Integrand size = 19, antiderivative size = 85 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/3}} \, dx=-\frac {3 d x}{5 b \left (a+b x^4\right )^{2/3}}+\frac {(5 b c+3 a d) x \left (1+\frac {b x^4}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{5 a b \left (a+b x^4\right )^{2/3}} \] Output:
-3/5*d*x/b/(b*x^4+a)^(2/3)+1/5*(3*a*d+5*b*c)*x*(1+b*x^4/a)^(2/3)*hypergeom ([1/4, 5/3],[5/4],-b*x^4/a)/a/b/(b*x^4+a)^(2/3)
Time = 10.05 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.84 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/3}} \, dx=\frac {-3 a d x+(5 b c+3 a d) x \left (1+\frac {b x^4}{a}\right )^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {5}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{5 a b \left (a+b x^4\right )^{2/3}} \] Input:
Integrate[(c + d*x^4)/(a + b*x^4)^(5/3),x]
Output:
(-3*a*d*x + (5*b*c + 3*a*d)*x*(1 + (b*x^4)/a)^(2/3)*Hypergeometric2F1[1/4, 5/3, 5/4, -((b*x^4)/a)])/(5*a*b*(a + b*x^4)^(2/3))
Time = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {910, 779, 778}
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.
\(\displaystyle \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/3}} \, dx\) |
\(\Big \downarrow \) 910 |
\(\displaystyle \frac {(3 a d+5 b c) \int \frac {1}{\left (b x^4+a\right )^{2/3}}dx}{8 a b}+\frac {3 x (b c-a d)}{8 a b \left (a+b x^4\right )^{2/3}}\) |
\(\Big \downarrow \) 779 |
\(\displaystyle \frac {\left (\frac {b x^4}{a}+1\right )^{2/3} (3 a d+5 b c) \int \frac {1}{\left (\frac {b x^4}{a}+1\right )^{2/3}}dx}{8 a b \left (a+b x^4\right )^{2/3}}+\frac {3 x (b c-a d)}{8 a b \left (a+b x^4\right )^{2/3}}\) |
\(\Big \downarrow \) 778 |
\(\displaystyle \frac {x \left (\frac {b x^4}{a}+1\right )^{2/3} (3 a d+5 b c) \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {2}{3},\frac {5}{4},-\frac {b x^4}{a}\right )}{8 a b \left (a+b x^4\right )^{2/3}}+\frac {3 x (b c-a d)}{8 a b \left (a+b x^4\right )^{2/3}}\) |
Input:
Int[(c + d*x^4)/(a + b*x^4)^(5/3),x]
Output:
(3*(b*c - a*d)*x)/(8*a*b*(a + b*x^4)^(2/3)) + ((5*b*c + 3*a*d)*x*(1 + (b*x ^4)/a)^(2/3)*Hypergeometric2F1[1/4, 2/3, 5/4, -((b*x^4)/a)])/(8*a*b*(a + b *x^4)^(2/3))
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*x*Hypergeometric2F 1[-p, 1/n, 1/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p , 0] && !IntegerQ[1/n] && !ILtQ[Simplify[1/n + p], 0] && (IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x ^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(1 + b*(x^n/a))^p, x], x ] /; FreeQ[{a, b, n, p}, x] && !IGtQ[p, 0] && !IntegerQ[1/n] && !ILtQ[Si mplify[1/n + p], 0] && !(IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Si mp[(-(b*c - a*d))*x*((a + b*x^n)^(p + 1)/(a*b*n*(p + 1))), x] - Simp[(a*d - b*c*(n*(p + 1) + 1))/(a*b*n*(p + 1)) Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/ n + p, 0])
\[\int \frac {d \,x^{4}+c}{\left (b \,x^{4}+a \right )^{\frac {5}{3}}}d x\]
Input:
int((d*x^4+c)/(b*x^4+a)^(5/3),x)
Output:
int((d*x^4+c)/(b*x^4+a)^(5/3),x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/3}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(5/3),x, algorithm="fricas")
Output:
integral((b*x^4 + a)^(1/3)*(d*x^4 + c)/(b^2*x^8 + 2*a*b*x^4 + a^2), x)
Result contains complex when optimal does not.
Time = 3.67 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.92 \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/3}} \, dx=\frac {c x \Gamma \left (\frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {5}{3} \\ \frac {5}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{3}} \Gamma \left (\frac {5}{4}\right )} + \frac {d x^{5} \Gamma \left (\frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{3} \\ \frac {9}{4} \end {matrix}\middle | {\frac {b x^{4} e^{i \pi }}{a}} \right )}}{4 a^{\frac {5}{3}} \Gamma \left (\frac {9}{4}\right )} \] Input:
integrate((d*x**4+c)/(b*x**4+a)**(5/3),x)
Output:
c*x*gamma(1/4)*hyper((1/4, 5/3), (5/4,), b*x**4*exp_polar(I*pi)/a)/(4*a**( 5/3)*gamma(5/4)) + d*x**5*gamma(5/4)*hyper((5/4, 5/3), (9/4,), b*x**4*exp_ polar(I*pi)/a)/(4*a**(5/3)*gamma(9/4))
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/3}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(5/3),x, algorithm="maxima")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(5/3), x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/3}} \, dx=\int { \frac {d x^{4} + c}{{\left (b x^{4} + a\right )}^{\frac {5}{3}}} \,d x } \] Input:
integrate((d*x^4+c)/(b*x^4+a)^(5/3),x, algorithm="giac")
Output:
integrate((d*x^4 + c)/(b*x^4 + a)^(5/3), x)
Timed out. \[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/3}} \, dx=\int \frac {d\,x^4+c}{{\left (b\,x^4+a\right )}^{5/3}} \,d x \] Input:
int((c + d*x^4)/(a + b*x^4)^(5/3),x)
Output:
int((c + d*x^4)/(a + b*x^4)^(5/3), x)
\[ \int \frac {c+d x^4}{\left (a+b x^4\right )^{5/3}} \, dx=\left (\int \frac {x^{4}}{\left (b \,x^{4}+a \right )^{\frac {2}{3}} a +\left (b \,x^{4}+a \right )^{\frac {2}{3}} b \,x^{4}}d x \right ) d +\left (\int \frac {1}{\left (b \,x^{4}+a \right )^{\frac {2}{3}} a +\left (b \,x^{4}+a \right )^{\frac {2}{3}} b \,x^{4}}d x \right ) c \] Input:
int((d*x^4+c)/(b*x^4+a)^(5/3),x)
Output:
int(x**4/((a + b*x**4)**(2/3)*a + (a + b*x**4)**(2/3)*b*x**4),x)*d + int(1 /((a + b*x**4)**(2/3)*a + (a + b*x**4)**(2/3)*b*x**4),x)*c