\(\int \frac {(e x)^{7/2} (A+B x^3)}{(a+b x^3)^{5/2}} \, dx\) [281]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 (A b-a B) e^2 (e x)^{3/2}}{9 b^2 \left (a+b x^3\right )^{3/2}}+\frac {2 (A b-4 a B) e^2 (e x)^{3/2}}{9 a b^2 \sqrt {a+b x^3}}+\frac {2 B e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{5/2}} \] Output:

-2/9*(A*b-B*a)*e^2*(e*x)^(3/2)/b^2/(b*x^3+a)^(3/2)+2/9*(A*b-4*B*a)*e^2*(e* 
x)^(3/2)/a/b^2/(b*x^3+a)^(1/2)+2/3*B*e^(7/2)*arctanh(b^(1/2)*(e*x)^(3/2)/e 
^(3/2)/(b*x^3+a)^(1/2))/b^(5/2)
 

Mathematica [A] (verified)

Time = 1.03 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.83 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {2 e^3 \sqrt {e x} \left (\frac {\sqrt {b} x^{3/2} \left (-3 a^2 B+A b^2 x^3-4 a b B x^3\right )}{a \left (a+b x^3\right )^{3/2}}+3 B \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )\right )}{9 b^{5/2} \sqrt {x}} \] Input:

Integrate[((e*x)^(7/2)*(A + B*x^3))/(a + b*x^3)^(5/2),x]
 

Output:

(2*e^3*Sqrt[e*x]*((Sqrt[b]*x^(3/2)*(-3*a^2*B + A*b^2*x^3 - 4*a*b*B*x^3))/( 
a*(a + b*x^3)^(3/2)) + 3*B*Log[Sqrt[b]*x^(3/2) + Sqrt[a + b*x^3]]))/(9*b^( 
5/2)*Sqrt[x])
 

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {954, 817, 851, 807, 224, 219}

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.

\(\displaystyle \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx\)

\(\Big \downarrow \) 954

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

\(\Big \downarrow \) 817

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

Int[((e*x)^(7/2)*(A + B*x^3))/(a + b*x^3)^(5/2),x]
 

Output:

(2*(A*b - a*B)*(e*x)^(9/2))/(9*a*b*e*(a + b*x^3)^(3/2)) + (B*((-2*e^2*(e*x 
)^(3/2))/(3*b*Sqrt[a + b*x^3]) + (2*e^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/ 
(e^(3/2)*Sqrt[a + (b*x)/e^2])])/(3*b^(3/2))))/b
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 807

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m 
+ 1, n]}, Simp[1/k   Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, 
x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
 

rule 817

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]
 

rule 851

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = 
 Denominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^ 
n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && 
FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]
 

rule 954

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*(m + 1))), x] + Simp[d/b   Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; Fre 
eQ[{a, b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 
 1, 0] && NeQ[m, -1]
 

Maple [A] (verified)

Time = 0.93 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.30

method	result	size
default	\(\frac {2 \left (A \,b^{2} x^{5} \sqrt {b e}-4 B a b \,x^{5} \sqrt {b e}+3 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a b \,x^{3} \sqrt {\left (b \,x^{3}+a \right ) e x}-3 B \,a^{2} x^{2} \sqrt {b e}+3 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{2} \sqrt {\left (b \,x^{3}+a \right ) e x}\right ) \sqrt {e x}\, e^{3}}{9 \sqrt {b e}\, b^{2} a \left (b \,x^{3}+a \right )^{\frac {3}{2}} x}\)	\(157\)
elliptic	\(\text {Expression too large to display}\)	\(1098\)

Input:

int((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

2/9*(A*b^2*x^5*(b*e)^(1/2)-4*B*a*b*x^5*(b*e)^(1/2)+3*B*arctanh(((b*x^3+a)* 
e*x)^(1/2)/x^2/(b*e)^(1/2))*a*b*x^3*((b*x^3+a)*e*x)^(1/2)-3*B*a^2*x^2*(b*e 
)^(1/2)+3*B*arctanh(((b*x^3+a)*e*x)^(1/2)/x^2/(b*e)^(1/2))*a^2*((b*x^3+a)* 
e*x)^(1/2))*(e*x)^(1/2)*e^3/(b*e)^(1/2)/b^2/a/(b*x^3+a)^(3/2)/x
 

Fricas [A] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 345, normalized size of antiderivative = 2.85 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (B a b^{2} e^{3} x^{6} + 2 \, B a^{2} b e^{3} x^{3} + B a^{3} e^{3}\right )} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right ) - 4 \, {\left ({\left (4 \, B a b - A b^{2}\right )} e^{3} x^{4} + 3 \, B a^{2} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{18 \, {\left (a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}}, -\frac {3 \, {\left (B a b^{2} e^{3} x^{6} + 2 \, B a^{2} b e^{3} x^{3} + B a^{3} e^{3}\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right ) + 2 \, {\left ({\left (4 \, B a b - A b^{2}\right )} e^{3} x^{4} + 3 \, B a^{2} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{9 \, {\left (a b^{4} x^{6} + 2 \, a^{2} b^{3} x^{3} + a^{3} b^{2}\right )}}\right ] \] Input:

integrate((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="fricas")
 

