Integrand size = 40, antiderivative size = 554 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right )}{\left (c+d x^2\right )^{5/2}} \, dx=-\frac {\left (b c \left (5 c^2 C d-5 B c d^2+5 A d^3-8 c^3 D\right )-5 a d \left (c^2 C d-B c d^2+A d^3-c^3 D\right )\right ) x \sqrt {a+b x^2}}{15 c d^4 \left (c+d x^2\right )^{3/2}}+\frac {b D x^7 \sqrt {a+b x^2}}{5 d \left (c+d x^2\right )^{3/2}}+\frac {\left (3 a^2 d^2 D+a b d (20 C d-47 c D)-b^2 \left (35 c C d-15 B d^2-56 c^2 D\right )\right ) x \sqrt {a+b x^2}}{15 b d^4 \sqrt {c+d x^2}}+\frac {(5 b C d-8 b c D+6 a d D) x^3 \sqrt {a+b x^2}}{15 d^3 \sqrt {c+d x^2}}-\frac {\left (3 a^2 c^2 d^2 D+a b d \left (40 c^2 C d-5 B c d^2-10 A d^3-88 c^3 D\right )-2 b^2 c \left (40 c^2 C d-20 B c d^2+5 A d^3-64 c^3 D\right )\right ) \sqrt {a+b x^2} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b c^{3/2} d^{9/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}}+\frac {\left (3 a c d (5 C d-12 c D)-b \left (40 c^2 C d-20 B c d^2+5 A d^3-64 c^3 D\right )\right ) \sqrt {a+b x^2} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{15 \sqrt {c} d^{9/2} \sqrt {\frac {c \left (a+b x^2\right )}{a \left (c+d x^2\right )}} \sqrt {c+d x^2}} \] Output:
-1/15*(b*c*(5*A*d^3-5*B*c*d^2+5*C*c^2*d-8*D*c^3)-5*a*d*(A*d^3-B*c*d^2+C*c^ 2*d-D*c^3))*x*(b*x^2+a)^(1/2)/c/d^4/(d*x^2+c)^(3/2)+1/5*b*D*x^7*(b*x^2+a)^ (1/2)/d/(d*x^2+c)^(3/2)+1/15*(3*a^2*d^2*D+a*b*d*(20*C*d-47*D*c)-b^2*(-15*B *d^2+35*C*c*d-56*D*c^2))*x*(b*x^2+a)^(1/2)/b/d^4/(d*x^2+c)^(1/2)+1/15*(5*C *b*d+6*D*a*d-8*D*b*c)*x^3*(b*x^2+a)^(1/2)/d^3/(d*x^2+c)^(1/2)-1/15*(3*a^2* c^2*d^2*D+a*b*d*(-10*A*d^3-5*B*c*d^2+40*C*c^2*d-88*D*c^3)-2*b^2*c*(5*A*d^3 -20*B*c*d^2+40*C*c^2*d-64*D*c^3))*(b*x^2+a)^(1/2)*EllipticE(d^(1/2)*x/c^(1 /2)/(1+d*x^2/c)^(1/2),(1-b*c/a/d)^(1/2))/b/c^(3/2)/d^(9/2)/(c*(b*x^2+a)/a/ (d*x^2+c))^(1/2)/(d*x^2+c)^(1/2)+1/15*(3*a*c*d*(5*C*d-12*D*c)-b*(5*A*d^3-2 0*B*c*d^2+40*C*c^2*d-64*D*c^3))*(b*x^2+a)^(1/2)*InverseJacobiAM(arctan(d^( 1/2)*x/c^(1/2)),(1-b*c/a/d)^(1/2))/c^(1/2)/d^(9/2)/(c*(b*x^2+a)/a/(d*x^2+c ))^(1/2)/(d*x^2+c)^(1/2)
Result contains complex when optimal does not.
