Integrand size = 21, antiderivative size = 169 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {d (2 b c-5 a d) (4 b c-3 a d) x \sqrt {a+b x^2}}{8 a b^3}-\frac {d (4 b c-5 a d) x \sqrt {a+b x^2} \left (c+d x^2\right )}{4 a b^2}+\frac {(b c-a d) x \left (c+d x^2\right )^2}{a b \sqrt {a+b x^2}}+\frac {3 d \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{8 b^{7/2}} \]
3/8*d*(5*a^2*d^2-12*a*b*c*d+8*b^2*c^2)*arctanh(x*b^(1/2)/(b*x^2+a)^(1/2))/ b^(7/2)+(-a*d+b*c)*x*(d*x^2+c)^2/a/b/(b*x^2+a)^(1/2)-1/8*d*(-5*a*d+2*b*c)* (-3*a*d+4*b*c)*x*(b*x^2+a)^(1/2)/a/b^3-1/4*d*(-5*a*d+4*b*c)*x*(d*x^2+c)*(b *x^2+a)^(1/2)/a/b^2
Time = 0.25 (sec) , antiderivative size = 139, normalized size of antiderivative = 0.82 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {x \left (8 b^3 c^3-15 a^3 d^3+a^2 b d^2 \left (36 c-5 d x^2\right )+2 a b^2 d \left (-12 c^2+6 c d x^2+d^2 x^4\right )\right )}{8 a b^3 \sqrt {a+b x^2}}-\frac {3 d \left (8 b^2 c^2-12 a b c d+5 a^2 d^2\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{8 b^{7/2}} \]
(x*(8*b^3*c^3 - 15*a^3*d^3 + a^2*b*d^2*(36*c - 5*d*x^2) + 2*a*b^2*d*(-12*c ^2 + 6*c*d*x^2 + d^2*x^4)))/(8*a*b^3*Sqrt[a + b*x^2]) - (3*d*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*Log[-(Sqrt[b]*x) + Sqrt[a + b*x^2]])/(8*b^(7/2))
Time = 0.33 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {315, 27, 403, 299, 224, 219}
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.
| \(\displaystyle \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{3/2}} \, dx\) |
| \(\Big \downarrow \) 315 |
| \(\displaystyle \frac {\int \frac {d \left (d x^2+c\right ) \left (a c-(4 b c-5 a d) x^2\right )}{\sqrt {b x^2+a}}dx}{a b}+\frac {x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt {a+b x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {d \int \frac {\left (d x^2+c\right ) \left (a c-(4 b c-5 a d) x^2\right )}{\sqrt {b x^2+a}}dx}{a b}+\frac {x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt {a+b x^2}}\) |
| \(\Big \downarrow \) 403 |
| \(\displaystyle \frac {d \left (\frac {\int \frac {a c (8 b c-5 a d)-(2 b c-5 a d) (4 b c-3 a d) x^2}{\sqrt {b x^2+a}}dx}{4 b}-\frac {x \sqrt {a+b x^2} \left (c+d x^2\right ) (4 b c-5 a d)}{4 b}\right )}{a b}+\frac {x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt {a+b x^2}}\) |
| \(\Big \downarrow \) 299 |
| \(\displaystyle \frac {d \left (\frac {\frac {3 a \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}-\frac {x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{2 b}}{4 b}-\frac {x \sqrt {a+b x^2} \left (c+d x^2\right ) (4 b c-5 a d)}{4 b}\right )}{a b}+\frac {x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt {a+b x^2}}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {d \left (\frac {\frac {3 a \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}-\frac {x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{2 b}}{4 b}-\frac {x \sqrt {a+b x^2} \left (c+d x^2\right ) (4 b c-5 a d)}{4 b}\right )}{a b}+\frac {x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt {a+b x^2}}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {d \left (\frac {\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (5 a^2 d^2-12 a b c d+8 b^2 c^2\right )}{2 b^{3/2}}-\frac {x \sqrt {a+b x^2} (2 b c-5 a d) (4 b c-3 a d)}{2 b}}{4 b}-\frac {x \sqrt {a+b x^2} \left (c+d x^2\right ) (4 b c-5 a d)}{4 b}\right )}{a b}+\frac {x \left (c+d x^2\right )^2 (b c-a d)}{a b \sqrt {a+b x^2}}\) |
