Integrand size = 22, antiderivative size = 185 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a-b x^2\right )^{5/4}} \, dx=\frac {2 (b c+a d)^3 x}{a b^3 \sqrt [4]{a-b x^2}}+\frac {2 d^2 (9 b c+5 a d) x \left (a-b x^2\right )^{3/4}}{15 b^3}+\frac {2 d^3 x^3 \left (a-b x^2\right )^{3/4}}{9 b^2}-\frac {2 \left (15 b^3 c^3+90 a b^2 c^2 d+108 a^2 b c d^2+40 a^3 d^3\right ) \sqrt [4]{1-\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{15 \sqrt {a} b^{7/2} \sqrt [4]{a-b x^2}} \] Output:
2*(a*d+b*c)^3*x/a/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)*EllipticE(sin(1/2*arcsin(b ^(1/2)*x/a^(1/2))),2^(1/2))/a^(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 = 162, normalized size of antiderivative = 0.88 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a-b x^2\right )^{5/4}} \, dx=\frac {2 x \left (45 b^3 c^3+60 a^3 d^3+2 a^2 b d^2 \left (81 c-5 d x^2\right )+a b^2 d \left (135 c^2-27 c d x^2-5 d^2 x^4\right )\right )-3 \left (15 b^3 c^3+90 a b^2 c^2 d+108 a^2 b c d^2+40 a^3 d^3\right ) x \sqrt [4]{1-\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {3}{2},\frac {b x^2}{a}\right )}{45 a b^3 \sqrt [4]{a-b x^2}} \] Input:
Integrate[(c + d*x^2)^3/(a - b*x^2)^(5/4),x]
Output:
(2*x*(45*b^3*c^3 + 60*a^3*d^3 + 2*a^2*b*d^2*(81*c - 5*d*x^2) + a*b^2*d*(13 5*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)*x*(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))
Time = 0.35 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.26, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {315, 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.
| \(\displaystyle \int \frac {\left (c+d x^2\right )^3}{\left (a-b x^2\right )^{5/4}} \, dx\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (a d+b c)}{a b \sqrt [4]{a-b x^2}}-\frac {2 \int \frac {\left (d x^2+c\right ) \left (d (9 b c+10 a d) x^2+c (b c+2 a d)\right )}{2 \sqrt [4]{a-b x^2}}dx}{a b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (a d+b c)}{a b \sqrt [4]{a-b x^2}}-\frac {\int \frac {\left (d x^2+c\right ) \left (d (9 b c+10 a d) x^2+c (b c+2 a d)\right )}{\sqrt [4]{a-b x^2}}dx}{a b}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (a d+b c)}{a b \sqrt [4]{a-b x^2}}-\frac {-\frac {2 \int -\frac {d \left (45 b^2 c^2+112 a b d c+60 a^2 d^2\right ) x^2+c (3 b c+2 a d) (3 b c+10 a d)}{2 \sqrt [4]{a-b x^2}}dx}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+9 b c)}{9 b}}{a b}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (a d+b c)}{a b \sqrt [4]{a-b x^2}}-\frac {\frac {\int \frac {d \left (45 b^2 c^2+112 a b d c+60 a^2 d^2\right ) x^2+c (3 b c+2 a d) (3 b c+10 a d)}{\sqrt [4]{a-b x^2}}dx}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+9 b c)}{9 b}}{a b}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (a d+b c)}{a b \sqrt [4]{a-b x^2}}-\frac {\frac {\frac {3 \left (40 a^3 d^3+108 a^2 b c d^2+90 a b^2 c^2 d+15 b^3 c^3\right ) \int \frac {1}{\sqrt [4]{a-b x^2}}dx}{5 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (60 a^2 d^2+112 a b c d+45 b^2 c^2\right )}{5 b}}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+9 b c)}{9 b}}{a b}\) |
| \(\Big \downarrow \) 227 |
| \(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (a d+b c)}{a b \sqrt [4]{a-b x^2}}-\frac {\frac {\frac {3 \sqrt [4]{1-\frac {b x^2}{a}} \left (40 a^3 d^3+108 a^2 b c d^2+90 a b^2 c^2 d+15 b^3 c^3\right ) \int \frac {1}{\sqrt [4]{1-\frac {b x^2}{a}}}dx}{5 b \sqrt [4]{a-b x^2}}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (60 a^2 d^2+112 a b c d+45 b^2 c^2\right )}{5 b}}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+9 b c)}{9 b}}{a b}\) |
| \(\Big \downarrow \) 226 |
| \(\displaystyle \frac {2 x \left (c+d x^2\right )^2 (a d+b c)}{a b \sqrt [4]{a-b x^2}}-\frac {\frac {\frac {6 \sqrt {a} \sqrt [4]{1-\frac {b x^2}{a}} \left (40 a^3 d^3+108 a^2 b c d^2+90 a b^2 c^2 d+15 b^3 c^3\right ) E\left (\left .\frac {1}{2} \arcsin \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 b^{3/2} \sqrt [4]{a-b x^2}}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (60 a^2 d^2+112 a b c d+45 b^2 c^2\right )}{5 b}}{9 b}-\frac {2 d x \left (a-b x^2\right )^{3/4} \left (c+d x^2\right ) (10 a d+9 b c)}{9 b}}{a b}\) |
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 + 10*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) + (6*Sqrt[a]*(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)*Ell ipticE[ArcSin[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(5*b^(3/2)*(a - b*x^2)^(1/4)))/( 9*b))/(a*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[(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]
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]
\[\int \frac {\left (x^{2} d +c \right )^{3}}{\left (-b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
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)
\[ \int \frac {\left (c+d x^2\right )^3}{\left (a-b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{3}}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] 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)
\[ \int \frac {\left (c+d x^2\right )^3}{\left (a-b x^2\right )^{5/4}} \, dx=\int \frac {\left (c + d x^{2}\right )^{3}}{\left (a - b x^{2}\right )^{\frac {5}{4}}}\, dx \] 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)
\[ \int \frac {\left (c+d x^2\right )^3}{\left (a-b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{3}}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] 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)
\[ \int \frac {\left (c+d x^2\right )^3}{\left (a-b x^2\right )^{5/4}} \, dx=\int { \frac {{\left (d x^{2} + c\right )}^{3}}{{\left (-b x^{2} + a\right )}^{\frac {5}{4}}} \,d x } \] 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)
Timed out. \[ \int \frac {\left (c+d x^2\right )^3}{\left (a-b x^2\right )^{5/4}} \, dx=\int \frac {{\left (d\,x^2+c\right )}^3}{{\left (a-b\,x^2\right )}^{5/4}} \,d x \] 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)
\[ \int \frac {\left (c+d x^2\right )^3}{\left (a-b x^2\right )^{5/4}} \, dx=\left (\int \frac {x^{6}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}} a -\left (-b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) d^{3}+3 \left (\int \frac {x^{4}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}} a -\left (-b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c \,d^{2}+3 \left (\int \frac {x^{2}}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}} a -\left (-b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c^{2} d +\left (\int \frac {1}{\left (-b \,x^{2}+a \right )^{\frac {1}{4}} a -\left (-b \,x^{2}+a \right )^{\frac {1}{4}} b \,x^{2}}d x \right ) c^{3} \] 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