Integrand size = 24, antiderivative size = 194 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=-\frac {c \left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) x \sqrt {c+d x^2}}{128 d^4}+\frac {\left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) x^3 \sqrt {c+d x^2}}{192 d^3}-\frac {b (7 b c-16 a d) x^5 \sqrt {c+d x^2}}{48 d^2}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}+\frac {c^2 \left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{128 d^{9/2}} \]
1/128*c^2*(48*a^2*d^2+5*b*c*(-16*a*d+7*b*c))*arctanh(x*d^(1/2)/(d*x^2+c)^( 1/2))/d^(9/2)-1/128*c*(48*a^2*d^2+5*b*c*(-16*a*d+7*b*c))*x*(d*x^2+c)^(1/2) /d^4+1/192*(48*a^2*d^2+5*b*c*(-16*a*d+7*b*c))*x^3*(d*x^2+c)^(1/2)/d^3-1/48 *b*(-16*a*d+7*b*c)*x^5*(d*x^2+c)^(1/2)/d^2+1/8*b^2*x^7*(d*x^2+c)^(1/2)/d
Time = 0.58 (sec) , antiderivative size = 166, normalized size of antiderivative = 0.86 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {d} x \sqrt {c+d x^2} \left (48 a^2 d^2 \left (-3 c+2 d x^2\right )+16 a b d \left (15 c^2-10 c d x^2+8 d^2 x^4\right )+b^2 \left (-105 c^3+70 c^2 d x^2-56 c d^2 x^4+48 d^3 x^6\right )\right )+6 c^2 \left (35 b^2 c^2-80 a b c d+48 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} x}{-\sqrt {c}+\sqrt {c+d x^2}}\right )}{384 d^{9/2}} \]
(Sqrt[d]*x*Sqrt[c + d*x^2]*(48*a^2*d^2*(-3*c + 2*d*x^2) + 16*a*b*d*(15*c^2 - 10*c*d*x^2 + 8*d^2*x^4) + b^2*(-105*c^3 + 70*c^2*d*x^2 - 56*c*d^2*x^4 + 48*d^3*x^6)) + 6*c^2*(35*b^2*c^2 - 80*a*b*c*d + 48*a^2*d^2)*ArcTanh[(Sqrt [d]*x)/(-Sqrt[c] + Sqrt[c + d*x^2])])/(384*d^(9/2))
Time = 0.28 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {367, 363, 262, 262, 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.
| \(\displaystyle \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx\) |
| \(\Big \downarrow \) 367 |
| \(\displaystyle \frac {\int \frac {x^4 \left (8 a^2 d-b (7 b c-16 a d) x^2\right )}{\sqrt {d x^2+c}}dx}{8 d}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}\) |
| \(\Big \downarrow \) 363 |
| \(\displaystyle \frac {\frac {\left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) \int \frac {x^4}{\sqrt {d x^2+c}}dx}{6 d}-\frac {b x^5 \sqrt {c+d x^2} (7 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {\frac {\left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) \left (\frac {x^3 \sqrt {c+d x^2}}{4 d}-\frac {3 c \int \frac {x^2}{\sqrt {d x^2+c}}dx}{4 d}\right )}{6 d}-\frac {b x^5 \sqrt {c+d x^2} (7 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}\) |
| \(\Big \downarrow \) 262 |
| \(\displaystyle \frac {\frac {\left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) \left (\frac {x^3 \sqrt {c+d x^2}}{4 d}-\frac {3 c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \int \frac {1}{\sqrt {d x^2+c}}dx}{2 d}\right )}{4 d}\right )}{6 d}-\frac {b x^5 \sqrt {c+d x^2} (7 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}\) |
| \(\Big \downarrow \) 224 |
