Integrand size = 21, antiderivative size = 300 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^5} \, dx=-\frac {(b c-a d) x \sqrt {a+b x^2}}{8 c d \left (c+d x^2\right )^4}+\frac {(2 b c+7 a d) x \sqrt {a+b x^2}}{48 c^2 d \left (c+d x^2\right )^3}+\frac {\left (8 b^2 c^2+24 a b c d-35 a^2 d^2\right ) x \sqrt {a+b x^2}}{192 c^3 d (b c-a d) \left (c+d x^2\right )^2}+\frac {\left (16 b^3 c^3+40 a b^2 c^2 d-170 a^2 b c d^2+105 a^3 d^3\right ) x \sqrt {a+b x^2}}{384 c^4 d (b c-a d)^2 \left (c+d x^2\right )}+\frac {a^2 \left (48 b^2 c^2-80 a b c d+35 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{128 c^{9/2} (b c-a d)^{5/2}} \]
1/128*a^2*(35*a^2*d^2-80*a*b*c*d+48*b^2*c^2)*arctanh(x*(-a*d+b*c)^(1/2)/c^ (1/2)/(b*x^2+a)^(1/2))/c^(9/2)/(-a*d+b*c)^(5/2)-1/8*(-a*d+b*c)*x*(b*x^2+a) ^(1/2)/c/d/(d*x^2+c)^4+1/48*(7*a*d+2*b*c)*x*(b*x^2+a)^(1/2)/c^2/d/(d*x^2+c )^3+1/192*(-35*a^2*d^2+24*a*b*c*d+8*b^2*c^2)*x*(b*x^2+a)^(1/2)/c^3/d/(-a*d +b*c)/(d*x^2+c)^2+1/384*(105*a^3*d^3-170*a^2*b*c*d^2+40*a*b^2*c^2*d+16*b^3 *c^3)*x*(b*x^2+a)^(1/2)/c^4/d/(-a*d+b*c)^2/(d*x^2+c)
Time = 10.91 (sec) , antiderivative size = 305, normalized size of antiderivative = 1.02 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^5} \, dx=\frac {a x \left (1+\frac {b x^2}{a}\right ) \left (c \left (a+b x^2\right ) \left (16 b^3 c^3 x^2 \left (6 c^2+4 c d x^2+d^2 x^4\right )+8 a b^2 c^2 \left (30 c^3+26 c^2 d x^2+19 c d^2 x^4+5 d^3 x^6\right )-2 a^2 b c d \left (264 c^3+421 c^2 d x^2+314 c d^2 x^4+85 d^3 x^6\right )+a^3 d^2 \left (279 c^3+511 c^2 d x^2+385 c d^2 x^4+105 d^3 x^6\right )\right )+\frac {3 a^2 \left (48 b^2 c^2-80 a b c d+35 a^2 d^2\right ) \left (c+d x^2\right )^4 \text {arctanh}\left (\sqrt {\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}}\right )}{\sqrt {\frac {(b c-a d) x^2}{c \left (a+b x^2\right )}}}\right )}{384 c^5 (b c-a d)^2 \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )^4} \]
(a*x*(1 + (b*x^2)/a)*(c*(a + b*x^2)*(16*b^3*c^3*x^2*(6*c^2 + 4*c*d*x^2 + d ^2*x^4) + 8*a*b^2*c^2*(30*c^3 + 26*c^2*d*x^2 + 19*c*d^2*x^4 + 5*d^3*x^6) - 2*a^2*b*c*d*(264*c^3 + 421*c^2*d*x^2 + 314*c*d^2*x^4 + 85*d^3*x^6) + a^3* d^2*(279*c^3 + 511*c^2*d*x^2 + 385*c*d^2*x^4 + 105*d^3*x^6)) + (3*a^2*(48* b^2*c^2 - 80*a*b*c*d + 35*a^2*d^2)*(c + d*x^2)^4*ArcTanh[Sqrt[((b*c - a*d) *x^2)/(c*(a + b*x^2))]])/Sqrt[((b*c - a*d)*x^2)/(c*(a + b*x^2))]))/(384*c^ 5*(b*c - a*d)^2*(a + b*x^2)^(3/2)*(c + d*x^2)^4)
Time = 0.47 (sec) , antiderivative size = 329, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {315, 402, 27, 402, 402, 27, 291, 221}
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.
