\(\int \frac {(a+b x^3)^{3/2} (A+B x^3)}{(e x)^{5/2}} \, dx\) [255]

Optimal result

Integrand size = 26, antiderivative size = 152 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\frac {(4 A b+a B) (e x)^{3/2} \sqrt {a+b x^3}}{4 e^4}+\frac {(4 A b+a B) (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 a e^4}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}+\frac {a (4 A b+a B) \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{4 \sqrt {b} e^{5/2}} \] Output:

1/4*(4*A*b+B*a)*(e*x)^(3/2)*(b*x^3+a)^(1/2)/e^4+1/6*(4*A*b+B*a)*(e*x)^(3/2 
)*(b*x^3+a)^(3/2)/a/e^4-2/3*A*(b*x^3+a)^(5/2)/a/e/(e*x)^(3/2)+1/4*a*(4*A*b 
+B*a)*arctanh(b^(1/2)*(e*x)^(3/2)/e^(3/2)/(b*x^3+a)^(1/2))/b^(1/2)/e^(5/2)
 

Mathematica [A] (verified)

Time = 0.58 (sec) , antiderivative size = 101, normalized size of antiderivative = 0.66 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\frac {x \left (\sqrt {b} \sqrt {a+b x^3} \left (-8 a A+4 A b x^3+5 a B x^3+2 b B x^6\right )+3 a (4 A b+a B) x^{3/2} \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )\right )}{12 \sqrt {b} (e x)^{5/2}} \] Input:

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

Output:

(x*(Sqrt[b]*Sqrt[a + b*x^3]*(-8*a*A + 4*A*b*x^3 + 5*a*B*x^3 + 2*b*B*x^6) + 
 3*a*(4*A*b + a*B)*x^(3/2)*Log[Sqrt[b]*x^(3/2) + Sqrt[a + b*x^3]]))/(12*Sq 
rt[b]*(e*x)^(5/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.52 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.269, Rules used = {955, 811, 811, 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.

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

\(\Big \downarrow \) 955

\(\displaystyle \frac {(a B+4 A b) \int \sqrt {e x} \left (b x^3+a\right )^{3/2}dx}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(a B+4 A b) \left (\frac {3}{4} a \int \sqrt {e x} \sqrt {b x^3+a}dx+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}\)

\(\Big \downarrow \) 811

\(\displaystyle \frac {(a B+4 A b) \left (\frac {3}{4} a \left (\frac {1}{2} a \int \frac {\sqrt {e x}}{\sqrt {b x^3+a}}dx+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {(a B+4 A b) \left (\frac {3}{4} a \left (\frac {a \int \frac {e x}{\sqrt {b x^3+a}}d\sqrt {e x}}{e}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(a B+4 A b) \left (\frac {3}{4} a \left (\frac {a \int \frac {1}{\sqrt {a+\frac {b x}{e^2}}}d(e x)^{3/2}}{3 e}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {(a B+4 A b) \left (\frac {3}{4} a \left (\frac {a \int \frac {1}{1-\frac {b x}{e^2}}d\frac {(e x)^{3/2}}{\sqrt {a+\frac {b x}{e^2}}}}{3 e}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(a B+4 A b) \left (\frac {3}{4} a \left (\frac {a \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 \sqrt {b}}+\frac {(e x)^{3/2} \sqrt {a+b x^3}}{3 e}\right )+\frac {(e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 e}\right )}{a e^3}-\frac {2 A \left (a+b x^3\right )^{5/2}}{3 a e (e x)^{3/2}}\)

Input:

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

Output:

(-2*A*(a + b*x^3)^(5/2))/(3*a*e*(e*x)^(3/2)) + ((4*A*b + a*B)*(((e*x)^(3/2 
)*(a + b*x^3)^(3/2))/(6*e) + (3*a*(((e*x)^(3/2)*Sqrt[a + b*x^3])/(3*e) + ( 
a*Sqrt[e]*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3/2)*Sqrt[a + (b*x)/e^2])])/(3 
*Sqrt[b])))/4))/(a*e^3)
 

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 811

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c* 
x)^(m + 1)*((a + b*x^n)^p/(c*(m + n*p + 1))), x] + Simp[a*n*(p/(m + n*p + 1 
))   Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && I 
GtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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 955

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

Maple [A] (verified)

Time = 1.41 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.77

method	result	size
risch	\(-\frac {\sqrt {b \,x^{3}+a}\, \left (-2 b B \,x^{6}-4 A b \,x^{3}-5 B a \,x^{3}+8 A a \right )}{12 x \,e^{2} \sqrt {e x}}+\frac {a \left (4 A b +B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) \sqrt {\left (b \,x^{3}+a \right ) e x}}{4 \sqrt {b e}\, e^{2} \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\)	\(117\)
default	\(\frac {\sqrt {b \,x^{3}+a}\, \left (2 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b \,x^{6}+12 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a b e \,x^{2}+4 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b \,x^{3}+3 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{2} e \,x^{2}+5 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a \,x^{3}-8 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a \right )}{12 x \,e^{2} \sqrt {e x}\, \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}}\)	\(197\)
elliptic	\(\text {Expression too large to display}\)	\(1140\)

Input:

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

Output:

