaystyle \frac {\frac {\frac {3 \sqrt [4]{1-\frac {b x^2}{a}} \left (40 a^3 d^3+156 a^2 b c d^2+234 a b^2 c^2 d+195 b^3 c^3\right ) \int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}}dx}{5 b \sqrt [4]{a-b x^2}}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (60 a^2 d^2+184 a b c d+201 b^2 c^2\right )}{5 b}}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+21 b c)}{9 b}}{13 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )^2}{13 b}\) |
\(\Big \downarrow \) 226 |
\(\displaystyle \frac {\frac {\frac {6 \sqrt {a} \sqrt [4]{1-\frac {b x^2}{a}} \left (40 a^3 d^3+156 a^2 b c d^2+234 a b^2 c^2 d+195 b^3 c^3\right ) E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a-b x^2}}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (60 a^2 d^2+184 a b c d+201 b^2 c^2\right )}{5 b}}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+21 b c)}{9 b}}{13 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right )^2}{13 b}\) |
Input:
Int[(c + d*x^2)^3/(a - b*x^2)^(1/4),x]
Output:
(-2*d*x*(a - b*x^2)^(3/4)*(c + d*x^2)^2)/(13*b) + ((-2*d*(21*b*c + 10*a*d) *x*(a - b*x^2)^(3/4)*(c + d*x^2))/(9*b) + ((-2*d*(201*b^2*c^2 + 184*a*b*c* d + 60*a^2*d^2)*x*(a - b*x^2)^(3/4))/(5*b) + (6*Sqrt[a]*(195*b^3*c^3 + 234 *a*b^2*c^2*d + 156*a^2*b*c*d^2 + 40*a^3*d^3)*(1 - (b*x^2)/a)^(1/4)*Ellipti cE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*b^(3/2)*(a - b*x^2)^(1/4)))/(9*b) )/(13*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2] ))*EllipticE[(1/2)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ [a, 0] && NegQ[b/a]
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]
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]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[d*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(b*(2*(p + q) + 1))), x] + S imp[1/(b*(2*(p + q) + 1)) Int[(a + b*x^2)^p*(c + d*x^2)^(q - 2)*Simp[c*(b *c*(2*(p + q) + 1) - a*d) + d*(b*c*(2*(p + 2*q - 1) + 1) - a*d*(2*(q - 1) + 1))*x^2, x], x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && G tQ[q, 1] && NeQ[2*(p + q) + 1, 0] && !IGtQ[p, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
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]
\[\int \frac {\left (x^{2} d +c \right )^{3}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}d x\]
Input:
int((d*x^2+c)^3/(-b*x^2+a)^(1/4),x)
Output:
int((d*x^2+c)^3/(-b*x^2+a)^(1/4),x)
\[ \int \frac {\left (c+d x^2\right )^3}{\sqrt [4]{a-b x^2}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{3}}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((d*x^2+c)^3/(-b*x^2+a)^(1/4),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*x^2 - a), x)
Result contains complex when optimal does not.
Time = 1.75 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.67 \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt [4]{a-b x^2}} \, dx=\frac {c^{3} x {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {1}{2} \\ \frac {3}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{\sqrt [4]{a}} + \frac {c^{2} d x^{3} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {3}{2} \\ \frac {5}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{\sqrt [4]{a}} + \frac {3 c d^{2} x^{5} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{5 \sqrt [4]{a}} + \frac {d^{3} x^{7} {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{2} \\ \frac {9}{2} \end {matrix}\middle | {\frac {b x^{2} e^{2 i \pi }}{a}} \right )}}{7 \sqrt [4]{a}} \] Input:
integrate((d*x**2+c)**3/(-b*x**2+a)**(1/4),x)
Output:
c**3*x*hyper((1/4, 1/2), (3/2,), b*x**2*exp_polar(2*I*pi)/a)/a**(1/4) + c* *2*d*x**3*hyper((1/4, 3/2), (5/2,), b*x**2*exp_polar(2*I*pi)/a)/a**(1/4) + 3*c*d**2*x**5*hyper((1/4, 5/2), (7/2,), b*x**2*exp_polar(2*I*pi)/a)/(5*a* *(1/4)) + d**3*x**7*hyper((1/4, 7/2), (9/2,), b*x**2*exp_polar(2*I*pi)/a)/ (7*a**(1/4))
\[ \int \frac {\left (c+d x^2\right )^3}{\sqrt [4]{a-b x^2}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{3}}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((d*x^2+c)^3/(-b*x^2+a)^(1/4),x, algorithm="maxima")
Output:
integrate((d*x^2 + c)^3/(-b*x^2 + a)^(1/4), x)
\[ \int \frac {\left (c+d x^2\right )^3}{\sqrt [4]{a-b x^2}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{3}}{{\left (-b x^{2} + a\right )}^{\frac {1}{4}}} \,d x } \] Input:
integrate((d*x^2+c)^3/(-b*x^2+a)^(1/4),x, algorithm="giac")
Output:
integrate((d*x^2 + c)^3/(-b*x^2 + a)^(1/4), x)
Timed out. \[ \int \frac {\left (c+d x^2\right )^3}{\sqrt [4]{a-b x^2}} \, dx=\int \frac {{\left (d\,x^2+c\right )}^3}{{\left (a-b\,x^2\right )}^{1/4}} \,d x \] Input:
int((c + d*x^2)^3/(a - b*x^2)^(1/4),x)
Output:
int((c + d*x^2)^3/(a - b*x^2)^(1/4), x)
\[ \int \frac {\left (c+d x^2\right )^3}{\sqrt [4]{a-b x^2}} \, dx=\left (\int \frac {x^{6}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) d^{3}+3 \left (\int \frac {x^{4}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) c \,d^{2}+3 \left (\int \frac {x^{2}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) c^{2} d +\left (\int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}d x \right ) c^{3} \] Input:
int((d*x^2+c)^3/(-b*x^2+a)^(1/4),x)
Output:
int(x**6/(a - b*x**2)**(1/4),x)*d**3 + 3*int(x**4/(a - b*x**2)**(1/4),x)*c *d**2 + 3*int(x**2/(a - b*x**2)**(1/4),x)*c**2*d + int(1/(a - b*x**2)**(1/ 4),x)*c**3