| \(\displaystyle \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^5} \, dx\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\int \frac {2 b (b c+3 a d) x^2+a (b c+7 a d)}{\sqrt {b x^2+a} \left (d x^2+c\right )^4}dx}{8 c d}-\frac {x \sqrt {a+b x^2} (b c-a d)}{8 c d \left (c+d x^2\right )^4}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\int \frac {(b c-a d) \left (4 b (2 b c+7 a d) x^2+a (4 b c+35 a d)\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )^3}dx}{6 c (b c-a d)}+\frac {x \sqrt {a+b x^2} (7 a d+2 b c)}{6 c \left (c+d x^2\right )^3}}{8 c d}-\frac {x \sqrt {a+b x^2} (b c-a d)}{8 c d \left (c+d x^2\right )^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\int \frac {4 b (2 b c+7 a d) x^2+a (4 b c+35 a d)}{\sqrt {b x^2+a} \left (d x^2+c\right )^3}dx}{6 c}+\frac {x \sqrt {a+b x^2} (7 a d+2 b c)}{6 c \left (c+d x^2\right )^3}}{8 c d}-\frac {x \sqrt {a+b x^2} (b c-a d)}{8 c d \left (c+d x^2\right )^4}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {2 b \left (8 b^2 c^2+24 a b d c-35 a^2 d^2\right ) x^2+a \left (8 b^2 c^2+100 a b d c-105 a^2 d^2\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )^2}dx}{4 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (-35 a^2 d^2+24 a b c d+8 b^2 c^2\right )}{4 c \left (c+d x^2\right )^2 (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2} (7 a d+2 b c)}{6 c \left (c+d x^2\right )^3}}{8 c d}-\frac {x \sqrt {a+b x^2} (b c-a d)}{8 c d \left (c+d x^2\right )^4}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\frac {\int \frac {3 a^2 d \left (48 b^2 c^2-80 a b d c+35 a^2 d^2\right )}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (105 a^3 d^3-170 a^2 b c d^2+40 a b^2 c^2 d+16 b^3 c^3\right )}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (-35 a^2 d^2+24 a b c d+8 b^2 c^2\right )}{4 c \left (c+d x^2\right )^2 (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2} (7 a d+2 b c)}{6 c \left (c+d x^2\right )^3}}{8 c d}-\frac {x \sqrt {a+b x^2} (b c-a d)}{8 c d \left (c+d x^2\right )^4}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 a^2 d \left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{2 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (105 a^3 d^3-170 a^2 b c d^2+40 a b^2 c^2 d+16 b^3 c^3\right )}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (-35 a^2 d^2+24 a b c d+8 b^2 c^2\right )}{4 c \left (c+d x^2\right )^2 (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2} (7 a d+2 b c)}{6 c \left (c+d x^2\right )^3}}{8 c d}-\frac {x \sqrt {a+b x^2} (b c-a d)}{8 c d \left (c+d x^2\right )^4}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {\frac {3 a^2 d \left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \int \frac {1}{c-\frac {(b c-a d) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (105 a^3 d^3-170 a^2 b c d^2+40 a b^2 c^2 d+16 b^3 c^3\right )}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}+\frac {x \sqrt {a+b x^2} \left (-35 a^2 d^2+24 a b c d+8 b^2 c^2\right )}{4 c \left (c+d x^2\right )^2 (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2} (7 a d+2 b c)}{6 c \left (c+d x^2\right )^3}}{8 c d}-\frac {x \sqrt {a+b x^2} (b c-a d)}{8 c d \left (c+d x^2\right )^4}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {x \sqrt {a+b x^2} \left (-35 a^2 d^2+24 a b c d+8 b^2 c^2\right )}{4 c \left (c+d x^2\right )^2 (b c-a d)}+\frac {\frac {3 a^2 d \left (35 a^2 d^2-80 a b c d+48 b^2 c^2\right ) \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{3/2}}+\frac {x \sqrt {a+b x^2} \left (105 a^3 d^3-170 a^2 b c d^2+40 a b^2 c^2 d+16 b^3 c^3\right )}{2 c \left (c+d x^2\right ) (b c-a d)}}{4 c (b c-a d)}}{6 c}+\frac {x \sqrt {a+b x^2} (7 a d+2 b c)}{6 c \left (c+d x^2\right )^3}}{8 c d}-\frac {x \sqrt {a+b x^2} (b c-a d)}{8 c d \left (c+d x^2\right )^4}\) |
-1/8*((b*c - a*d)*x*Sqrt[a + b*x^2])/(c*d*(c + d*x^2)^4) + (((2*b*c + 7*a* d)*x*Sqrt[a + b*x^2])/(6*c*(c + d*x^2)^3) + (((8*b^2*c^2 + 24*a*b*c*d - 35 *a^2*d^2)*x*Sqrt[a + b*x^2])/(4*c*(b*c - a*d)*(c + d*x^2)^2) + (((16*b^3*c ^3 + 40*a*b^2*c^2*d - 170*a^2*b*c*d^2 + 105*a^3*d^3)*x*Sqrt[a + b*x^2])/(2 *c*(b*c - a*d)*(c + d*x^2)) + (3*a^2*d*(48*b^2*c^2 - 80*a*b*c*d + 35*a^2*d ^2)*ArcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(2*c^(3/2)*(b* c - a*d)^(3/2)))/(4*c*(b*c - a*d)))/(6*c))/(8*c*d)
3.1.61.3.1 Defintions of rubi rules used
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), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[1/(Sqrt[(a_) + (b_.)*(x_)^2]*((c_) + (d_.)*(x_)^2)), x_Symbol] :> Subst [Int[1/(c - (b*c - a*d)*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 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[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Time = 2.60 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.90
| method | result | size |
| pseudoelliptic | \(\frac {93 x \left (d^{2} \left (\frac {35}{93} d^{3} x^{6}+\frac {385}{279} c \,d^{2} x^{4}+\frac {511}{279} c^{2} d \,x^{2}+c^{3}\right ) a^{3}-\frac {176 b d \left (\frac {85}{264} d^{3} x^{6}+\frac {157}{132} c \,d^{2} x^{4}+\frac {421}{264} c^{2} d \,x^{2}+c^{3}\right ) c \,a^{2}}{93}+\frac {80 b^{2} \left (\frac {1}{6} d^{3} x^{6}+\frac {19}{30} c \,d^{2} x^{4}+\frac {13}{15} c^{2} d \,x^{2}+c^{3}\right ) c^{2} a}{93}+\frac {32 x^{2} \left (\frac {1}{6} d^{2} x^{4}+\frac {2}{3} c d \,x^{2}+c^{2}\right ) b^{3} c^{3}}{93}\right ) \sqrt {\left (a d -b c \right ) c}\, \sqrt {b \,x^{2}+a}-35 \left (a^{2} d^{2}-\frac {16}{7} a b c d +\frac {48}{35} b^{2} c^{2}\right ) \left (d \,x^{2}+c \right )^{4} a^{2} \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )}{128 \sqrt {\left (a d -b c \right ) c}\, \left (d \,x^{2}+c \right )^{4} \left (a d -b c \right )^{2} c^{4}}\) | \(270\) |
| default | \(\text {Expression too large to display}\) | \(22502\) |
1/128/((a*d-b*c)*c)^(1/2)*(93*x*(d^2*(35/93*d^3*x^6+385/279*c*d^2*x^4+511/ 279*c^2*d*x^2+c^3)*a^3-176/93*b*d*(85/264*d^3*x^6+157/132*c*d^2*x^4+421/26 4*c^2*d*x^2+c^3)*c*a^2+80/93*b^2*(1/6*d^3*x^6+19/30*c*d^2*x^4+13/15*c^2*d* x^2+c^3)*c^2*a+32/93*x^2*(1/6*d^2*x^4+2/3*c*d*x^2+c^2)*b^3*c^3)*((a*d-b*c) *c)^(1/2)*(b*x^2+a)^(1/2)-35*(a^2*d^2-16/7*a*b*c*d+48/35*b^2*c^2)*(d*x^2+c )^4*a^2*arctan(c*(b*x^2+a)^(1/2)/x/((a*d-b*c)*c)^(1/2)))/(d*x^2+c)^4/(a*d- b*c)^2/c^4
Leaf count of result is larger than twice the leaf count of optimal. 782 vs. \(2 (272) = 544\).