((b*c - a*d)*x*(c + d*x^2)^2)/(a*b*Sqrt[a + b*x^2]) + (d*(-1/4*((4*b*c - 5 *a*d)*x*Sqrt[a + b*x^2]*(c + d*x^2))/b + (-1/2*((2*b*c - 5*a*d)*(4*b*c - 3 *a*d)*x*Sqrt[a + b*x^2])/b + (3*a*(8*b^2*c^2 - 12*a*b*c*d + 5*a^2*d^2)*Arc Tanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2)))/(4*b)))/(a*b)
3.1.82.3.1 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[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
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]
Time = 2.43 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.81
| method | result | size |
| pseudoelliptic | \(\frac {\frac {15 \sqrt {b \,x^{2}+a}\, a d \left (a^{2} d^{2}-\frac {12}{5} a b c d +\frac {8}{5} b^{2} c^{2}\right ) \operatorname {arctanh}\left (\frac {\sqrt {b \,x^{2}+a}}{x \sqrt {b}}\right )}{8}-\frac {15 x \left (\frac {8 d \left (-\frac {1}{12} d^{2} x^{4}-\frac {1}{2} c d \,x^{2}+c^{2}\right ) a \,b^{\frac {5}{2}}}{5}-\frac {12 \left (-\frac {5 d \,x^{2}}{36}+c \right ) d^{2} a^{2} b^{\frac {3}{2}}}{5}+\sqrt {b}\, a^{3} d^{3}-\frac {8 b^{\frac {7}{2}} c^{3}}{15}\right )}{8}}{a \,b^{\frac {7}{2}} \sqrt {b \,x^{2}+a}}\) | \(137\) |
| risch | \(-\frac {x \,d^{2} \left (-2 b d \,x^{2}+7 a d -12 b c \right ) \sqrt {b \,x^{2}+a}}{8 b^{3}}+\frac {\frac {7 a^{2} d^{3} x}{\sqrt {b \,x^{2}+a}}+\frac {8 b^{3} c^{3} x}{a \sqrt {b \,x^{2}+a}}-\frac {12 a b c \,d^{2} x}{\sqrt {b \,x^{2}+a}}+\left (15 a^{2} b \,d^{3}-36 a \,b^{2} c \,d^{2}+24 b^{3} c^{2} d \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{8 b^{3}}\) | \(165\) |
| default | \(\frac {c^{3} x}{a \sqrt {b \,x^{2}+a}}+d^{3} \left (\frac {x^{5}}{4 b \sqrt {b \,x^{2}+a}}-\frac {5 a \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )}{4 b}\right )+3 c \,d^{2} \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )+3 c^{2} d \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (x \sqrt {b}+\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )\) | \(215\) |
15/8/(b*x^2+a)^(1/2)*((b*x^2+a)^(1/2)*a*d*(a^2*d^2-12/5*a*b*c*d+8/5*b^2*c^ 2)*arctanh((b*x^2+a)^(1/2)/x/b^(1/2))-x*(8/5*d*(-1/12*d^2*x^4-1/2*c*d*x^2+ c^2)*a*b^(5/2)-12/5*(-5/36*d*x^2+c)*d^2*a^2*b^(3/2)+b^(1/2)*a^3*d^3-8/15*b ^(7/2)*c^3))/b^(7/2)/a
Time = 0.29 (sec) , antiderivative size = 416, normalized size of antiderivative = 2.46 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left (8 \, a^{2} b^{2} c^{2} d - 12 \, a^{3} b c d^{2} + 5 \, a^{4} d^{3} + {\left (8 \, a b^{3} c^{2} d - 12 \, a^{2} b^{2} c d^{2} + 5 \, a^{3} b d^{3}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (2 \, a b^{3} d^{3} x^{5} + {\left (12 \, a b^{3} c d^{2} - 5 \, a^{2} b^{2} d^{3}\right )} x^{3} + {\left (8 \, b^{4} c^{3} - 24 \, a b^{3} c^{2} d + 36 \, a^{2} b^{2} c d^{2} - 15 \, a^{3} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{16 \, {\left (a b^{5} x^{2} + a^{2} b^{4}\right )}}, -\frac {3 \, {\left (8 \, a^{2} b^{2} c^{2} d - 12 \, a^{3} b c d^{2} + 5 \, a^{4} d^{3} + {\left (8 \, a b^{3} c^{2} d - 12 \, a^{2} b^{2} c d^{2} + 5 \, a^{3} b d^{3}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (2 \, a b^{3} d^{3} x^{5} + {\left (12 \, a b^{3} c d^{2} - 5 \, a^{2} b^{2} d^{3}\right )} x^{3} + {\left (8 \, b^{4} c^{3} - 24 \, a b^{3} c^{2} d + 36 \, a^{2} b^{2} c d^{2} - 15 \, a^{3} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{8 \, {\left (a b^{5} x^{2} + a^{2} b^{4}\right )}}\right ] \]