| \(\displaystyle \frac {\frac {\left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) \left (\frac {x^3 \sqrt {c+d x^2}}{4 d}-\frac {3 c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \int \frac {1}{1-\frac {d x^2}{d x^2+c}}d\frac {x}{\sqrt {d x^2+c}}}{2 d}\right )}{4 d}\right )}{6 d}-\frac {b x^5 \sqrt {c+d x^2} (7 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {\frac {\left (48 a^2 d^2+5 b c (7 b c-16 a d)\right ) \left (\frac {x^3 \sqrt {c+d x^2}}{4 d}-\frac {3 c \left (\frac {x \sqrt {c+d x^2}}{2 d}-\frac {c \text {arctanh}\left (\frac {\sqrt {d} x}{\sqrt {c+d x^2}}\right )}{2 d^{3/2}}\right )}{4 d}\right )}{6 d}-\frac {b x^5 \sqrt {c+d x^2} (7 b c-16 a d)}{6 d}}{8 d}+\frac {b^2 x^7 \sqrt {c+d x^2}}{8 d}\) |
(b^2*x^7*Sqrt[c + d*x^2])/(8*d) + (-1/6*(b*(7*b*c - 16*a*d)*x^5*Sqrt[c + d *x^2])/d + ((48*a^2*d^2 + 5*b*c*(7*b*c - 16*a*d))*((x^3*Sqrt[c + d*x^2])/( 4*d) - (3*c*((x*Sqrt[c + d*x^2])/(2*d) - (c*ArcTanh[(Sqrt[d]*x)/Sqrt[c + d *x^2]])/(2*d^(3/2))))/(4*d)))/(6*d))/(8*d)
3.7.36.3.1 Defintions of rubi rules used
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x _Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(b*e*(m + 2*p + 3))), x] - Simp[(a*d*(m + 1) - b*c*(m + 2*p + 3))/(b*(m + 2*p + 3)) Int[(e*x)^ m*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d , 0] && NeQ[m + 2*p + 3, 0]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^2, x_Symbol] :> Simp[d^2*(e*x)^(m + 3)*((a + b*x^2)^(p + 1)/(b*e^3*(m + 2*p + 5))), x] + Simp[1/(b*(m + 2*p + 5)) Int[(e*x)^m*(a + b*x^2)^p*Simp[b*c^2* (m + 2*p + 5) - d*(a*d*(m + 3) - 2*b*c*(m + 2*p + 5))*x^2, x], x], x] /; Fr eeQ[{a, b, c, d, e, m, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + 2*p + 5, 0]
Time = 2.99 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.74
| method | result | size |
| pseudoelliptic | \(-\frac {3 \left (\left (-a^{2} c^{2} d^{2}+\frac {5}{3} a b \,c^{3} d -\frac {35}{48} b^{2} c^{4}\right ) \operatorname {arctanh}\left (\frac {\sqrt {d \,x^{2}+c}}{x \sqrt {d}}\right )+\left (c \left (\frac {7}{18} b^{2} x^{4}+\frac {10}{9} a b \,x^{2}+a^{2}\right ) d^{\frac {5}{2}}+\frac {\left (-b^{2} x^{6}-\frac {8}{3} a b \,x^{4}-2 a^{2} x^{2}\right ) d^{\frac {7}{2}}}{3}-\frac {5 b \,c^{2} \left (\left (\frac {7 b \,x^{2}}{24}+a \right ) d^{\frac {3}{2}}-\frac {7 b \sqrt {d}\, c}{16}\right )}{3}\right ) x \sqrt {d \,x^{2}+c}\right )}{8 d^{\frac {9}{2}}}\) | \(144\) |
| risch | \(-\frac {x \left (-48 b^{2} d^{3} x^{6}-128 a b \,d^{3} x^{4}+56 b^{2} c \,d^{2} x^{4}-96 a^{2} d^{3} x^{2}+160 a b c \,d^{2} x^{2}-70 b^{2} c^{2} d \,x^{2}+144 c \,a^{2} d^{2}-240 a b \,c^{2} d +105 b^{2} c^{3}\right ) \sqrt {d \,x^{2}+c}}{384 d^{4}}+\frac {c^{2} \left (48 a^{2} d^{2}-80 a b c d +35 b^{2} c^{2}\right ) \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{128 d^{\frac {9}{2}}}\) | \(157\) |
| default | \(b^{2} \left (\frac {x^{7} \sqrt {d \,x^{2}+c}}{8 d}-\frac {7 c \left (\frac {x^{5} \sqrt {d \,x^{2}+c}}{6 d}-\frac {5 c \left (\frac {x^{3} \sqrt {d \,x^{2}+c}}{4 d}-\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )}{4 d}\right )}{6 d}\right )}{8 d}\right )+a^{2} \left (\frac {x^{3} \sqrt {d \,x^{2}+c}}{4 d}-\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )}{4 d}\right )+2 a b \left (\frac {x^{5} \sqrt {d \,x^{2}+c}}{6 d}-\frac {5 c \left (\frac {x^{3} \sqrt {d \,x^{2}+c}}{4 d}-\frac {3 c \left (\frac {x \sqrt {d \,x^{2}+c}}{2 d}-\frac {c \ln \left (x \sqrt {d}+\sqrt {d \,x^{2}+c}\right )}{2 d^{\frac {3}{2}}}\right )}{4 d}\right )}{6 d}\right )\) | \(272\) |
