Integrand size = 22, antiderivative size = 182 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=-\frac {(b c-a d) x}{10 b^2 \left (a+b x^4\right )^{5/2}}+\frac {(b c-11 a d) x}{60 a b^2 \left (a+b x^4\right )^{3/2}}+\frac {(b c+a d) x}{24 a^2 b^2 \sqrt {a+b x^4}}+\frac {(b c+a d) \left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{48 a^{9/4} b^{9/4} \sqrt {a+b x^4}} \] Output:
-1/10*(-a*d+b*c)*x/b^2/(b*x^4+a)^(5/2)+1/60*(-11*a*d+b*c)*x/a/b^2/(b*x^4+a )^(3/2)+1/24*(a*d+b*c)*x/a^2/b^2/(b*x^4+a)^(1/2)+1/48*(a*d+b*c)*(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*ar ctan(b^(1/4)*x/a^(1/4)),1/2*2^(1/2))/a^(9/4)/b^(9/4)/(b*x^4+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.10 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.69 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\frac {-5 a^3 d x+5 b^3 c x^9+a b^2 x^5 \left (12 c+5 d x^4\right )-a^2 b x \left (5 c+12 d x^4\right )+5 (b c+a d) x \left (a+b x^4\right )^2 \sqrt {1+\frac {b x^4}{a}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {b x^4}{a}\right )}{120 a^2 b^2 \left (a+b x^4\right )^{5/2}} \] Input:
Integrate[(x^4*(c + d*x^4))/(a + b*x^4)^(7/2),x]
Output:
(-5*a^3*d*x + 5*b^3*c*x^9 + a*b^2*x^5*(12*c + 5*d*x^4) - a^2*b*x*(5*c + 12 *d*x^4) + 5*(b*c + a*d)*x*(a + b*x^4)^2*Sqrt[1 + (b*x^4)/a]*Hypergeometric 2F1[1/4, 1/2, 5/4, -((b*x^4)/a)])/(120*a^2*b^2*(a + b*x^4)^(5/2))
Time = 0.45 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {957, 817, 749, 761}
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.
| \(\displaystyle \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx\) |
| \(\Big \downarrow \) 957 |
| \(\displaystyle \frac {(a d+b c) \int \frac {x^4}{\left (b x^4+a\right )^{5/2}}dx}{2 a b}+\frac {x^5 (b c-a d)}{10 a b \left (a+b x^4\right )^{5/2}}\) |
| \(\Big \downarrow \) 817 |
| \(\displaystyle \frac {(a d+b c) \left (\frac {\int \frac {1}{\left (b x^4+a\right )^{3/2}}dx}{6 b}-\frac {x}{6 b \left (a+b x^4\right )^{3/2}}\right )}{2 a b}+\frac {x^5 (b c-a d)}{10 a b \left (a+b x^4\right )^{5/2}}\) |
| \(\Big \downarrow \) 749 |
| \(\displaystyle \frac {(a d+b c) \left (\frac {\frac {\int \frac {1}{\sqrt {b x^4+a}}dx}{2 a}+\frac {x}{2 a \sqrt {a+b x^4}}}{6 b}-\frac {x}{6 b \left (a+b x^4\right )^{3/2}}\right )}{2 a b}+\frac {x^5 (b c-a d)}{10 a b \left (a+b x^4\right )^{5/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {(a d+b c) \left (\frac {\frac {\left (\sqrt {a}+\sqrt {b} x^2\right ) \sqrt {\frac {a+b x^4}{\left (\sqrt {a}+\sqrt {b} x^2\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} x}{\sqrt [4]{a}}\right ),\frac {1}{2}\right )}{4 a^{5/4} \sqrt [4]{b} \sqrt {a+b x^4}}+\frac {x}{2 a \sqrt {a+b x^4}}}{6 b}-\frac {x}{6 b \left (a+b x^4\right )^{3/2}}\right )}{2 a b}+\frac {x^5 (b c-a d)}{10 a b \left (a+b x^4\right )^{5/2}}\) |
Input:
Int[(x^4*(c + d*x^4))/(a + b*x^4)^(7/2),x]
Output:
((b*c - a*d)*x^5)/(10*a*b*(a + b*x^4)^(5/2)) + ((b*c + a*d)*(-1/6*x/(b*(a + b*x^4)^(3/2)) + (x/(2*a*Sqrt[a + b*x^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])/(4*a^(5/4)*b^(1/4)*Sqrt[a + b*x^4]))/(6*b)))/(2*a*b)
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Simp[(n*(p + 1) + 1)/(a*n*(p + 1)) Int[(a + b*x^ n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (Inte gerQ[2*p] || Denominator[p + 1/n] < Denominator[p])
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[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n _)), x_Symbol] :> Simp[(-(b*c - a*d))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a *b*e*n*(p + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*b*n* (p + 1)) Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && (( !IntegerQ[p + 1/2] && N eQ[p, -5/4]) || !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0] && LeQ[-1 , m, (-n)*(p + 1)]))
Result contains complex when optimal does not.