[1/16*(3*(8*a^2*b^2*c^2*d - 12*a^3*b*c*d^2 + 5*a^4*d^3 + (8*a*b^3*c^2*d - 12*a^2*b^2*c*d^2 + 5*a^3*b*d^3)*x^2)*sqrt(b)*log(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(2*a*b^3*d^3*x^5 + (12*a*b^3*c*d^2 - 5*a^2*b^2*d^3) *x^3 + (8*b^4*c^3 - 24*a*b^3*c^2*d + 36*a^2*b^2*c*d^2 - 15*a^3*b*d^3)*x)*s qrt(b*x^2 + a))/(a*b^5*x^2 + a^2*b^4), -1/8*(3*(8*a^2*b^2*c^2*d - 12*a^3*b *c*d^2 + 5*a^4*d^3 + (8*a*b^3*c^2*d - 12*a^2*b^2*c*d^2 + 5*a^3*b*d^3)*x^2) *sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (2*a*b^3*d^3*x^5 + (12*a*b^ 3*c*d^2 - 5*a^2*b^2*d^3)*x^3 + (8*b^4*c^3 - 24*a*b^3*c^2*d + 36*a^2*b^2*c* d^2 - 15*a^3*b*d^3)*x)*sqrt(b*x^2 + a))/(a*b^5*x^2 + a^2*b^4)]
\[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {\left (c + d x^{2}\right )^{3}}{\left (a + b x^{2}\right )^{\frac {3}{2}}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.17 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {d^{3} x^{5}}{4 \, \sqrt {b x^{2} + a} b} + \frac {3 \, c d^{2} x^{3}}{2 \, \sqrt {b x^{2} + a} b} - \frac {5 \, a d^{3} x^{3}}{8 \, \sqrt {b x^{2} + a} b^{2}} + \frac {c^{3} x}{\sqrt {b x^{2} + a} a} - \frac {3 \, c^{2} d x}{\sqrt {b x^{2} + a} b} + \frac {9 \, a c d^{2} x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {15 \, a^{2} d^{3} x}{8 \, \sqrt {b x^{2} + a} b^{3}} + \frac {3 \, c^{2} d \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} - \frac {9 \, a c d^{2} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} + \frac {15 \, a^{2} d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{8 \, b^{\frac {7}{2}}} \]
1/4*d^3*x^5/(sqrt(b*x^2 + a)*b) + 3/2*c*d^2*x^3/(sqrt(b*x^2 + a)*b) - 5/8* a*d^3*x^3/(sqrt(b*x^2 + a)*b^2) + c^3*x/(sqrt(b*x^2 + a)*a) - 3*c^2*d*x/(s qrt(b*x^2 + a)*b) + 9/2*a*c*d^2*x/(sqrt(b*x^2 + a)*b^2) - 15/8*a^2*d^3*x/( sqrt(b*x^2 + a)*b^3) + 3*c^2*d*arcsinh(b*x/sqrt(a*b))/b^(3/2) - 9/2*a*c*d^ 2*arcsinh(b*x/sqrt(a*b))/b^(5/2) + 15/8*a^2*d^3*arcsinh(b*x/sqrt(a*b))/b^( 7/2)
Time = 0.29 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.93 \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {2 \, d^{3} x^{2}}{b} + \frac {12 \, a b^{4} c d^{2} - 5 \, a^{2} b^{3} d^{3}}{a b^{5}}\right )} x^{2} + \frac {8 \, b^{5} c^{3} - 24 \, a b^{4} c^{2} d + 36 \, a^{2} b^{3} c d^{2} - 15 \, a^{3} b^{2} d^{3}}{a b^{5}}\right )} x}{8 \, \sqrt {b x^{2} + a}} - \frac {3 \, {\left (8 \, b^{2} c^{2} d - 12 \, a b c d^{2} + 5 \, a^{2} d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{8 \, b^{\frac {7}{2}}} \]
1/8*((2*d^3*x^2/b + (12*a*b^4*c*d^2 - 5*a^2*b^3*d^3)/(a*b^5))*x^2 + (8*b^5 *c^3 - 24*a*b^4*c^2*d + 36*a^2*b^3*c*d^2 - 15*a^3*b^2*d^3)/(a*b^5))*x/sqrt (b*x^2 + a) - 3/8*(8*b^2*c^2*d - 12*a*b*c*d^2 + 5*a^2*d^3)*log(abs(-sqrt(b )*x + sqrt(b*x^2 + a)))/b^(7/2)
Timed out. \[ \int \frac {\left (c+d x^2\right )^3}{\left (a+b x^2\right )^{3/2}} \, dx=\int \frac {{\left (d\,x^2+c\right )}^3}{{\left (b\,x^2+a\right )}^{3/2}} \,d x \]