Time = 8.70 (sec) , antiderivative size = 511, normalized size of antiderivative = 0.92 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right )}{\left (c+d x^2\right )^{5/2}} \, dx=\frac {\sqrt {\frac {b}{a}} d x \left (a+b x^2\right ) \left (a d \left (36 c^4 D+10 A d^4 x^2+5 c d^3 \left (3 A+B x^2\right )+2 c^2 d^2 x^2 \left (-10 C+3 D x^2\right )+c^3 \left (-15 C d+47 d D x^2\right )\right )+b c \left (-64 c^4 D+10 A d^4 x^2+40 c^3 d \left (C-2 D x^2\right )-2 c^2 d^2 \left (10 B-25 C x^2+4 D x^4\right )+c d^3 \left (5 A-25 B x^2+5 C x^4+3 D x^6\right )\right )\right )-i c \left (3 a^2 c^2 d^2 D+2 b^2 c \left (-40 c^2 C d+20 B c d^2-5 A d^3+64 c^3 D\right )-a b d \left (-40 c^2 C d+5 B c d^2+10 A d^3+88 c^3 D\right )\right ) \sqrt {1+\frac {b x^2}{a}} \left (c+d x^2\right ) \sqrt {1+\frac {d x^2}{c}} E\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right )|\frac {a d}{b c}\right )+i c \left (3 a^2 c d^2 (-5 C d+13 c D)+2 b^2 c \left (-40 c^2 C d+20 B c d^2-5 A d^3+64 c^3 D\right )-a b d \left (-80 c^2 C d+25 B c d^2+5 A d^3+152 c^3 D\right )\right ) \sqrt {1+\frac {b x^2}{a}} \left (c+d x^2\right ) \sqrt {1+\frac {d x^2}{c}} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {\frac {b}{a}} x\right ),\frac {a d}{b c}\right )}{15 \sqrt {\frac {b}{a}} c^2 d^5 \sqrt {a+b x^2} \left (c+d x^2\right )^{3/2}} \] Input:
Integrate[((a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4 + D*x^6))/(c + d*x^2)^(5/2 ),x]
Output:
(Sqrt[b/a]*d*x*(a + b*x^2)*(a*d*(36*c^4*D + 10*A*d^4*x^2 + 5*c*d^3*(3*A + B*x^2) + 2*c^2*d^2*x^2*(-10*C + 3*D*x^2) + c^3*(-15*C*d + 47*d*D*x^2)) + b *c*(-64*c^4*D + 10*A*d^4*x^2 + 40*c^3*d*(C - 2*D*x^2) - 2*c^2*d^2*(10*B - 25*C*x^2 + 4*D*x^4) + c*d^3*(5*A - 25*B*x^2 + 5*C*x^4 + 3*D*x^6))) - I*c*( 3*a^2*c^2*d^2*D + 2*b^2*c*(-40*c^2*C*d + 20*B*c*d^2 - 5*A*d^3 + 64*c^3*D) - a*b*d*(-40*c^2*C*d + 5*B*c*d^2 + 10*A*d^3 + 88*c^3*D))*Sqrt[1 + (b*x^2)/ a]*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*EllipticE[I*ArcSinh[Sqrt[b/a]*x], (a*d) /(b*c)] + I*c*(3*a^2*c*d^2*(-5*C*d + 13*c*D) + 2*b^2*c*(-40*c^2*C*d + 20*B *c*d^2 - 5*A*d^3 + 64*c^3*D) - a*b*d*(-80*c^2*C*d + 25*B*c*d^2 + 5*A*d^3 + 152*c^3*D))*Sqrt[1 + (b*x^2)/a]*(c + d*x^2)*Sqrt[1 + (d*x^2)/c]*EllipticF [I*ArcSinh[Sqrt[b/a]*x], (a*d)/(b*c)])/(15*Sqrt[b/a]*c^2*d^5*Sqrt[a + b*x^ 2]*(c + d*x^2)^(3/2))
Leaf count is larger than twice the leaf count of optimal. \(1314\) vs. \(2(554)=1108\).
Time = 2.40 (sec) , antiderivative size = 1314, normalized size of antiderivative = 2.37, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.050, Rules used = {7293, 2009}
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.