Output:

[1/18*(3*(B*a*b^2*e^3*x^6 + 2*B*a^2*b*e^3*x^3 + B*a^3*e^3)*sqrt(e/b)*log(- 
8*b^2*e*x^6 - 8*a*b*e*x^3 - a^2*e - 4*(2*b^2*x^4 + a*b*x)*sqrt(b*x^3 + a)* 
sqrt(e*x)*sqrt(e/b)) - 4*((4*B*a*b - A*b^2)*e^3*x^4 + 3*B*a^2*e^3*x)*sqrt( 
b*x^3 + a)*sqrt(e*x))/(a*b^4*x^6 + 2*a^2*b^3*x^3 + a^3*b^2), -1/9*(3*(B*a* 
b^2*e^3*x^6 + 2*B*a^2*b*e^3*x^3 + B*a^3*e^3)*sqrt(-e/b)*arctan(2*sqrt(b*x^ 
3 + a)*sqrt(e*x)*b*x*sqrt(-e/b)/(2*b*e*x^3 + a*e)) + 2*((4*B*a*b - A*b^2)* 
e^3*x^4 + 3*B*a^2*e^3*x)*sqrt(b*x^3 + a)*sqrt(e*x))/(a*b^4*x^6 + 2*a^2*b^3 
*x^3 + a^3*b^2)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((e*x)**(7/2)*(B*x**3+A)/(b*x**3+a)**(5/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int { \frac {{\left (B x^{3} + A\right )} \left (e x\right )^{\frac {7}{2}}}{{\left (b x^{3} + a\right )}^{\frac {5}{2}}} \,d x } \] Input:

integrate((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)*(e*x)^(7/2)/(b*x^3 + a)^(5/2), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.04 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=-\frac {2 \, B e^{6} \log \left ({\left | -\sqrt {b e} \sqrt {e x} e x + \sqrt {b e^{4} x^{3} + a e^{4}} \right |}\right )}{3 \, \sqrt {b e} b^{2} {\left | e \right |}^{2}} - \frac {2 \, {\left (\frac {3 \, B a e^{8}}{b^{2}} + \frac {{\left (4 \, B a^{5} b^{6} e^{24} - A a^{4} b^{7} e^{24}\right )} x^{3}}{a^{5} b^{7} e^{16}}\right )} \sqrt {e x} e x}{9 \, {\left (b e^{4} x^{3} + a e^{4}\right )}^{\frac {3}{2}}} \] Input:

integrate((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(5/2),x, algorithm="giac")
 

Output:

-2/3*B*e^6*log(abs(-sqrt(b*e)*sqrt(e*x)*e*x + sqrt(b*e^4*x^3 + a*e^4)))/(s 
qrt(b*e)*b^2*abs(e)^2) - 2/9*(3*B*a*e^8/b^2 + (4*B*a^5*b^6*e^24 - A*a^4*b^ 
7*e^24)*x^3/(a^5*b^7*e^16))*sqrt(e*x)*e*x/(b*e^4*x^3 + a*e^4)^(3/2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\int \frac {\left (B\,x^3+A\right )\,{\left (e\,x\right )}^{7/2}}{{\left (b\,x^3+a\right )}^{5/2}} \,d x \] Input:

int(((A + B*x^3)*(e*x)^(7/2))/(a + b*x^3)^(5/2),x)
 

Output:

int(((A + B*x^3)*(e*x)^(7/2))/(a + b*x^3)^(5/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{5/2}} \, dx=\frac {\sqrt {e}\, e^{3} \left (-2 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b x -\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a -\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) b \,x^{3}+\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a +\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) b \,x^{3}\right )}{3 b^{2} \left (b \,x^{3}+a \right )} \] Input:

int((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(5/2),x)
 

Output:

(sqrt(e)*e**3*( - 2*sqrt(x)*sqrt(a + b*x**3)*b*x - sqrt(b)*log(sqrt(a + b* 
x**3) - sqrt(x)*sqrt(b)*x)*a - sqrt(b)*log(sqrt(a + b*x**3) - sqrt(x)*sqrt 
(b)*x)*b*x**3 + sqrt(b)*log(sqrt(a + b*x**3) + sqrt(x)*sqrt(b)*x)*a + sqrt 
(b)*log(sqrt(a + b*x**3) + sqrt(x)*sqrt(b)*x)*b*x**3))/(3*b**2*(a + b*x**3 
))