Integrand size = 21, antiderivative size = 202 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2} \, dx=\frac {b (2 b c+3 a d) x}{6 a c (b c-a d)^2 \left (a+b x^2\right )^{3/2}}+\frac {b \left (4 b^2 c^2-16 a b c d-3 a^2 d^2\right ) x}{6 a^2 c (b c-a d)^3 \sqrt {a+b x^2}}-\frac {d x}{2 c (b c-a d) \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}+\frac {d^2 (6 b c-a d) \text {arctanh}\left (\frac {\sqrt {b c-a d} x}{\sqrt {c} \sqrt {a+b x^2}}\right )}{2 c^{3/2} (b c-a d)^{7/2}} \]
1/6*b*(3*a*d+2*b*c)*x/a/c/(-a*d+b*c)^2/(b*x^2+a)^(3/2)-1/2*d*x/c/(-a*d+b*c )/(b*x^2+a)^(3/2)/(d*x^2+c)+1/2*d^2*(-a*d+6*b*c)*arctanh(x*(-a*d+b*c)^(1/2 )/c^(1/2)/(b*x^2+a)^(1/2))/c^(3/2)/(-a*d+b*c)^(7/2)+1/6*b*(-3*a^2*d^2-16*a *b*c*d+4*b^2*c^2)*x/a^2/c/(-a*d+b*c)^3/(b*x^2+a)^(1/2)
Time = 1.26 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2} \, dx=\frac {x \left (3 a^4 d^3+6 a^3 b d^3 x^2-4 b^4 c^2 x^2 \left (c+d x^2\right )+3 a^2 b^2 d \left (6 c^2+6 c d x^2+d^2 x^4\right )+2 a b^3 c \left (-3 c^2+5 c d x^2+8 d^2 x^4\right )\right )}{6 a^2 c (-b c+a d)^3 \left (a+b x^2\right )^{3/2} \left (c+d x^2\right )}+\frac {d^2 (6 b c-a d) \arctan \left (\frac {-d x \sqrt {a+b x^2}+\sqrt {b} \left (c+d x^2\right )}{\sqrt {c} \sqrt {-b c+a d}}\right )}{2 c^{3/2} (-b c+a d)^{7/2}} \]
(x*(3*a^4*d^3 + 6*a^3*b*d^3*x^2 - 4*b^4*c^2*x^2*(c + d*x^2) + 3*a^2*b^2*d* (6*c^2 + 6*c*d*x^2 + d^2*x^4) + 2*a*b^3*c*(-3*c^2 + 5*c*d*x^2 + 8*d^2*x^4) ))/(6*a^2*c*(-(b*c) + a*d)^3*(a + b*x^2)^(3/2)*(c + d*x^2)) + (d^2*(6*b*c - a*d)*ArcTan[(-(d*x*Sqrt[a + b*x^2]) + Sqrt[b]*(c + d*x^2))/(Sqrt[c]*Sqrt [-(b*c) + a*d])])/(2*c^(3/2)*(-(b*c) + a*d)^(7/2))
Time = 0.36 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.13, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {316, 402, 25, 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.
