\(\int \frac {(e x)^m (A+B x^3)}{(a+b x^3)^{5/2}} \, dx\) [388]

Optimal result

Integrand size = 24, antiderivative size = 123 \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 B (e x)^{1+m}}{b e (7-2 m) \left (a+b x^3\right )^{3/2}}+\frac {\left (\frac {A}{1+m}+\frac {2 a B}{7 b-2 b m}\right ) (e x)^{1+m} \sqrt {1+\frac {b x^3}{a}} \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )}{a^2 e \sqrt {a+b x^3}} \] Output:

-2*B*(e*x)^(1+m)/b/e/(7-2*m)/(b*x^3+a)^(3/2)+(A/(1+m)+2*a*B/(-2*b*m+7*b))* 
(e*x)^(1+m)*(1+b*x^3/a)^(1/2)*hypergeom([5/2, 1/3+1/3*m],[4/3+1/3*m],-b*x^ 
3/a)/a^2/e/(b*x^3+a)^(1/2)
 

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.92 \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {x (e x)^m \sqrt {1+\frac {b x^3}{a}} \left (A (4+m) \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {1+m}{3},\frac {4+m}{3},-\frac {b x^3}{a}\right )+B (1+m) x^3 \operatorname {Hypergeometric2F1}\left (\frac {5}{2},\frac {4+m}{3},\frac {7+m}{3},-\frac {b x^3}{a}\right )\right )}{a^2 (1+m) (4+m) \sqrt {a+b x^3}} \] Input:

Integrate[((e*x)^m*(A + B*x^3))/(a + b*x^3)^(5/2),x]
 

Output:

(x*(e*x)^m*Sqrt[1 + (b*x^3)/a]*(A*(4 + m)*Hypergeometric2F1[5/2, (1 + m)/3 
, (4 + m)/3, -((b*x^3)/a)] + B*(1 + m)*x^3*Hypergeometric2F1[5/2, (4 + m)/ 
3, (7 + m)/3, -((b*x^3)/a)]))/(a^2*(1 + m)*(4 + m)*Sqrt[a + b*x^3])
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.08, 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.

\(\displaystyle \int \frac {\left (A+B x^3\right ) (e x)^m}{\left (a+b x^3\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 957

\(\displaystyle \frac {(2 a B (m+1)+A b (7-2 m)) \int \frac {(e x)^m}{\left (b x^3+a\right )^{3/2}}dx}{9 a b}+\frac {2 (e x)^{m+1} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 889

\(\displaystyle \frac {\sqrt {\frac {b x^3}{a}+1} (2 a B (m+1)+A b (7-2 m)) \int \frac {(e x)^m}{\left (\frac {b x^3}{a}+1\right )^{3/2}}dx}{9 a^2 b \sqrt {a+b x^3}}+\frac {2 (e x)^{m+1} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}}\)

\(\Big \downarrow \) 888

\(\displaystyle \frac {\sqrt {\frac {b x^3}{a}+1} (e x)^{m+1} (2 a B (m+1)+A b (7-2 m)) \operatorname {Hypergeometric2F1}\left (\frac {3}{2},\frac {m+1}{3},\frac {m+4}{3},-\frac {b x^3}{a}\right )}{9 a^2 b e (m+1) \sqrt {a+b x^3}}+\frac {2 (e x)^{m+1} (A b-a B)}{9 a b e \left (a+b x^3\right )^{3/2}}\)

Input:

Int[((e*x)^m*(A + B*x^3))/(a + b*x^3)^(5/2),x]
 

Output:

(2*(A*b - a*B)*(e*x)^(1 + m))/(9*a*b*e*(a + b*x^3)^(3/2)) + ((A*b*(7 - 2*m 
) + 2*a*B*(1 + m))*(e*x)^(1 + m)*Sqrt[1 + (b*x^3)/a]*Hypergeometric2F1[3/2 
, (1 + m)/3, (4 + m)/3, -((b*x^3)/a)])/(9*a^2*b*e*(1 + m)*Sqrt[a + b*x^3])
 

Defintions of rubi rules used

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 957

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)]))
 

Maple [F]

\[\int \frac {\left (e x \right )^{m} \left (B \,x^{3}+A \right )}{\left (b \,x^{3}+a \right )^{\frac {5}{2}}}d x\]

Input:

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(5/2),x)
 

Output:

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(5/2),x)
 

Fricas [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x)^m*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="fricas")
 

Output:

integral((B*x^3 + A)*sqrt(b*x^3 + a)*(e*x)^m/(b^3*x^9 + 3*a*b^2*x^6 + 3*a^ 
2*b*x^3 + a^3), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((e*x)**m*(B*x**3+A)/(b*x**3+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x)^m*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)*(e*x)^m/(b*x^3 + a)^(5/2), x)
 

Giac [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{m}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x)^m*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="giac")
 

Output:

integrate((B*x^3 + A)*(e*x)^m/(b*x^3 + a)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int \frac {\left (B\,x^3+A\right )\,{\left (e\,x\right )}^m}{{\left (b\,x^3+a\right )}^{5/2}} \,d x \] Input:

int(((A + B*x^3)*(e*x)^m)/(a + b*x^3)^(5/2),x)
 

Output:

int(((A + B*x^3)*(e*x)^m)/(a + b*x^3)^(5/2), x)
 

Reduce [F]

\[ \int \frac {(e x)^m \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=e^{m} \left (\int \frac {x^{m} \sqrt {b \,x^{3}+a}}{b^{2} x^{6}+2 a b \,x^{3}+a^{2}}d x \right ) \] Input:

int((e*x)^m*(B*x^3+A)/(b*x^3+a)^(5/2),x)
 

Output:

e**m*int((x**m*sqrt(a + b*x**3))/(a**2 + 2*a*b*x**3 + b**2*x**6),x)