Time = 1.37 (sec) , antiderivative size = 1604, normalized size of antiderivative = 5.35 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^5} \, dx=\text {Too large to display} \]
[1/1536*(3*(48*a^2*b^2*c^6 - 80*a^3*b*c^5*d + 35*a^4*c^4*d^2 + (48*a^2*b^2 *c^2*d^4 - 80*a^3*b*c*d^5 + 35*a^4*d^6)*x^8 + 4*(48*a^2*b^2*c^3*d^3 - 80*a ^3*b*c^2*d^4 + 35*a^4*c*d^5)*x^6 + 6*(48*a^2*b^2*c^4*d^2 - 80*a^3*b*c^3*d^ 3 + 35*a^4*c^2*d^4)*x^4 + 4*(48*a^2*b^2*c^5*d - 80*a^3*b*c^4*d^2 + 35*a^4* c^3*d^3)*x^2)*sqrt(b*c^2 - a*c*d)*log(((8*b^2*c^2 - 8*a*b*c*d + a^2*d^2)*x ^4 + a^2*c^2 + 2*(4*a*b*c^2 - 3*a^2*c*d)*x^2 + 4*((2*b*c - a*d)*x^3 + a*c* x)*sqrt(b*c^2 - a*c*d)*sqrt(b*x^2 + a))/(d^2*x^4 + 2*c*d*x^2 + c^2)) + 4*( (16*b^4*c^5*d^2 + 24*a*b^3*c^4*d^3 - 210*a^2*b^2*c^3*d^4 + 275*a^3*b*c^2*d ^5 - 105*a^4*c*d^6)*x^7 + (64*b^4*c^6*d + 88*a*b^3*c^5*d^2 - 780*a^2*b^2*c ^4*d^3 + 1013*a^3*b*c^3*d^4 - 385*a^4*c^2*d^5)*x^5 + (96*b^4*c^7 + 112*a*b ^3*c^6*d - 1050*a^2*b^2*c^5*d^2 + 1353*a^3*b*c^4*d^3 - 511*a^4*c^3*d^4)*x^ 3 + 3*(80*a*b^3*c^7 - 256*a^2*b^2*c^6*d + 269*a^3*b*c^5*d^2 - 93*a^4*c^4*d ^3)*x)*sqrt(b*x^2 + a))/(b^3*c^12 - 3*a*b^2*c^11*d + 3*a^2*b*c^10*d^2 - a^ 3*c^9*d^3 + (b^3*c^8*d^4 - 3*a*b^2*c^7*d^5 + 3*a^2*b*c^6*d^6 - a^3*c^5*d^7 )*x^8 + 4*(b^3*c^9*d^3 - 3*a*b^2*c^8*d^4 + 3*a^2*b*c^7*d^5 - a^3*c^6*d^6)* x^6 + 6*(b^3*c^10*d^2 - 3*a*b^2*c^9*d^3 + 3*a^2*b*c^8*d^4 - a^3*c^7*d^5)*x ^4 + 4*(b^3*c^11*d - 3*a*b^2*c^10*d^2 + 3*a^2*b*c^9*d^3 - a^3*c^8*d^4)*x^2 ), -1/768*(3*(48*a^2*b^2*c^6 - 80*a^3*b*c^5*d + 35*a^4*c^4*d^2 + (48*a^2*b ^2*c^2*d^4 - 80*a^3*b*c*d^5 + 35*a^4*d^6)*x^8 + 4*(48*a^2*b^2*c^3*d^3 - 80 *a^3*b*c^2*d^4 + 35*a^4*c*d^5)*x^6 + 6*(48*a^2*b^2*c^4*d^2 - 80*a^3*b*c...
Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^5} \, dx=\text {Timed out} \]
\[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^5} \, dx=\int { \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{{\left (d x^{2} + c\right )}^{5}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 1557 vs. \(2 (272) = 544\).
Time = 4.02 (sec) , antiderivative size = 1557, normalized size of antiderivative = 5.19 \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^5} \, dx=\text {Too large to display} \]
-1/128*(48*a^2*b^(5/2)*c^2 - 80*a^3*b^(3/2)*c*d + 35*a^4*sqrt(b)*d^2)*arct an(1/2*((sqrt(b)*x - sqrt(b*x^2 + a))^2*d + 2*b*c - a*d)/sqrt(-b^2*c^2 + a *b*c*d))/((b^2*c^6 - 2*a*b*c^5*d + a^2*c^4*d^2)*sqrt(-b^2*c^2 + a*b*c*d)) - 1/192*(144*(sqrt(b)*x - sqrt(b*x^2 + a))^14*a^2*b^(5/2)*c^2*d^5 - 240*(s qrt(b)*x - sqrt(b*x^2 + a))^14*a^3*b^(3/2)*c*d^6 + 105*(sqrt(b)*x - sqrt(b *x^2 + a))^14*a^4*sqrt(b)*d^7 + 2016*(sqrt(b)*x - sqrt(b*x^2 + a))^12*a^2* b^(7/2)*c^3*d^4 - 4368*(sqrt(b)*x - sqrt(b*x^2 + a))^12*a^3*b^(5/2)*c^2*d^ 5 + 3150*(sqrt(b)*x - sqrt(b*x^2 + a))^12*a^4*b^(3/2)*c*d^6 - 735*(sqrt(b) *x - sqrt(b*x^2 + a))^12*a^5*sqrt(b)*d^7 - 2048*(sqrt(b)*x - sqrt(b*x^2 + a))^10*b^(13/2)*c^6*d + 4096*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a*b^(11/2)*c ^5*d^2 + 7936*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^2*b^(9/2)*c^4*d^3 - 26624 *(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^3*b^(7/2)*c^3*d^4 + 26944*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^4*b^(5/2)*c^2*d^5 - 12320*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^5*b^(3/2)*c*d^6 + 2205*(sqrt(b)*x - sqrt(b*x^2 + a))^10*a^6*sqrt (b)*d^7 - 2048*(sqrt(b)*x - sqrt(b*x^2 + a))^8*b^(15/2)*c^7 - 1024*(sqrt(b )*x - sqrt(b*x^2 + a))^8*a*b^(13/2)*c^6*d + 27392*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^2*b^(11/2)*c^5*d^2 - 65920*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^3*b ^(9/2)*c^4*d^3 + 81680*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^4*b^(7/2)*c^3*d^4 - 58840*(sqrt(b)*x - sqrt(b*x^2 + a))^8*a^5*b^(5/2)*c^2*d^5 + 22750*(sqrt (b)*x - sqrt(b*x^2 + a))^8*a^6*b^(3/2)*c*d^6 - 3675*(sqrt(b)*x - sqrt(b...
Timed out. \[ \int \frac {\left (a+b x^2\right )^{3/2}}{\left (c+d x^2\right )^5} \, dx=\int \frac {{\left (b\,x^2+a\right )}^{3/2}}{{\left (d\,x^2+c\right )}^5} \,d x \]