Integrand size = 15, antiderivative size = 282 \[ \int \frac {x^{14}}{\left (a+b x^4\right )^{3/2}} \, dx=-\frac {x^{11}}{2 b \sqrt {a+b x^4}}-\frac {77 a x^3 \sqrt {a+b x^4}}{90 b^3}+\frac {11 x^7 \sqrt {a+b x^4}}{18 b^2}+\frac {77 a^2 x \sqrt {a+b x^4}}{30 b^{7/2} \left (\sqrt {a}+\sqrt {b} x^2\right )}-\frac {77 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{30 b^{15/4} \sqrt {a+b x^4}}+\frac {77 a^{9/4} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{60 b^{15/4} \sqrt {a+b x^4}} \] Output:
-1/2*x^11/b/(b*x^4+a)^(1/2)-77/90*a*x^3*(b*x^4+a)^(1/2)/b^3+11/18*x^7*(b*x ^4+a)^(1/2)/b^2+77/30*a^2*x*(b*x^4+a)^(1/2)/b^(7/2)/(a^(1/2)+b^(1/2)*x^2)- 77/30*a^(9/4)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2)*x^2)^2)^(1 /2)*EllipticE(sin(2*arctan(b^(1/4)*x/a^(1/4))),1/2*2^(1/2))/b^(15/4)/(b*x^ 4+a)^(1/2)+77/60*a^(9/4)*(a^(1/2)+b^(1/2)*x^2)*((b*x^4+a)/(a^(1/2)+b^(1/2) *x^2)^2)^(1/2)*InverseJacobiAM(2*arctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/b^ (15/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.04 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.28 \[ \int \frac {x^{14}}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x^3 \left (77 a^2-11 a b x^4+5 b^2 x^8-77 a^2 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {3}{4},\frac {3}{2},\frac {7}{4},-\frac {b x^4}{a}\right )\right )}{45 b^3 \sqrt {a+b x^4}} \] Input:
Integrate[x^14/(a + b*x^4)^(3/2),x]
Output:
(x^3*(77*a^2 - 11*a*b*x^4 + 5*b^2*x^8 - 77*a^2*Sqrt[1 + (b*x^4)/a]*Hyperge ometric2F1[3/4, 3/2, 7/4, -((b*x^4)/a)]))/(45*b^3*Sqrt[a + b*x^4])
Time = 0.61 (sec) , antiderivative size = 302, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.467, Rules used = {817, 843, 843, 834, 27, 761, 1510}
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.
\(\displaystyle \int \frac {x^{14}}{\left (a+b x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 817 |
\(\displaystyle \frac {11 \int \frac {x^{10}}{\sqrt {b x^4+a}}dx}{2 b}-\frac {x^{11}}{2 b \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {11 \left (\frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \int \frac {x^6}{\sqrt {b x^4+a}}dx}{9 b}\right )}{2 b}-\frac {x^{11}}{2 b \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {11 \left (\frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \int \frac {x^2}{\sqrt {b x^4+a}}dx}{5 b}\right )}{9 b}\right )}{2 b}-\frac {x^{11}}{2 b \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 834 |
\(\displaystyle \frac {11 \left (\frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\sqrt {a} \int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {a} \sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\right )}{2 b}-\frac {x^{11}}{2 b \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {11 \left (\frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt {a} \int \frac {1}{\sqrt {b x^4+a}}dx}{\sqrt {b}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\right )}{2 b}-\frac {x^{11}}{2 b \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {11 \left (\frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\int \frac {\sqrt {a}-\sqrt {b} x^2}{\sqrt {b x^4+a}}dx}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\right )}{2 b}-\frac {x^{11}}{2 b \sqrt {a+b x^4}}\) |