\(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right )}{\left (c+d x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 7293 |
\(\displaystyle \int \left (\frac {A \left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^{5/2}}+\frac {B x^2 \left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^{5/2}}+\frac {C x^4 \left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^{5/2}}+\frac {D x^6 \left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^{5/2}}\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {(8 b c-5 a d) D \sqrt {b x^2+a} x^5}{3 c d^2 \sqrt {d x^2+c}}-\frac {D \left (b x^2+a\right )^{3/2} x^5}{3 d \left (d x^2+c\right )^{3/2}}+\frac {(48 b c-25 a d) D \sqrt {b x^2+a} \sqrt {d x^2+c} x^3}{15 c d^3}-\frac {C (2 b c-a d) \sqrt {b x^2+a} x^3}{c d^2 \sqrt {d x^2+c}}-\frac {C \left (b x^2+a\right )^{3/2} x^3}{3 d \left (d x^2+c\right )^{3/2}}+\frac {C (8 b c-3 a d) \sqrt {b x^2+a} \sqrt {d x^2+c} x}{3 c d^3}-\frac {4 (16 b c-9 a d) D \sqrt {b x^2+a} \sqrt {d x^2+c} x}{15 d^4}-\frac {8 C (2 b c-a d) \sqrt {b x^2+a} x}{3 d^3 \sqrt {d x^2+c}}-\frac {B (4 b c-a d) \sqrt {b x^2+a} x}{3 c d^2 \sqrt {d x^2+c}}+\frac {B (8 b c-a d) \sqrt {b x^2+a} x}{3 c d^2 \sqrt {d x^2+c}}+\frac {\left (128 b^2 c^2-88 a b d c+3 a^2 d^2\right ) D \sqrt {b x^2+a} x}{15 b d^4 \sqrt {d x^2+c}}-\frac {B \left (b x^2+a\right )^{3/2} x}{3 d \left (d x^2+c\right )^{3/2}}-\frac {A (b c-a d) \sqrt {b x^2+a} x}{3 c d \left (d x^2+c\right )^{3/2}}+\frac {8 \sqrt {c} C (2 b c-a d) \sqrt {b x^2+a} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 d^{7/2} \sqrt {\frac {c \left (b x^2+a\right )}{a \left (d x^2+c\right )}} \sqrt {d x^2+c}}-\frac {B (8 b c-a d) \sqrt {b x^2+a} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 \sqrt {c} d^{5/2} \sqrt {\frac {c \left (b x^2+a\right )}{a \left (d x^2+c\right )}} \sqrt {d x^2+c}}+\frac {2 A (b c+a d) \sqrt {b x^2+a} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{3 c^{3/2} d^{3/2} \sqrt {\frac {c \left (b x^2+a\right )}{a \left (d x^2+c\right )}} \sqrt {d x^2+c}}-\frac {\sqrt {c} \left (128 b^2 c^2-88 a b d c+3 a^2 d^2\right ) D \sqrt {b x^2+a} E\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right )|1-\frac {b c}{a d}\right )}{15 b d^{9/2} \sqrt {\frac {c \left (b x^2+a\right )}{a \left (d x^2+c\right )}} \sqrt {d x^2+c}}-\frac {\sqrt {c} C (8 b c-3 a d) \sqrt {b x^2+a} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{3 