-1/12*(b*x^3+a)^(1/2)*(-2*B*b*x^6-4*A*b*x^3-5*B*a*x^3+8*A*a)/x/e^2/(e*x)^( 
1/2)+1/4*a*(4*A*b+B*a)/(b*e)^(1/2)*arctanh(((b*x^3+a)*e*x)^(1/2)/x^2/(b*e) 
^(1/2))/e^2*((b*x^3+a)*e*x)^(1/2)/(e*x)^(1/2)/(b*x^3+a)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.68 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\left [\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {b e} x^{2} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b x^{4} + a x\right )} \sqrt {b x^{3} + a} \sqrt {b e} \sqrt {e x}\right ) + 4 \, {\left (2 \, B b^{2} x^{6} + {\left (5 \, B a b + 4 \, A b^{2}\right )} x^{3} - 8 \, A a b\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{48 \, b e^{3} x^{2}}, -\frac {3 \, {\left (B a^{2} + 4 \, A a b\right )} \sqrt {-b e} x^{2} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {-b e} \sqrt {e x} x}{2 \, b e x^{3} + a e}\right ) - 2 \, {\left (2 \, B b^{2} x^{6} + {\left (5 \, B a b + 4 \, A b^{2}\right )} x^{3} - 8 \, A a b\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{24 \, b e^{3} x^{2}}\right ] \] Input:

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

Output:

[1/48*(3*(B*a^2 + 4*A*a*b)*sqrt(b*e)*x^2*log(-8*b^2*e*x^6 - 8*a*b*e*x^3 - 
a^2*e - 4*(2*b*x^4 + a*x)*sqrt(b*x^3 + a)*sqrt(b*e)*sqrt(e*x)) + 4*(2*B*b^ 
2*x^6 + (5*B*a*b + 4*A*b^2)*x^3 - 8*A*a*b)*sqrt(b*x^3 + a)*sqrt(e*x))/(b*e 
^3*x^2), -1/24*(3*(B*a^2 + 4*A*a*b)*sqrt(-b*e)*x^2*arctan(2*sqrt(b*x^3 + a 
)*sqrt(-b*e)*sqrt(e*x)*x/(2*b*e*x^3 + a*e)) - 2*(2*B*b^2*x^6 + (5*B*a*b + 
4*A*b^2)*x^3 - 8*A*a*b)*sqrt(b*x^3 + a)*sqrt(e*x))/(b*e^3*x^2)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 289 vs. \(2 (138) = 276\).

Time = 20.53 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.90 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=- \frac {2 A a^{\frac {3}{2}}}{3 e^{\frac {5}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {A \sqrt {a} b x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e^{\frac {5}{2}}} - \frac {2 A \sqrt {a} b x^{\frac {3}{2}}}{3 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {A a \sqrt {b} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{e^{\frac {5}{2}}} + \frac {B a^{\frac {3}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 e^{\frac {5}{2}}} + \frac {B a^{\frac {3}{2}} x^{\frac {3}{2}}}{12 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B \sqrt {a} b x^{\frac {9}{2}}}{4 e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {B a^{2} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{4 \sqrt {b} e^{\frac {5}{2}}} + \frac {B b^{2} x^{\frac {15}{2}}}{6 \sqrt {a} e^{\frac {5}{2}} \sqrt {1 + \frac {b x^{3}}{a}}} \] Input:

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

Output:

-2*A*a**(3/2)/(3*e**(5/2)*x**(3/2)*sqrt(1 + b*x**3/a)) + A*sqrt(a)*b*x**(3 
/2)*sqrt(1 + b*x**3/a)/(3*e**(5/2)) - 2*A*sqrt(a)*b*x**(3/2)/(3*e**(5/2)*s 
qrt(1 + b*x**3/a)) + A*a*sqrt(b)*asinh(sqrt(b)*x**(3/2)/sqrt(a))/e**(5/2) 
+ B*a**(3/2)*x**(3/2)*sqrt(1 + b*x**3/a)/(3*e**(5/2)) + B*a**(3/2)*x**(3/2 
)/(12*e**(5/2)*sqrt(1 + b*x**3/a)) + B*sqrt(a)*b*x**(9/2)/(4*e**(5/2)*sqrt 
(1 + b*x**3/a)) + B*a**2*asinh(sqrt(b)*x**(3/2)/sqrt(a))/(4*sqrt(b)*e**(5/ 
2)) + B*b**2*x**(15/2)/(6*sqrt(a)*e**(5/2)*sqrt(1 + b*x**3/a))
 

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: NotImplementedError >> unable to parse Giac output: Recu 
rsive assumption sageVARa>=(-sageVARb*sageVARe/(sageVARe^4*t_nostep^6)) ig 
nored1/sageVARe^3/((1/sageVARe)^2)*2*(120*sageVARb^5*sageVARe^6*sageVARB/1 
440/sageVARb^4/sage
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 112, normalized size of antiderivative = 0.74 \[ \int \frac {\left (a+b x^3\right )^{3/2} \left (A+B x^3\right )}{(e x)^{5/2}} \, dx=\frac {\sqrt {e}\, \left (-16 \sqrt {b \,x^{3}+a}\, a^{2}+18 \sqrt {b \,x^{3}+a}\, a b \,x^{3}+4 \sqrt {b \,x^{3}+a}\, b^{2} x^{6}-15 \sqrt {x}\, \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a^{2} x +15 \sqrt {x}\, \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a^{2} x \right )}{24 \sqrt {x}\, e^{3} x} \] Input:

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

Output:

(sqrt(e)*( - 16*sqrt(a + b*x**3)*a**2 + 18*sqrt(a + b*x**3)*a*b*x**3 + 4*s 
qrt(a + b*x**3)*b**2*x**6 - 15*sqrt(x)*sqrt(b)*log(sqrt(a + b*x**3) - sqrt 
(x)*sqrt(b)*x)*a**2*x + 15*sqrt(x)*sqrt(b)*log(sqrt(a + b*x**3) + sqrt(x)* 
sqrt(b)*x)*a**2*x))/(24*sqrt(x)*e**3*x)