| \(\displaystyle \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2} \, dx\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\int \frac {-4 b d x^2+2 b c-a d}{\left (b x^2+a\right )^{5/2} \left (d x^2+c\right )}dx}{2 c (b c-a d)}-\frac {d x}{2 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {b x (3 a d+2 b c)}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)}-\frac {\int -\frac {4 b^2 c^2-12 a b d c+3 a^2 d^2+2 b d (2 b c+3 a d) x^2}{\left (b x^2+a\right )^{3/2} \left (d x^2+c\right )}dx}{3 a (b c-a d)}}{2 c (b c-a d)}-\frac {d x}{2 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\frac {\int \frac {4 b^2 c^2-12 a b d c+3 a^2 d^2+2 b d (2 b c+3 a d) x^2}{\left (b x^2+a\right )^{3/2} \left (d x^2+c\right )}dx}{3 a (b c-a d)}+\frac {b x (3 a d+2 b c)}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)}}{2 c (b c-a d)}-\frac {d x}{2 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {b x \left (-3 a^2 d^2-16 a b c d+4 b^2 c^2\right )}{a \sqrt {a+b x^2} (b c-a d)}-\frac {\int -\frac {3 a^2 d^2 (6 b c-a d)}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{a (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+2 b c)}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)}}{2 c (b c-a d)}-\frac {d x}{2 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\frac {\frac {3 a d^2 (6 b c-a d) \int \frac {1}{\sqrt {b x^2+a} \left (d x^2+c\right )}dx}{b c-a d}+\frac {b x \left (-3 a^2 d^2-16 a b c d+4 b^2 c^2\right )}{a \sqrt {a+b x^2} (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+2 b c)}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)}}{2 c (b c-a d)}-\frac {d x}{2 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 291 |
| \(\displaystyle \frac {\frac {\frac {3 a d^2 (6 b c-a d) \int \frac {1}{c-\frac {(b c-a d) x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{b c-a d}+\frac {b x \left (-3 a^2 d^2-16 a b c d+4 b^2 c^2\right )}{a \sqrt {a+b x^2} (b c-a d)}}{3 a (b c-a d)}+\frac {b x (3 a d+2 b c)}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)}}{2 c (b c-a d)}-\frac {d x}{2 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {\frac {b x \left (-3 a^2 d^2-16 a b c d+4 b^2 c^2\right )}{a \sqrt {a+b x^2} (b c-a d)}+\frac {3 a d^2 (6 b c-a d) \text {arctanh}\left (\frac {x \sqrt {b c-a d}}{\sqrt {c} \sqrt {a+b x^2}}\right )}{\sqrt {c} (b c-a d)^{3/2}}}{3 a (b c-a d)}+\frac {b x (3 a d+2 b c)}{3 a \left (a+b x^2\right )^{3/2} (b c-a d)}}{2 c (b c-a d)}-\frac {d x}{2 c \left (a+b x^2\right )^{3/2} \left (c+d x^2\right ) (b c-a d)}\) |
-1/2*(d*x)/(c*(b*c - a*d)*(a + b*x^2)^(3/2)*(c + d*x^2)) + ((b*(2*b*c + 3* a*d)*x)/(3*a*(b*c - a*d)*(a + b*x^2)^(3/2)) + ((b*(4*b^2*c^2 - 16*a*b*c*d - 3*a^2*d^2)*x)/(a*(b*c - a*d)*Sqrt[a + b*x^2]) + (3*a*d^2*(6*b*c - a*d)*A rcTanh[(Sqrt[b*c - a*d]*x)/(Sqrt[c]*Sqrt[a + b*x^2])])/(Sqrt[c]*(b*c - a*d )^(3/2)))/(3*a*(b*c - a*d)))/(2*c*(b*c - a*d))
3.1.95.3.1 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[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[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 = 2.63 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.03
| method | result | size |
| pseudoelliptic | \(\frac {-\left (a d -6 b c \right ) \left (d \,x^{2}+c \right ) d^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{2} \arctan \left (\frac {c \sqrt {b \,x^{2}+a}}{x \sqrt {\left (a d -b c \right ) c}}\right )+x \sqrt {\left (a d -b c \right ) c}\, \left (\left (-\frac {4}{3} b^{4} x^{2}-2 a \,b^{3}\right ) c^{3}+6 b^{2} d \left (-\frac {2}{9} b^{2} x^{4}+\frac {5}{9} a b \,x^{2}+a^{2}\right ) c^{2}+6 x^{2} b^{2} d^{2} a \left (\frac {8 b \,x^{2}}{9}+a \right ) c +a^{2} d^{3} \left (b \,x^{2}+a \right )^{2}\right )}{2 \sqrt {\left (a d -b c \right ) c}\, \left (b \,x^{2}+a \right )^{\frac {3}{2}} c \left (d \,x^{2}+c \right ) \left (a d -b c \right )^{3} a^{2}}\) | \(208\) |
| default | \(\text {Expression too large to display}\) | \(3449\) |
1/2/((a*d-b*c)*c)^(1/2)/(b*x^2+a)^(3/2)*(-(a*d-6*b*c)*(d*x^2+c)*d^2*(b*x^2 +a)^(3/2)*a^2*arctan(c*(b*x^2+a)^(1/2)/x/((a*d-b*c)*c)^(1/2))+x*((a*d-b*c) *c)^(1/2)*((-4/3*b^4*x^2-2*a*b^3)*c^3+6*b^2*d*(-2/9*b^2*x^4+5/9*a*b*x^2+a^ 2)*c^2+6*x^2*b^2*d^2*a*(8/9*b*x^2+a)*c+a^2*d^3*(b*x^2+a)^2))/c/(d*x^2+c)/( a*d-b*c)^3/a^2
Leaf count of result is larger than twice the leaf count of optimal. 700 vs. \(2 (178) = 356\).
