Integrand size = 24, antiderivative size = 132 \[ \int \frac {(e x)^m \left (a+b x^4\right )}{\left (c+d x^4\right )^{3/2}} \, dx=-\frac {(b c-a d) (e x)^{1+m}}{2 c d e \sqrt {c+d x^4}}+\frac {(a d (1-m)+b c (1+m)) (e x)^{1+m} \sqrt {1+\frac {d x^4}{c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{4},\frac {5+m}{4},-\frac {d x^4}{c}\right )}{2 c d e (1+m) \sqrt {c+d x^4}} \]
-1/2*(-a*d+b*c)*(e*x)^(1+m)/c/d/e/(d*x^4+c)^(1/2)+1/2*(a*d*(-m+1)+b*c*(1+m ))*(e*x)^(1+m)*hypergeom([1/2, 1/4+1/4*m],[5/4+1/4*m],-d*x^4/c)*(1+d*x^4/c )^(1/2)/c/d/e/(1+m)/(d*x^4+c)^(1/2)
Time = 5.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.86 \[ \int \frac {(e x)^m \left (a+b x^4\right )}{\left (c+d x^4\right )^{3/2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {d x^4}{c}} \left (a (5+m) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{4},\frac {5+m}{4},-\frac {d x^4}{c}\right )+b (1+m) x^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5+m}{4},\frac {9+m}{4},-\frac {d x^4}{c}\right )\right )}{c (1+m) (5+m) \sqrt {c+d x^4}} \]
(x*(e*x)^m*Sqrt[1 + (d*x^4)/c]*(a*(5 + m)*Hypergeometric2F1[3/2, (1 + m)/4 , (5 + m)/4, -((d*x^4)/c)] + b*(1 + m)*x^4*Hypergeometric2F1[3/2, (5 + m)/ 4, (9 + m)/4, -((d*x^4)/c)]))/(c*(1 + m)*(5 + m)*Sqrt[c + d*x^4])
Time = 0.25 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {957, 889, 888}
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.
\(\displaystyle \int \frac {\left (a+b x^4\right ) (e x)^m}{\left (c+d x^4\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 957 |
\(\displaystyle \frac {(a d (1-m)+b c (m+1)) \int \frac {(e x)^m}{\sqrt {d x^4+c}}dx}{2 c d}-\frac {(e x)^{m+1} (b c-a d)}{2 c d e \sqrt {c+d x^4}}\) |
\(\Big \downarrow \) 889 |
\(\displaystyle \frac {\sqrt {\frac {d x^4}{c}+1} (a d (1-m)+b c (m+1)) \int \frac {(e x)^m}{\sqrt {\frac {d x^4}{c}+1}}dx}{2 c d \sqrt {c+d x^4}}-\frac {(e x)^{m+1} (b c-a d)}{2 c d e \sqrt {c+d x^4}}\) |
\(\Big \downarrow \) 888 |
\(\displaystyle \frac {\sqrt {\frac {d x^4}{c}+1} (e x)^{m+1} (a d (1-m)+b c (m+1)) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{4},\frac {m+5}{4},-\frac {d x^4}{c}\right )}{2 c d e (m+1) \sqrt {c+d x^4}}-\frac {(e x)^{m+1} (b c-a d)}{2 c d e \sqrt {c+d x^4}}\) |
-1/2*((b*c - a*d)*(e*x)^(1 + m))/(c*d*e*Sqrt[c + d*x^4]) + ((a*d*(1 - m) + b*c*(1 + m))*(e*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*Hypergeometric2F1[1/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)])/(2*c*d*e*(1 + m)*Sqrt[c + d*x^4])
3.9.48.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p *((c*x)^(m + 1)/(c*(m + 1)))*Hypergeometric2F1[-p, (m + 1)/n, (m + 1)/n + 1 , (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0] && (ILt Q[p, 0] || GtQ[a, 0])
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^I ntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]) Int[(c*x) ^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] && !IGtQ[p, 0 ] && !(ILtQ[p, 0] || GtQ[a, 0])
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)]))
\[\int \frac {\left (e x \right )^{m} \left (b \,x^{4}+a \right )}{\left (d \,x^{4}+c \right )^{\frac {3}{2}}}d x\]
\[ \int \frac {(e x)^m \left (a+b x^4\right )}{\left (c+d x^4\right )^{3/2}} \, dx=\int { \frac {{\left (b x^{4} + a\right )} \left (e x\right )^{m}}{{\left (d x^{4} + c\right )}^{\frac {3}{2}}} \,d x } \]
Result contains complex when optimal does not.
Time = 17.87 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.89 \[ \int \frac {(e x)^m \left (a+b x^4\right )}{\left (c+d x^4\right )^{3/2}} \, dx=\frac {a e^{m} x^{m + 1} \Gamma \left (\frac {m}{4} + \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{4} + \frac {1}{4} \\ \frac {m}{4} + \frac {5}{4} \end {matrix}\middle | {\frac {d x^{4} e^{i \pi }}{c}} \right )}}{4 c^{\frac {3}{2}} \Gamma \left (\frac {m}{4} + \frac {5}{4}\right )} + \frac {b e^{m} x^{m + 5} \Gamma \left (\frac {m}{4} + \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{2}, \frac {m}{4} + \frac {5}{4} \\ \frac {m}{4} + \frac {9}{4} \end {matrix}\middle | {\frac {d x^{4} e^{i \pi }}{c}} \right )}}{4 c^{\frac {3}{2}} \Gamma \left (\frac {m}{4} + \frac {9}{4}\right )} \]
a*e**m*x**(m + 1)*gamma(m/4 + 1/4)*hyper((3/2, m/4 + 1/4), (m/4 + 5/4,), d *x**4*exp_polar(I*pi)/c)/(4*c**(3/2)*gamma(m/4 + 5/4)) + b*e**m*x**(m + 5) *gamma(m/4 + 5/4)*hyper((3/2, m/4 + 5/4), (m/4 + 9/4,), d*x**4*exp_polar(I *pi)/c)/(4*c**(3/2)*gamma(m/4 + 9/4))
\[ \int \frac {(e x)^m \left (a+b x^4\right )}{\left (c+d x^4\right )^{3/2}} \, dx=\int { \frac {{\left (b x^{4} + a\right )} \left (e x\right )^{m}}{{\left (d x^{4} + c\right )}^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(e x)^m \left (a+b x^4\right )}{\left (c+d x^4\right )^{3/2}} \, dx=\int { \frac {{\left (b x^{4} + a\right )} \left (e x\right )^{m}}{{\left (d x^{4} + c\right )}^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(e x)^m \left (a+b x^4\right )}{\left (c+d x^4\right )^{3/2}} \, dx=\int \frac {{\left (e\,x\right )}^m\,\left (b\,x^4+a\right )}{{\left (d\,x^4+c\right )}^{3/2}} \,d x \]