-3/8/d^(9/2)*((-a^2*c^2*d^2+5/3*a*b*c^3*d-35/48*b^2*c^4)*arctanh((d*x^2+c) ^(1/2)/x/d^(1/2))+(c*(7/18*b^2*x^4+10/9*a*b*x^2+a^2)*d^(5/2)+1/3*(-b^2*x^6 -8/3*a*b*x^4-2*a^2*x^2)*d^(7/2)-5/3*b*c^2*((7/24*b*x^2+a)*d^(3/2)-7/16*b*d ^(1/2)*c))*x*(d*x^2+c)^(1/2))
Time = 0.30 (sec) , antiderivative size = 344, normalized size of antiderivative = 1.77 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\left [\frac {3 \, {\left (35 \, b^{2} c^{4} - 80 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {d} \log \left (-2 \, d x^{2} - 2 \, \sqrt {d x^{2} + c} \sqrt {d} x - c\right ) + 2 \, {\left (48 \, b^{2} d^{4} x^{7} - 8 \, {\left (7 \, b^{2} c d^{3} - 16 \, a b d^{4}\right )} x^{5} + 2 \, {\left (35 \, b^{2} c^{2} d^{2} - 80 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{3} - 3 \, {\left (35 \, b^{2} c^{3} d - 80 \, a b c^{2} d^{2} + 48 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{768 \, d^{5}}, -\frac {3 \, {\left (35 \, b^{2} c^{4} - 80 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \sqrt {-d} \arctan \left (\frac {\sqrt {-d} x}{\sqrt {d x^{2} + c}}\right ) - {\left (48 \, b^{2} d^{4} x^{7} - 8 \, {\left (7 \, b^{2} c d^{3} - 16 \, a b d^{4}\right )} x^{5} + 2 \, {\left (35 \, b^{2} c^{2} d^{2} - 80 \, a b c d^{3} + 48 \, a^{2} d^{4}\right )} x^{3} - 3 \, {\left (35 \, b^{2} c^{3} d - 80 \, a b c^{2} d^{2} + 48 \, a^{2} c d^{3}\right )} x\right )} \sqrt {d x^{2} + c}}{384 \, d^{5}}\right ] \]
[1/768*(3*(35*b^2*c^4 - 80*a*b*c^3*d + 48*a^2*c^2*d^2)*sqrt(d)*log(-2*d*x^ 2 - 2*sqrt(d*x^2 + c)*sqrt(d)*x - c) + 2*(48*b^2*d^4*x^7 - 8*(7*b^2*c*d^3 - 16*a*b*d^4)*x^5 + 2*(35*b^2*c^2*d^2 - 80*a*b*c*d^3 + 48*a^2*d^4)*x^3 - 3 *(35*b^2*c^3*d - 80*a*b*c^2*d^2 + 48*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/d^5, - 1/384*(3*(35*b^2*c^4 - 80*a*b*c^3*d + 48*a^2*c^2*d^2)*sqrt(-d)*arctan(sqrt (-d)*x/sqrt(d*x^2 + c)) - (48*b^2*d^4*x^7 - 8*(7*b^2*c*d^3 - 16*a*b*d^4)*x ^5 + 2*(35*b^2*c^2*d^2 - 80*a*b*c*d^3 + 48*a^2*d^4)*x^3 - 3*(35*b^2*c^3*d - 80*a*b*c^2*d^2 + 48*a^2*c*d^3)*x)*sqrt(d*x^2 + c))/d^5]
Time = 0.38 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.13 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\begin {cases} \frac {3 c^{2} \left (a^{2} - \frac {5 c \left (2 a b - \frac {7 b^{2} c}{8 d}\right )}{6 d}\right ) \left (\begin {cases} \frac {\log {\left (2 \sqrt {d} \sqrt {c + d x^{2}} + 2 d x \right )}}{\sqrt {d}} & \text {for}\: c \neq 0 \\\frac {x \log {\left (x \right )}}{\sqrt {d x^{2}}} & \text {otherwise} \end {cases}\right )}{8 d^{2}} + \sqrt {c + d x^{2}} \left (\frac {b^{2} x^{7}}{8 d} - \frac {3 c x \left (a^{2} - \frac {5 c \left (2 a b - \frac {7 b^{2} c}{8 d}\right )}{6 d}\right )}{8 d^{2}} + \frac {x^{5} \cdot \left (2 a b - \frac {7 b^{2} c}{8 d}\right )}{6 d} + \frac {x^{3} \left (a^{2} - \frac {5 c \left (2 a b - \frac {7 b^{2} c}{8 d}\right )}{6 d}\right )}{4 d}\right ) & \text {for}\: d \neq 0 \\\frac {\frac {a^{2} x^{5}}{5} + \frac {2 a b x^{7}}{7} + \frac {b^{2} x^{9}}{9}}{\sqrt {c}} & \text {otherwise} \end {cases} \]