d^{7/2} \sqrt {\frac {c \left (b x^2+a\right )}{a \left (d x^2+c\right )}} \sqrt {d x^2+c}}+\frac {4 c^{3/2} (16 b c-9 a d) D \sqrt {b x^2+a} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{15 d^{9/2} \sqrt {\frac {c \left (b x^2+a\right )}{a \left (d x^2+c\right )}} \sqrt {d x^2+c}}-\frac {A b \sqrt {b x^2+a} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{3 \sqrt {c} d^{3/2} \sqrt {\frac {c \left (b x^2+a\right )}{a \left (d x^2+c\right )}} \sqrt {d x^2+c}}+\frac {4 b B \sqrt {c} \sqrt {b x^2+a} \operatorname {EllipticF}\left (\arctan \left (\frac {\sqrt {d} x}{\sqrt {c}}\right ),1-\frac {b c}{a d}\right )}{3 d^{5/2} \sqrt {\frac {c \left (b x^2+a\right )}{a \left (d x^2+c\right )}} \sqrt {d x^2+c}}\) |
Input:
Int[((a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4 + D*x^6))/(c + d*x^2)^(5/2),x]
Output:
-1/3*(A*(b*c - a*d)*x*Sqrt[a + b*x^2])/(c*d*(c + d*x^2)^(3/2)) - (B*x*(a + b*x^2)^(3/2))/(3*d*(c + d*x^2)^(3/2)) - (C*x^3*(a + b*x^2)^(3/2))/(3*d*(c + d*x^2)^(3/2)) - (D*x^5*(a + b*x^2)^(3/2))/(3*d*(c + d*x^2)^(3/2)) - (8* C*(2*b*c - a*d)*x*Sqrt[a + b*x^2])/(3*d^3*Sqrt[c + d*x^2]) - (B*(4*b*c - a *d)*x*Sqrt[a + b*x^2])/(3*c*d^2*Sqrt[c + d*x^2]) + (B*(8*b*c - a*d)*x*Sqrt [a + b*x^2])/(3*c*d^2*Sqrt[c + d*x^2]) + ((128*b^2*c^2 - 88*a*b*c*d + 3*a^ 2*d^2)*D*x*Sqrt[a + b*x^2])/(15*b*d^4*Sqrt[c + d*x^2]) - (C*(2*b*c - a*d)* x^3*Sqrt[a + b*x^2])/(c*d^2*Sqrt[c + d*x^2]) - ((8*b*c - 5*a*d)*D*x^5*Sqrt [a + b*x^2])/(3*c*d^2*Sqrt[c + d*x^2]) + (C*(8*b*c - 3*a*d)*x*Sqrt[a + b*x ^2]*Sqrt[c + d*x^2])/(3*c*d^3) - (4*(16*b*c - 9*a*d)*D*x*Sqrt[a + b*x^2]*S qrt[c + d*x^2])/(15*d^4) + ((48*b*c - 25*a*d)*D*x^3*Sqrt[a + b*x^2]*Sqrt[c + d*x^2])/(15*c*d^3) + (8*Sqrt[c]*C*(2*b*c - a*d)*Sqrt[a + b*x^2]*Ellipti cE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*d^(7/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) - (B*(8*b*c - a*d)*Sqrt[a + b*x^ 2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(3*Sqrt[c]*d^( 5/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^2]) + (2*A*(b*c + a*d)*Sqrt[a + b*x^2]*EllipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d )])/(3*c^(3/2)*d^(3/2)*Sqrt[(c*(a + b*x^2))/(a*(c + d*x^2))]*Sqrt[c + d*x^ 2]) - (Sqrt[c]*(128*b^2*c^2 - 88*a*b*c*d + 3*a^2*d^2)*D*Sqrt[a + b*x^2]*El lipticE[ArcTan[(Sqrt[d]*x)/Sqrt[c]], 1 - (b*c)/(a*d)])/(15*b*d^(9/2)*Sq...
Leaf count of result is larger than twice the leaf count of optimal. \(1119\) vs. \(2(513)=1026\).
