3.9.47 \(\int \frac {(e x)^m (a+b x^4)^2}{(c+d x^4)^{3/2}} \, dx\) [847]

3.9.47.1 Optimal result

Integrand size = 26, antiderivative size = 198 \[ \int \frac {(e x)^m \left (a+b x^4\right )^2}{\left (c+d x^4\right )^{3/2}} \, dx=\frac {(b c-a d)^2 (e x)^{1+m}}{2 c d^2 e \sqrt {c+d x^4}}+\frac {b^2 (e x)^{1+m} \sqrt {c+d x^4}}{d^2 e (3+m)}-\frac {\left (2 b^2 c^2 (1+m)-(3+m) \left (2 a^2 d^2-(b c-a d)^2 (1+m)\right )\right ) (e x)^{1+m} \sqrt {1+\frac {d x^4}{c}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1+m}{4},\frac {5+m}{4},-\frac {d x^4}{c}\right )}{2 c d^2 e (1+m) (3+m) \sqrt {c+d x^4}} \]

output

1/2*(-a*d+b*c)^2*(e*x)^(1+m)/c/d^2/e/(d*x^4+c)^(1/2)+b^2*(e*x)^(1+m)*(d*x^ 
4+c)^(1/2)/d^2/e/(3+m)-1/2*(2*b^2*c^2*(1+m)-(3+m)*(2*a^2*d^2-(-a*d+b*c)^2* 
(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^2/e/(1+m)/(3+m)/(d*x^4+c)^(1/2)
 

3.9.47.2 Mathematica [A] (verified)

Time = 11.19 (sec) , antiderivative size = 167, normalized size of antiderivative = 0.84 \[ \int \frac {(e x)^m \left (a+b x^4\right )^2}{\left (c+d x^4\right )^{3/2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {d x^4}{c}} \left (a^2 \left (45+14 m+m^2\right ) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {1+m}{4},\frac {5+m}{4},-\frac {d x^4}{c}\right )+b (1+m) x^4 \left (2 a (9+m) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {5+m}{4},\frac {9+m}{4},-\frac {d x^4}{c}\right )+b (5+m) x^4 \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {9+m}{4},\frac {13+m}{4},-\frac {d x^4}{c}\right )\right )\right )}{c (1+m) (5+m) (9+m) \sqrt {c+d x^4}} \]

input

Integrate[((e*x)^m*(a + b*x^4)^2)/(c + d*x^4)^(3/2),x]
 

output

(x*(e*x)^m*Sqrt[1 + (d*x^4)/c]*(a^2*(45 + 14*m + m^2)*Hypergeometric2F1[3/ 
2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)] + b*(1 + m)*x^4*(2*a*(9 + m)*Hyperg 
eometric2F1[3/2, (5 + m)/4, (9 + m)/4, -((d*x^4)/c)] + b*(5 + m)*x^4*Hyper 
geometric2F1[3/2, (9 + m)/4, (13 + m)/4, -((d*x^4)/c)])))/(c*(1 + m)*(5 + 
m)*(9 + m)*Sqrt[c + d*x^4])
 

3.9.47.3 Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.01, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {963, 25, 959, 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.

\(\displaystyle \int \frac {\left (a+b x^4\right )^2 (e x)^m}{\left (c+d x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 963

\(\displaystyle \frac {(e x)^{m+1} (b c-a d)^2}{2 c d^2 e \sqrt {c+d x^4}}-\frac {\int -\frac {(e x)^m \left (2 b^2 c d x^4+2 a^2 d^2-(b c-a d)^2 (m+1)\right )}{\sqrt {d x^4+c}}dx}{2 c d^2}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {(e x)^m \left (2 b^2 c d x^4+2 a^2 d^2-(b c-a d)^2 (m+1)\right )}{\sqrt {d x^4+c}}dx}{2 c d^2}+\frac {(e x)^{m+1} (b c-a d)^2}{2 c d^2 e \sqrt {c+d x^4}}\)

\(\Big \downarrow \) 959

\(\displaystyle \frac {\frac {2 b^2 c \sqrt {c+d x^4} (e x)^{m+1}}{e (m+3)}-\frac {\left (2 b^2 c^2 (m+1)-(m+3) \left (2 a^2 d^2-(m+1) (b c-a d)^2\right )\right ) \int \frac {(e x)^m}{\sqrt {d x^4+c}}dx}{m+3}}{2 c d^2}+\frac {(e x)^{m+1} (b c-a d)^2}{2 c d^2 e \sqrt {c+d x^4}}\)

\(\Big \downarrow \) 889

\(\displaystyle \frac {\frac {2 b^2 c \sqrt {c+d x^4} (e x)^{m+1}}{e (m+3)}-\frac {\sqrt {\frac {d x^4}{c}+1} \left (2 b^2 c^2 (m+1)-(m+3) \left (2 a^2 d^2-(m+1) (b c-a d)^2\right )\right ) \int \frac {(e x)^m}{\sqrt {\frac {d x^4}{c}+1}}dx}{(m+3) \sqrt {c+d x^4}}}{2 c d^2}+\frac {(e x)^{m+1} (b c-a d)^2}{2 c d^2 e \sqrt {c+d x^4}}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {\frac {2 b^2 c \sqrt {c+d x^4} (e x)^{m+1}}{e (m+3)}-\frac {\sqrt {\frac {d x^4}{c}+1} (e x)^{m+1} \left (2 b^2 c^2 (m+1)-(m+3) \left (2 a^2 d^2-(m+1) (b c-a d)^2\right )\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m+1}{4},\frac {m+5}{4},-\frac {d x^4}{c}\right )}{e (m+1) (m+3) \sqrt {c+d x^4}}}{2 c d^2}+\frac {(e x)^{m+1} (b c-a d)^2}{2 c d^2 e \sqrt {c+d x^4}}\)