Time = 0.95 (sec) , antiderivative size = 1440, normalized size of antiderivative = 7.13 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2} \, dx=\text {Too large to display} \]
[1/24*(3*(6*a^4*b*c^2*d^2 - a^5*c*d^3 + (6*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^ 6 + (6*a^2*b^3*c^2*d^2 + 11*a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^4 + (12*a^3*b^2 *c^2*d^2 + 4*a^4*b*c*d^3 - a^5*d^4)*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*((4*b^5*c^4*d - 20*a*b^4*c^3*d^2 + 13*a^2*b^3*c^ 2*d^3 + 3*a^3*b^2*c*d^4)*x^5 + 2*(2*b^5*c^5 - 7*a*b^4*c^4*d - 4*a^2*b^3*c^ 3*d^2 + 6*a^3*b^2*c^2*d^3 + 3*a^4*b*c*d^4)*x^3 + 3*(2*a*b^4*c^5 - 8*a^2*b^ 3*c^4*d + 6*a^3*b^2*c^3*d^2 - a^4*b*c^2*d^3 + a^5*c*d^4)*x)*sqrt(b*x^2 + a ))/(a^4*b^4*c^7 - 4*a^5*b^3*c^6*d + 6*a^6*b^2*c^5*d^2 - 4*a^7*b*c^4*d^3 + a^8*c^3*d^4 + (a^2*b^6*c^6*d - 4*a^3*b^5*c^5*d^2 + 6*a^4*b^4*c^4*d^3 - 4*a ^5*b^3*c^3*d^4 + a^6*b^2*c^2*d^5)*x^6 + (a^2*b^6*c^7 - 2*a^3*b^5*c^6*d - 2 *a^4*b^4*c^5*d^2 + 8*a^5*b^3*c^4*d^3 - 7*a^6*b^2*c^3*d^4 + 2*a^7*b*c^2*d^5 )*x^4 + (2*a^3*b^5*c^7 - 7*a^4*b^4*c^6*d + 8*a^5*b^3*c^5*d^2 - 2*a^6*b^2*c ^4*d^3 - 2*a^7*b*c^3*d^4 + a^8*c^2*d^5)*x^2), -1/12*(3*(6*a^4*b*c^2*d^2 - a^5*c*d^3 + (6*a^2*b^3*c*d^3 - a^3*b^2*d^4)*x^6 + (6*a^2*b^3*c^2*d^2 + 11* a^3*b^2*c*d^3 - 2*a^4*b*d^4)*x^4 + (12*a^3*b^2*c^2*d^2 + 4*a^4*b*c*d^3 - a ^5*d^4)*x^2)*sqrt(-b*c^2 + a*c*d)*arctan(1/2*sqrt(-b*c^2 + a*c*d)*((2*b*c - a*d)*x^2 + a*c)*sqrt(b*x^2 + a)/((b^2*c^2 - a*b*c*d)*x^3 + (a*b*c^2 - a^ 2*c*d)*x)) - 2*((4*b^5*c^4*d - 20*a*b^4*c^3*d^2 + 13*a^2*b^3*c^2*d^3 + ...
\[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2} \, dx=\int \frac {1}{\left (a + b x^{2}\right )^{\frac {5}{2}} \left (c + d x^{2}\right )^{2}}\, dx \]
\[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} {\left (d x^{2} + c\right )}^{2}} \,d x } \]
Leaf count of result is larger than twice the leaf count of optimal. 620 vs. \(2 (178) = 356\).