Time = 1.52 (sec) , antiderivative size = 186, normalized size of antiderivative = 1.02
| method | result | size |
| elliptic | \(\frac {x \left (a d -c b \right ) \sqrt {b \,x^{4}+a}}{10 b^{5} \left (x^{4}+\frac {a}{b}\right )^{3}}-\frac {x \left (11 a d -c b \right ) \sqrt {b \,x^{4}+a}}{60 a \,b^{4} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x \left (a d +c b \right )}{24 b^{2} a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\left (a d +c b \right ) \sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{24 b^{2} a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\) | \(186\) |
| default | \(c \left (-\frac {x \sqrt {b \,x^{4}+a}}{10 b^{4} \left (x^{4}+\frac {a}{b}\right )^{3}}+\frac {x \sqrt {b \,x^{4}+a}}{60 a \,b^{3} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x}{24 b \,a^{2} \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{24 b \,a^{2} \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )+d \left (\frac {a x \sqrt {b \,x^{4}+a}}{10 b^{5} \left (x^{4}+\frac {a}{b}\right )^{3}}-\frac {11 x \sqrt {b \,x^{4}+a}}{60 b^{4} \left (x^{4}+\frac {a}{b}\right )^{2}}+\frac {x}{24 b^{2} a \sqrt {\left (x^{4}+\frac {a}{b}\right ) b}}+\frac {\sqrt {1-\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \sqrt {1+\frac {i \sqrt {b}\, x^{2}}{\sqrt {a}}}\, \operatorname {EllipticF}\left (x \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}, i\right )}{24 b^{2} a \sqrt {\frac {i \sqrt {b}}{\sqrt {a}}}\, \sqrt {b \,x^{4}+a}}\right )\) | \(312\) |
Input:
int(x^4*(d*x^4+c)/(b*x^4+a)^(7/2),x,method=_RETURNVERBOSE)
Output:
1/10*x/b^5*(a*d-b*c)*(b*x^4+a)^(1/2)/(x^4+a/b)^3-1/60/a*x*(11*a*d-b*c)/b^4 *(b*x^4+a)^(1/2)/(x^4+a/b)^2+1/24/b^2/a^2*x*(a*d+b*c)/((x^4+a/b)*b)^(1/2)+ 1/24/b^2/a^2*(a*d+b*c)/(I/a^(1/2)*b^(1/2))^(1/2)*(1-I/a^(1/2)*b^(1/2)*x^2) ^(1/2)*(1+I/a^(1/2)*b^(1/2)*x^2)^(1/2)/(b*x^4+a)^(1/2)*EllipticF(x*(I/a^(1 /2)*b^(1/2))^(1/2),I)
Time = 0.11 (sec) , antiderivative size = 207, normalized size of antiderivative = 1.14 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=-\frac {5 \, {\left ({\left (b^{4} c + a b^{3} d\right )} x^{12} + 3 \, {\left (a b^{3} c + a^{2} b^{2} d\right )} x^{8} + a^{3} b c + a^{4} d + 3 \, {\left (a^{2} b^{2} c + a^{3} b d\right )} x^{4}\right )} \sqrt {a} \left (-\frac {b}{a}\right )^{\frac {3}{4}} F(\arcsin \left (x \left (-\frac {b}{a}\right )^{\frac {1}{4}}\right )\,|\,-1) - {\left (5 \, {\left (b^{4} c + a b^{3} d\right )} x^{9} + 12 \, {\left (a b^{3} c - a^{2} b^{2} d\right )} x^{5} - 5 \, {\left (a^{2} b^{2} c + a^{3} b d\right )} x\right )} \sqrt {b x^{4} + a}}{120 \, {\left (a^{2} b^{6} x^{12} + 3 \, a^{3} b^{5} x^{8} + 3 \, a^{4} b^{4} x^{4} + a^{5} b^{3}\right )}} \] Input:
integrate(x^4*(d*x^4+c)/(b*x^4+a)^(7/2),x, algorithm="fricas")
Output:
-1/120*(5*((b^4*c + a*b^3*d)*x^12 + 3*(a*b^3*c + a^2*b^2*d)*x^8 + a^3*b*c + a^4*d + 3*(a^2*b^2*c + a^3*b*d)*x^4)*sqrt(a)*(-b/a)^(3/4)*elliptic_f(arc sin(x*(-b/a)^(1/4)), -1) - (5*(b^4*c + a*b^3*d)*x^9 + 12*(a*b^3*c - a^2*b^ 2*d)*x^5 - 5*(a^2*b^2*c + a^3*b*d)*x)*sqrt(b*x^4 + a))/(a^2*b^6*x^12 + 3*a ^3*b^5*x^8 + 3*a^4*b^4*x^4 + a^5*b^3)
Result contains complex when optimal does not.