input

Int[((e*x)^m*(a + b*x^4)^2)/(c + d*x^4)^(3/2),x]
 

output

((b*c - a*d)^2*(e*x)^(1 + m))/(2*c*d^2*e*Sqrt[c + d*x^4]) + ((2*b^2*c*(e*x 
)^(1 + m)*Sqrt[c + d*x^4])/(e*(3 + m)) - ((2*b^2*c^2*(1 + m) - (3 + m)*(2* 
a^2*d^2 - (b*c - a*d)^2*(1 + m)))*(e*x)^(1 + m)*Sqrt[1 + (d*x^4)/c]*Hyperg 
eometric2F1[1/2, (1 + m)/4, (5 + m)/4, -((d*x^4)/c)])/(e*(1 + m)*(3 + m)*S 
qrt[c + d*x^4]))/(2*c*d^2)
 

3.9.47.3.1 Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 888

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])
 

rule 889

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])
 

rule 959

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_)), x_Symbol] :> Simp[d*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(b*e*(m + n*(p 
+ 1) + 1))), x] - Simp[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(b*(m + n*(p 
 + 1) + 1))   Int[(e*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, m, 
 n, p}, x] && NeQ[b*c - a*d, 0] && NeQ[m + n*(p + 1) + 1, 0]
 

rule 963

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^2, x_Symbol] :> Simp[(-(b*c - a*d)^2)*(e*x)^(m + 1)*((a + b*x^n)^(p + 1) 
/(a*b^2*e*n*(p + 1))), x] + Simp[1/(a*b^2*n*(p + 1))   Int[(e*x)^m*(a + b*x 
^n)^(p + 1)*Simp[(b*c - a*d)^2*(m + 1) + b^2*c^2*n*(p + 1) + a*b*d^2*n*(p + 
 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] 
 && IGtQ[n, 0] && LtQ[p, -1]
 

3.9.47.4 Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (b \,x^{4}+a \right )^{2}}{\left (d \,x^{4}+c \right )^{\frac {3}{2}}}d x\]

input

int((e*x)^m*(b*x^4+a)^2/(d*x^4+c)^(3/2),x)
 

output

int((e*x)^m*(b*x^4+a)^2/(d*x^4+c)^(3/2),x)
 

3.9.47.5 Fricas [F]

\[ \int \frac {(e x)^m \left (a+b x^4\right )^2}{\left (c+d x^4\right )^{3/2}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{4} + c\right )}^{\frac {3}{2}}} \,d x } \]

input

integrate((e*x)^m*(b*x^4+a)^2/(d*x^4+c)^(3/2),x, algorithm="fricas")
 

output

integral((b^2*x^8 + 2*a*b*x^4 + a^2)*sqrt(d*x^4 + c)*(e*x)^m/(d^2*x^8 + 2* 
c*d*x^4 + c^2), x)
 

3.9.47.6 Sympy [F]

\[ \int \frac {(e x)^m \left (a+b x^4\right )^2}{\left (c+d x^4\right )^{3/2}} \, dx=\int \frac {\left (e x\right )^{m} \left (a + b x^{4}\right )^{2}}{\left (c + d x^{4}\right )^{\frac {3}{2}}}\, dx \]

input

integrate((e*x)**m*(b*x**4+a)**2/(d*x**4+c)**(3/2),x)
 

output

Integral((e*x)**m*(a + b*x**4)**2/(c + d*x**4)**(3/2), x)
 

3.9.47.7 Maxima [F]

\[ \int \frac {(e x)^m \left (a+b x^4\right )^2}{\left (c+d x^4\right )^{3/2}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{4} + c\right )}^{\frac {3}{2}}} \,d x } \]

input

integrate((e*x)^m*(b*x^4+a)^2/(d*x^4+c)^(3/2),x, algorithm="maxima")
 

output

integrate((b*x^4 + a)^2*(e*x)^m/(d*x^4 + c)^(3/2), x)
 

3.9.47.8 Giac [F]

\[ \int \frac {(e x)^m \left (a+b x^4\right )^2}{\left (c+d x^4\right )^{3/2}} \, dx=\int { \frac {{\left (b x^{4} + a\right )}^{2} \left (e x\right )^{m}}{{\left (d x^{4} + c\right )}^{\frac {3}{2}}} \,d x } \]

input

integrate((e*x)^m*(b*x^4+a)^2/(d*x^4+c)^(3/2),x, algorithm="giac")
 

output

integrate((b*x^4 + a)^2*(e*x)^m/(d*x^4 + c)^(3/2), x)
 

3.9.47.9 Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (a+b x^4\right )^2}{\left (c+d x^4\right )^{3/2}} \, dx=\int \frac {{\left (e\,x\right )}^m\,{\left (b\,x^4+a\right )}^2}{{\left (d\,x^4+c\right )}^{3/2}} \,d x \]

input

int(((e*x)^m*(a + b*x^4)^2)/(c + d*x^4)^(3/2),x)
 

output

int(((e*x)^m*(a + b*x^4)^2)/(c + d*x^4)^(3/2), x)