Time = 0.89 (sec) , antiderivative size = 620, normalized size of antiderivative = 3.07 \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2} \, dx=\frac {{\left (\frac {2 \, {\left (b^{8} c^{4} - 7 \, a b^{7} c^{3} d + 15 \, a^{2} b^{6} c^{2} d^{2} - 13 \, a^{3} b^{5} c d^{3} + 4 \, a^{4} b^{4} d^{4}\right )} x^{2}}{a^{2} b^{7} c^{6} - 6 \, a^{3} b^{6} c^{5} d + 15 \, a^{4} b^{5} c^{4} d^{2} - 20 \, a^{5} b^{4} c^{3} d^{3} + 15 \, a^{6} b^{3} c^{2} d^{4} - 6 \, a^{7} b^{2} c d^{5} + a^{8} b d^{6}} + \frac {3 \, {\left (a b^{7} c^{4} - 6 \, a^{2} b^{6} c^{3} d + 12 \, a^{3} b^{5} c^{2} d^{2} - 10 \, a^{4} b^{4} c d^{3} + 3 \, a^{5} b^{3} d^{4}\right )}}{a^{2} b^{7} c^{6} - 6 \, a^{3} b^{6} c^{5} d + 15 \, a^{4} b^{5} c^{4} d^{2} - 20 \, a^{5} b^{4} c^{3} d^{3} + 15 \, a^{6} b^{3} c^{2} d^{4} - 6 \, a^{7} b^{2} c d^{5} + a^{8} b d^{6}}\right )} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} + \frac {{\left (6 \, b^{\frac {3}{2}} c d^{2} - a \sqrt {b} d^{3}\right )} \arctan \left (-\frac {{\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} d + 2 \, b c - a d}{2 \, \sqrt {-b^{2} c^{2} + a b c d}}\right )}{2 \, {\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} \sqrt {-b^{2} c^{2} + a b c d}} - \frac {2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b^{\frac {3}{2}} c d^{2} - {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a \sqrt {b} d^{3} + a^{2} \sqrt {b} d^{3}}{{\left (b^{3} c^{4} - 3 \, a b^{2} c^{3} d + 3 \, a^{2} b c^{2} d^{2} - a^{3} c d^{3}\right )} {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} d + 4 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} b c - 2 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} a d + a^{2} d\right )}} \]
1/3*(2*(b^8*c^4 - 7*a*b^7*c^3*d + 15*a^2*b^6*c^2*d^2 - 13*a^3*b^5*c*d^3 + 4*a^4*b^4*d^4)*x^2/(a^2*b^7*c^6 - 6*a^3*b^6*c^5*d + 15*a^4*b^5*c^4*d^2 - 2 0*a^5*b^4*c^3*d^3 + 15*a^6*b^3*c^2*d^4 - 6*a^7*b^2*c*d^5 + a^8*b*d^6) + 3* (a*b^7*c^4 - 6*a^2*b^6*c^3*d + 12*a^3*b^5*c^2*d^2 - 10*a^4*b^4*c*d^3 + 3*a ^5*b^3*d^4)/(a^2*b^7*c^6 - 6*a^3*b^6*c^5*d + 15*a^4*b^5*c^4*d^2 - 20*a^5*b ^4*c^3*d^3 + 15*a^6*b^3*c^2*d^4 - 6*a^7*b^2*c*d^5 + a^8*b*d^6))*x/(b*x^2 + a)^(3/2) + 1/2*(6*b^(3/2)*c*d^2 - a*sqrt(b)*d^3)*arctan(-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^3*c^4 - 3*a*b^2*c^3*d + 3*a^2*b*c^2*d^2 - a^3*c*d^3)*sqrt(-b^2*c^2 + a*b*c*d)) - (2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*b^(3/2)*c*d^2 - (sqrt(b)*x - sqrt(b*x^ 2 + a))^2*a*sqrt(b)*d^3 + a^2*sqrt(b)*d^3)/((b^3*c^4 - 3*a*b^2*c^3*d + 3*a ^2*b*c^2*d^2 - a^3*c*d^3)*((sqrt(b)*x - sqrt(b*x^2 + a))^4*d + 4*(sqrt(b)* x - sqrt(b*x^2 + a))^2*b*c - 2*(sqrt(b)*x - sqrt(b*x^2 + a))^2*a*d + a^2*d ))
Timed out. \[ \int \frac {1}{\left (a+b x^2\right )^{5/2} \left (c+d x^2\right )^2} \, dx=\int \frac {1}{{\left (b\,x^2+a\right )}^{5/2}\,{\left (d\,x^2+c\right )}^2} \,d x \]