Time = 9.72 (sec) , antiderivative size = 1120, normalized size of antiderivative = 2.02
method | result | size |
elliptic | \(\text {Expression too large to display}\) | \(1120\) |
default | \(\text {Expression too large to display}\) | \(3024\) |
Input:
int((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A)/(d*x^2+c)^(5/2),x,method=_RETURN VERBOSE)
Output:
((b*x^2+a)*(d*x^2+c))^(1/2)/(b*x^2+a)^(1/2)/(d*x^2+c)^(1/2)*(1/3*(A*a*d^4- A*b*c*d^3-B*a*c*d^3+B*b*c^2*d^2+C*a*c^2*d^2-C*b*c^3*d-D*a*c^3*d+D*b*c^4)/c /d^6*x*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)/(x^2+c/d)^2+1/3*(b*d*x^2+a*d)*( 2*A*a*d^4+2*A*b*c*d^3+B*a*c*d^3-5*B*b*c^2*d^2-4*C*a*c^2*d^2+8*C*b*c^3*d+7* D*a*c^3*d-11*D*b*c^4)/c^2/d^5*x/((x^2+c/d)*(b*d*x^2+a*d))^(1/2)+1/5*D*b/d^ 3*x^3*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+1/3*(b/d^3*(C*b*d+2*D*a*d-2*D*b* c)-1/5*D*b/d^3*(4*a*d+4*b*c))/b/d*x*(b*d*x^4+a*d*x^2+b*c*x^2+a*c)^(1/2)+(( A*b^2*d^3+2*B*a*b*d^3-2*B*b^2*c*d^2+C*a^2*d^3-4*C*a*b*c*d^2+3*C*b^2*c^2*d- 2*D*a^2*c*d^2+6*D*a*b*c^2*d-4*D*b^2*c^3)/d^5+1/3*(A*a*d^4-A*b*c*d^3-B*a*c* d^3+B*b*c^2*d^2+C*a*c^2*d^2-C*b*c^3*d-D*a*c^3*d+D*b*c^4)/d^5*b/c+1/3*(2*A* a*d^4+2*A*b*c*d^3+B*a*c*d^3-5*B*b*c^2*d^2-4*C*a*c^2*d^2+8*C*b*c^3*d+7*D*a* c^3*d-11*D*b*c^4)/d^5*(a*d-b*c)/c^2-1/3*a/d^4*(2*A*a*d^4+2*A*b*c*d^3+B*a*c *d^3-5*B*b*c^2*d^2-4*C*a*c^2*d^2+8*C*b*c^3*d+7*D*a*c^3*d-11*D*b*c^4)/c^2-1 /3*(b/d^3*(C*b*d+2*D*a*d-2*D*b*c)-1/5*D*b/d^3*(4*a*d+4*b*c))/b/d*a*c)/(-b/ a)^(1/2)*(1+b*x^2/a)^(1/2)*(1+d*x^2/c)^(1/2)/(b*d*x^4+a*d*x^2+b*c*x^2+a*c) ^(1/2)*EllipticF(x*(-b/a)^(1/2),(-1+(a*d+b*c)/c/b)^(1/2))-(1/d^4*(B*b^2*d^ 2+2*C*a*b*d^2-2*C*b^2*c*d+D*a^2*d^2-4*D*a*b*c*d+3*D*b^2*c^2)-1/3*(2*A*a*d^ 4+2*A*b*c*d^3+B*a*c*d^3-5*B*b*c^2*d^2-4*C*a*c^2*d^2+8*C*b*c^3*d+7*D*a*c^3* d-11*D*b*c^4)/d^4*b/c^2-3/5*D*b/d^3*a*c-1/3*(b/d^3*(C*b*d+2*D*a*d-2*D*b*c) -1/5*D*b/d^3*(4*a*d+4*b*c))/b/d*(2*a*d+2*b*c))*c/(-b/a)^(1/2)*(1+b*x^2/...
Leaf count of result is larger than twice the leaf count of optimal. 1077 vs. \(2 (514) = 1028\).