\(\Big \downarrow \) 1510 |
\(\displaystyle \frac {11 \left (\frac {x^7 \sqrt {a+b x^4}}{9 b}-\frac {7 a \left (\frac {x^3 \sqrt {a+b x^4}}{5 b}-\frac {3 a \left (\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{2 b^{3/4} \sqrt {a+b x^4}}-\frac {\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{\sqrt [4]{b} \sqrt {a+b x^4}}-\frac {x \sqrt {a+b x^4}}{\sqrt {a}+\sqrt {b} x^2}}{\sqrt {b}}\right )}{5 b}\right )}{9 b}\right )}{2 b}-\frac {x^{11}}{2 b \sqrt {a+b x^4}}\) |
Input:
Int[x^14/(a + b*x^4)^(3/2),x]
Output:
-1/2*x^11/(b*Sqrt[a + b*x^4]) + (11*((x^7*Sqrt[a + b*x^4])/(9*b) - (7*a*(( x^3*Sqrt[a + b*x^4])/(5*b) - (3*a*(-((-((x*Sqrt[a + b*x^4])/(Sqrt[a] + Sqr t[b]*x^2)) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticE[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2])/(b^(1/4)*Sq rt[a + b*x^4]))/Sqrt[b]) + (a^(1/4)*(Sqrt[a] + Sqrt[b]*x^2)*Sqrt[(a + b*x^ 4)/(Sqrt[a] + Sqrt[b]*x^2)^2]*EllipticF[2*ArcTan[(b^(1/4)*x)/a^(1/4)], 1/2 ])/(2*b^(3/4)*Sqrt[a + b*x^4])))/(5*b)))/(9*b)))/(2*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^( n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*n*(p + 1))), x] - Simp[c^n *((m - n + 1)/(b*n*(p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x], x ] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && ! ILtQ[(m + n*(p + 1) + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + c*x^4]/(a*(1 + q^2*x^2))), x] + Simp[d* (1 + q^2*x^2)*(Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]/(q*Sqrt[a + c*x^4]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Result contains complex when optimal does not.
Time = 2.57 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.56
method | result | size |
default | \(-\frac {x^{3} a^{2}}{2 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{7} \sqrt {b \,x^{4}+a}}{9 b^{2}}-\frac {16 a \,x^{3} \sqrt {b \,x^{4}+a}}{45 b^{3}}+\frac {77 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{30 b^{\frac {7}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(157\) |
elliptic | \(-\frac {x^{3} a^{2}}{2 b^{3} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {x^{7} \sqrt {b \,x^{4}+a}}{9 b^{2}}-\frac {16 a \,x^{3} \sqrt {b \,x^{4}+a}}{45 b^{3}}+\frac {77 i a^{\frac {5}{2}} \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{30 b^{\frac {7}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(157\) |
risch | \(-\frac {x^{3} \left (-5 b \,x^{4}+16 a \right ) \sqrt {b \,x^{4}+a}}{45 b^{3}}+\frac {a^{2} \left (16 a \left (\frac {x^{3}}{2 a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}-\frac {i \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 \sqrt {a}\, \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}\, \sqrt {b}}\right )+31 b \left (-\frac {x^{3}}{2 b \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {3 i \sqrt {a}\, \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \left (\operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )-\operatorname {EllipticE}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )\right )}{2 b^{\frac {3}{2}} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\right )}{15 b^{3}}\) | \(280\) |
Input:
int(x^14/(b*x^4+a)^(3/2),x,method=_RETURNVERBOSE)
Output:
-1/2/b^3*x^3*a^2/((x^4+a/b)*b)^(1/2)+1/9*x^7*(b*x^4+a)^(1/2)/b^2-16/45*a*x ^3*(b*x^4+a)^(1/2)/b^3+77/30*I*a^(5/2)/b^(7/2)/(I/a^(1/2)*b^(1/2))^(1/2)*( 1-I*b^(1/2)*x^2/a^(1/2))^(1/2)*(1+I*b^(1/2)*x^2/a^(1/2))^(1/2)/(b*x^4+a)^( 1/2)*(EllipticF(x*(I/a^(1/2)*b^(1/2))^(1/2),I)-EllipticE(x*(I/a^(1/2)*b^(1 /2))^(1/2),I))
Time = 0.08 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.51 \[ \int \frac {x^{14}}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {231 \, {\left (a^{2} b x^{5} + a^{3} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} E(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) - 231 \, {\left (a^{2} b x^{5} + a^{3} x\right )} \sqrt {b} \left (-\frac {a}{b}\right )^{\frac {3}{4}} F(\arcsin \left (\frac {\left (-\frac {a}{b}\right )^{\frac {1}{4}}}{x}\right )\,|\,-1) + {\left (10 \, b^{3} x^{12} - 22 \, a b^{2} x^{8} + 154 \, a^{2} b x^{4} + 231 \, a^{3}\right )} \sqrt {b x^{4} + a}}{90 \, {\left (b^{5} x^{5} + a b^{4} x\right )}} \] Input:
integrate(x^14/(b*x^4+a)^(3/2),x, algorithm="fricas")
Output:
1/90*(231*(a^2*b*x^5 + a^3*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_e(arcsin((-a/b )^(1/4)/x), -1) - 231*(a^2*b*x^5 + a^3*x)*sqrt(b)*(-a/b)^(3/4)*elliptic_f( arcsin((-a/b)^(1/4)/x), -1) + (10*b^3*x^12 - 22*a*b^2*x^8 + 154*a^2*b*x^4 + 231*a^3)*sqrt(b*x^4 + a))/(b^5*x^5 + a*b^4*x)
Result contains complex when optimal does not.
Time = 0.80 (sec) , antiderivative size = 37, normalized size of antiderivative = 0.13 \[ \int \frac {x^{14}}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {x^{15} \Gamma \left (\frac {15}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {15}{4} \\ \frac {19}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {3}{2}} \Gamma \left (\frac {19}{4}\right )} \] Input:
integrate(x**14/(b*x**4+a)**(3/2),x)
Output:
x**15*gamma(15/4)*hyper((3/2, 15/4), (19/4,), b*x**4*exp_polar(I*pi)/a)/(4 *a**(3/2)*gamma(19/4))
\[ \int \frac {x^{14}}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{14}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^14/(b*x^4+a)^(3/2),x, algorithm="maxima")
Output:
integrate(x^14/(b*x^4 + a)^(3/2), x)
\[ \int \frac {x^{14}}{\left (a+b x^4\right )^{3/2}} \, dx=\int { \frac {x^{14}}{{\left (b x^{4} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate(x^14/(b*x^4+a)^(3/2),x, algorithm="giac")
Output:
integrate(x^14/(b*x^4 + a)^(3/2), x)
Timed out. \[ \int \frac {x^{14}}{\left (a+b x^4\right )^{3/2}} \, dx=\int \frac {x^{14}}{{\left (b\,x^4+a\right )}^{3/2}} \,d x \] Input:
int(x^14/(a + b*x^4)^(3/2),x)
Output:
int(x^14/(a + b*x^4)^(3/2), x)
\[ \int \frac {x^{14}}{\left (a+b x^4\right )^{3/2}} \, dx=\frac {77 \sqrt {b \,x^{4}+a}\, a^{2} x^{3}-11 \sqrt {b \,x^{4}+a}\, a b \,x^{7}+5 \sqrt {b \,x^{4}+a}\, b^{2} x^{11}-231 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{4}-231 \left (\int \frac {\sqrt {b \,x^{4}+a}\, x^{2}}{b^{2} x^{8}+2 a b \,x^{4}+a^{2}}d x \right ) a^{3} b \,x^{4}}{45 b^{3} \left (b \,x^{4}+a \right )} \] Input:
int(x^14/(b*x^4+a)^(3/2),x)
Output:
(77*sqrt(a + b*x**4)*a**2*x**3 - 11*sqrt(a + b*x**4)*a*b*x**7 + 5*sqrt(a + b*x**4)*b**2*x**11 - 231*int((sqrt(a + b*x**4)*x**2)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**4 - 231*int((sqrt(a + b*x**4)*x**2)/(a**2 + 2*a*b*x**4 + b**2*x**8),x)*a**3*b*x**4)/(45*b**3*(a + b*x**4))