Piecewise((3*c**2*(a**2 - 5*c*(2*a*b - 7*b**2*c/(8*d))/(6*d))*Piecewise((l og(2*sqrt(d)*sqrt(c + d*x**2) + 2*d*x)/sqrt(d), Ne(c, 0)), (x*log(x)/sqrt( d*x**2), True))/(8*d**2) + sqrt(c + d*x**2)*(b**2*x**7/(8*d) - 3*c*x*(a**2 - 5*c*(2*a*b - 7*b**2*c/(8*d))/(6*d))/(8*d**2) + x**5*(2*a*b - 7*b**2*c/( 8*d))/(6*d) + x**3*(a**2 - 5*c*(2*a*b - 7*b**2*c/(8*d))/(6*d))/(4*d)), Ne( d, 0)), ((a**2*x**5/5 + 2*a*b*x**7/7 + b**2*x**9/9)/sqrt(c), True))
Time = 0.21 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.25 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {\sqrt {d x^{2} + c} b^{2} x^{7}}{8 \, d} - \frac {7 \, \sqrt {d x^{2} + c} b^{2} c x^{5}}{48 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a b x^{5}}{3 \, d} + \frac {35 \, \sqrt {d x^{2} + c} b^{2} c^{2} x^{3}}{192 \, d^{3}} - \frac {5 \, \sqrt {d x^{2} + c} a b c x^{3}}{12 \, d^{2}} + \frac {\sqrt {d x^{2} + c} a^{2} x^{3}}{4 \, d} - \frac {35 \, \sqrt {d x^{2} + c} b^{2} c^{3} x}{128 \, d^{4}} + \frac {5 \, \sqrt {d x^{2} + c} a b c^{2} x}{8 \, d^{3}} - \frac {3 \, \sqrt {d x^{2} + c} a^{2} c x}{8 \, d^{2}} + \frac {35 \, b^{2} c^{4} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{128 \, d^{\frac {9}{2}}} - \frac {5 \, a b c^{3} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {7}{2}}} + \frac {3 \, a^{2} c^{2} \operatorname {arsinh}\left (\frac {d x}{\sqrt {c d}}\right )}{8 \, d^{\frac {5}{2}}} \]
1/8*sqrt(d*x^2 + c)*b^2*x^7/d - 7/48*sqrt(d*x^2 + c)*b^2*c*x^5/d^2 + 1/3*s qrt(d*x^2 + c)*a*b*x^5/d + 35/192*sqrt(d*x^2 + c)*b^2*c^2*x^3/d^3 - 5/12*s qrt(d*x^2 + c)*a*b*c*x^3/d^2 + 1/4*sqrt(d*x^2 + c)*a^2*x^3/d - 35/128*sqrt (d*x^2 + c)*b^2*c^3*x/d^4 + 5/8*sqrt(d*x^2 + c)*a*b*c^2*x/d^3 - 3/8*sqrt(d *x^2 + c)*a^2*c*x/d^2 + 35/128*b^2*c^4*arcsinh(d*x/sqrt(c*d))/d^(9/2) - 5/ 8*a*b*c^3*arcsinh(d*x/sqrt(c*d))/d^(7/2) + 3/8*a^2*c^2*arcsinh(d*x/sqrt(c* d))/d^(5/2)
Time = 0.34 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.92 \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\frac {1}{384} \, {\left (2 \, {\left (4 \, {\left (\frac {6 \, b^{2} x^{2}}{d} - \frac {7 \, b^{2} c d^{5} - 16 \, a b d^{6}}{d^{7}}\right )} x^{2} + \frac {35 \, b^{2} c^{2} d^{4} - 80 \, a b c d^{5} + 48 \, a^{2} d^{6}}{d^{7}}\right )} x^{2} - \frac {3 \, {\left (35 \, b^{2} c^{3} d^{3} - 80 \, a b c^{2} d^{4} + 48 \, a^{2} c d^{5}\right )}}{d^{7}}\right )} \sqrt {d x^{2} + c} x - \frac {{\left (35 \, b^{2} c^{4} - 80 \, a b c^{3} d + 48 \, a^{2} c^{2} d^{2}\right )} \log \left ({\left | -\sqrt {d} x + \sqrt {d x^{2} + c} \right |}\right )}{128 \, d^{\frac {9}{2}}} \]
1/384*(2*(4*(6*b^2*x^2/d - (7*b^2*c*d^5 - 16*a*b*d^6)/d^7)*x^2 + (35*b^2*c ^2*d^4 - 80*a*b*c*d^5 + 48*a^2*d^6)/d^7)*x^2 - 3*(35*b^2*c^3*d^3 - 80*a*b* c^2*d^4 + 48*a^2*c*d^5)/d^7)*sqrt(d*x^2 + c)*x - 1/128*(35*b^2*c^4 - 80*a* b*c^3*d + 48*a^2*c^2*d^2)*log(abs(-sqrt(d)*x + sqrt(d*x^2 + c)))/d^(9/2)
Timed out. \[ \int \frac {x^4 \left (a+b x^2\right )^2}{\sqrt {c+d x^2}} \, dx=\int \frac {x^4\,{\left (b\,x^2+a\right )}^2}{\sqrt {d\,x^2+c}} \,d x \]