Time = 0.11 (sec) , antiderivative size = 1077, normalized size of antiderivative = 1.94 \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right )}{\left (c+d x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:
integrate((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A)/(d*x^2+c)^(5/2),x, algorit hm="fricas")
Output:
-1/15*(((128*D*b^2*c^5*d^2 - 10*A*a*b*c*d^6 - 8*(11*D*a*b + 10*C*b^2)*c^4* d^3 + (3*D*a^2 + 40*C*a*b + 40*B*b^2)*c^3*d^4 - 5*(B*a*b + 2*A*b^2)*c^2*d^ 5)*x^5 + 2*(128*D*b^2*c^6*d - 10*A*a*b*c^2*d^5 - 8*(11*D*a*b + 10*C*b^2)*c ^5*d^2 + (3*D*a^2 + 40*C*a*b + 40*B*b^2)*c^4*d^3 - 5*(B*a*b + 2*A*b^2)*c^3 *d^4)*x^3 + (128*D*b^2*c^7 - 10*A*a*b*c^3*d^4 - 8*(11*D*a*b + 10*C*b^2)*c^ 6*d + (3*D*a^2 + 40*C*a*b + 40*B*b^2)*c^5*d^2 - 5*(B*a*b + 2*A*b^2)*c^4*d^ 3)*x)*sqrt(b*d)*sqrt(-c/d)*elliptic_e(arcsin(sqrt(-c/d)/x), a*d/(b*c)) - ( (128*D*b^2*c^5*d^2 - 5*A*a*b*d^7 - 8*(11*D*a*b + 10*C*b^2)*c^4*d^3 + (3*D* a^2 + 8*(5*C + 8*D)*a*b + 40*B*b^2)*c^3*d^4 - (36*D*a^2 + 5*(B + 8*C)*a*b + 10*A*b^2)*c^2*d^5 + 5*(3*C*a^2 - 2*(A - 2*B)*a*b)*c*d^6)*x^5 + 2*(128*D* b^2*c^6*d - 5*A*a*b*c*d^6 - 8*(11*D*a*b + 10*C*b^2)*c^5*d^2 + (3*D*a^2 + 8 *(5*C + 8*D)*a*b + 40*B*b^2)*c^4*d^3 - (36*D*a^2 + 5*(B + 8*C)*a*b + 10*A* b^2)*c^3*d^4 + 5*(3*C*a^2 - 2*(A - 2*B)*a*b)*c^2*d^5)*x^3 + (128*D*b^2*c^7 - 5*A*a*b*c^2*d^5 - 8*(11*D*a*b + 10*C*b^2)*c^6*d + (3*D*a^2 + 8*(5*C + 8 *D)*a*b + 40*B*b^2)*c^5*d^2 - (36*D*a^2 + 5*(B + 8*C)*a*b + 10*A*b^2)*c^4* d^3 + 5*(3*C*a^2 - 2*(A - 2*B)*a*b)*c^3*d^4)*x)*sqrt(b*d)*sqrt(-c/d)*ellip tic_f(arcsin(sqrt(-c/d)/x), a*d/(b*c)) - (3*D*b^2*c^2*d^5*x^8 + 128*D*b^2* c^6*d - 10*A*a*b*c^2*d^5 - 8*(11*D*a*b + 10*C*b^2)*c^5*d^2 + (3*D*a^2 + 40 *C*a*b + 40*B*b^2)*c^4*d^3 - 5*(B*a*b + 2*A*b^2)*c^3*d^4 - (8*D*b^2*c^3*d^ 4 - (6*D*a*b + 5*C*b^2)*c^2*d^5)*x^6 + (48*D*b^2*c^4*d^3 - (41*D*a*b + ...