Time = 141.58 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.44 \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\frac {c x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {7}{2} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {7}{2}} \Gamma \left (\frac {9}{4}\right )} + \frac {d x^{9} \Gamma \left (\frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{4}, \frac {7}{2} \\ \frac {13}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {7}{2}} \Gamma \left (\frac {13}{4}\right )} \] Input:
integrate(x**4*(d*x**4+c)/(b*x**4+a)**(7/2),x)
Output:
c*x**5*gamma(5/4)*hyper((5/4, 7/2), (9/4,), b*x**4*exp_polar(I*pi)/a)/(4*a **(7/2)*gamma(9/4)) + d*x**9*gamma(9/4)*hyper((9/4, 7/2), (13/4,), b*x**4* exp_polar(I*pi)/a)/(4*a**(7/2)*gamma(13/4))
\[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{4}}{{\left (b x^{4} + a\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(x^4*(d*x^4+c)/(b*x^4+a)^(7/2),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)*x^4/(b*x^4 + a)^(7/2), x)
\[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\int { \frac {{\left (d x^{4} + c\right )} x^{4}}{{\left (b x^{4} + a\right )}^{\frac {7}{2}}} \,d x } \] Input:
integrate(x^4*(d*x^4+c)/(b*x^4+a)^(7/2),x, algorithm="giac")
Output:
integrate((d*x^4 + c)*x^4/(b*x^4 + a)^(7/2), x)
Timed out. \[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\int \frac {x^4\,\left (d\,x^4+c\right )}{{\left (b\,x^4+a\right )}^{7/2}} \,d x \] Input:
int((x^4*(c + d*x^4))/(a + b*x^4)^(7/2),x)
Output:
int((x^4*(c + d*x^4))/(a + b*x^4)^(7/2), x)
\[ \int \frac {x^4 \left (c+d x^4\right )}{\left (a+b x^4\right )^{7/2}} \, dx=\frac {-5 \sqrt {b \,x^{4}+a}\, a d x -5 \sqrt {b \,x^{4}+a}\, b c x -9 \sqrt {b \,x^{4}+a}\, b d \,x^{5}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{5} d +5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{4} b c +15 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{4} b d \,x^{4}+15 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{3} b^{2} c \,x^{4}+15 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{3} b^{2} d \,x^{8}+15 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{2} b^{3} c \,x^{8}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a^{2} b^{3} d \,x^{12}+5 \left (\int \frac {\sqrt {b \,x^{4}+a}}{b^{4} x^{16}+4 a \,b^{3} x^{12}+6 a^{2} b^{2} x^{8}+4 a^{3} b \,x^{4}+a^{4}}d x \right ) a \,b^{4} c \,x^{12}}{45 b^{2} \left (b^{3} x^{12}+3 a \,b^{2} x^{8}+3 a^{2} b \,x^{4}+a^{3}\right )} \] Input:
int(x^4*(d*x^4+c)/(b*x^4+a)^(7/2),x)
Output:
( - 5*sqrt(a + b*x**4)*a*d*x - 5*sqrt(a + b*x**4)*b*c*x - 9*sqrt(a + b*x** 4)*b*d*x**5 + 5*int(sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x **8 + 4*a*b**3*x**12 + b**4*x**16),x)*a**5*d + 5*int(sqrt(a + b*x**4)/(a** 4 + 4*a**3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)*a** 4*b*c + 15*int(sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)*a**4*b*d*x**4 + 15*int(sqrt(a + b*x**4)/( a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)* a**3*b**2*c*x**4 + 15*int(sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6*a**2* b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)*a**3*b**2*d*x**8 + 15*int(sqrt (a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**12 + b **4*x**16),x)*a**2*b**3*c*x**8 + 5*int(sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x **4 + 6*a**2*b**2*x**8 + 4*a*b**3*x**12 + b**4*x**16),x)*a**2*b**3*d*x**12 + 5*int(sqrt(a + b*x**4)/(a**4 + 4*a**3*b*x**4 + 6*a**2*b**2*x**8 + 4*a*b **3*x**12 + b**4*x**16),x)*a*b**4*c*x**12)/(45*b**2*(a**3 + 3*a**2*b*x**4 + 3*a*b**2*x**8 + b**3*x**12))