\[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right )}{\left (c+d x^2\right )^{5/2}} \, dx=\int \frac {\left (a + b x^{2}\right )^{\frac {3}{2}} \left (A + B x^{2} + C x^{4} + D x^{6}\right )}{\left (c + d x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((b*x**2+a)**(3/2)*(D*x**6+C*x**4+B*x**2+A)/(d*x**2+c)**(5/2),x)
Output:
Integral((a + b*x**2)**(3/2)*(A + B*x**2 + C*x**4 + D*x**6)/(c + d*x**2)** (5/2), x)
\[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right )}{\left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (D x^{6} + C x^{4} + B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}}}{{\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A)/(d*x^2+c)^(5/2),x, algorit hm="maxima")
Output:
integrate((D*x^6 + C*x^4 + B*x^2 + A)*(b*x^2 + a)^(3/2)/(d*x^2 + c)^(5/2), x)
\[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right )}{\left (c+d x^2\right )^{5/2}} \, dx=\int { \frac {{\left (D x^{6} + C x^{4} + B x^{2} + A\right )} {\left (b x^{2} + a\right )}^{\frac {3}{2}}}{{\left (d x^{2} + c\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A)/(d*x^2+c)^(5/2),x, algorit hm="giac")
Output:
integrate((D*x^6 + C*x^4 + B*x^2 + A)*(b*x^2 + a)^(3/2)/(d*x^2 + c)^(5/2), x)
Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right )}{\left (c+d x^2\right )^{5/2}} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/2}\,\left (A+B\,x^2+C\,x^4+x^6\,D\right )}{{\left (d\,x^2+c\right )}^{5/2}} \,d x \] Input:
int(((a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4 + x^6*D))/(c + d*x^2)^(5/2),x)
Output:
int(((a + b*x^2)^(3/2)*(A + B*x^2 + C*x^4 + x^6*D))/(c + d*x^2)^(5/2), x)
\[ \int \frac {\left (a+b x^2\right )^{3/2} \left (A+B x^2+C x^4+D x^6\right )}{\left (c+d x^2\right )^{5/2}} \, dx=\text {too large to display} \] Input:
int((b*x^2+a)^(3/2)*(D*x^6+C*x^4+B*x^2+A)/(d*x^2+c)^(5/2),x)
Output:
(3*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a**2*c*d*x + 2*sqrt(c + d*x**2)*sqrt( a + b*x**2)*a**2*d**2*x**3 - 15*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*b**2*d *x - 18*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*b*c**2*x - 14*sqrt(c + d*x**2) *sqrt(a + b*x**2)*a*b*c*d*x**3 + 4*sqrt(c + d*x**2)*sqrt(a + b*x**2)*a*b*d **2*x**5 + 10*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**3*d*x**3 + 12*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**2*c**2*x**3 - 2*sqrt(c + d*x**2)*sqrt(a + b*x* *2)*b**2*c*d*x**5 + 2*sqrt(c + d*x**2)*sqrt(a + b*x**2)*b**2*d**2*x**7 - 1 5*int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a*c**3 + 3*a*c**2*d*x**2 + 3*a*c*d**2*x**4 + a*d**3*x**6 + b*c**3*x**2 + 3*b*c**2*d*x**4 + 3*b*c*d** 2*x**6 + b*d**3*x**8),x)*a**2*b*c**3*d**2 - 30*int((sqrt(c + d*x**2)*sqrt( a + b*x**2)*x**4)/(a*c**3 + 3*a*c**2*d*x**2 + 3*a*c*d**2*x**4 + a*d**3*x** 6 + b*c**3*x**2 + 3*b*c**2*d*x**4 + 3*b*c*d**2*x**6 + b*d**3*x**8),x)*a**2 *b*c**2*d**3*x**2 - 15*int((sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a*c** 3 + 3*a*c**2*d*x**2 + 3*a*c*d**2*x**4 + a*d**3*x**6 + b*c**3*x**2 + 3*b*c* *2*d*x**4 + 3*b*c*d**2*x**6 + b*d**3*x**8),x)*a**2*b*c*d**4*x**4 + 15*int( (sqrt(c + d*x**2)*sqrt(a + b*x**2)*x**4)/(a*c**3 + 3*a*c**2*d*x**2 + 3*a*c *d**2*x**4 + a*d**3*x**6 + b*c**3*x**2 + 3*b*c**2*d*x**4 + 3*b*c*d**2*x**6 + b*d**3*x**8),x)*a*b**3*c**2*d**2 + 30*int((sqrt(c + d*x**2)*sqrt(a + b* x**2)*x**4)/(a*c**3 + 3*a*c**2*d*x**2 + 3*a*c*d**2*x**4 + a*d**3*x**6 + b* c**3*x**2 + 3*b*c**2*d*x**4 + 3*b*c*d**2*x**6 + b*d**3*x**8),x)*a*b**3*...