\(\int \sqrt {e x} \sqrt {a+b x^3} (A+B x^3) \, dx\) [241]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \sqrt {e x} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {(4 A b-a B) (e x)^{3/2} \sqrt {a+b x^3}}{12 b e}+\frac {B (e x)^{3/2} \left (a+b x^3\right )^{3/2}}{6 b e}+\frac {a (4 A b-a B) \sqrt {e} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{12 b^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.40 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79 \[ \int \sqrt {e x} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {x \sqrt {e x} \sqrt {a+b x^3} \left (4 A b+a B+2 b B x^3\right )}{12 b}-\frac {a (-4 A b+a B) \sqrt {e x} \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )}{12 b^{3/2} \sqrt {x}} \] Input:

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

Output:

(x*Sqrt[e*x]*Sqrt[a + b*x^3]*(4*A*b + a*B + 2*b*B*x^3))/(12*b) - (a*(-4*A* 
b + a*B)*Sqrt[e*x]*Log[Sqrt[b]*x^(3/2) + Sqrt[a + b*x^3]])/(12*b^(3/2)*Sqr 
t[x])
 

Rubi [A] (warning: unable to verify)

Time = 0.45 (sec) , antiderivative size = 118, 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 = {959, 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.

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 811

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(B*(e*x)^(3/2)*(a + b*x^3)^(3/2))/(6*b*e) + ((4*A*b - a*B)*(((e*x)^(3/2)*S 
qrt[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*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 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 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]
 

Maple [A] (verified)

Time = 1.20 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.90

method	result	size
risch	\(\frac {x^{2} \left (2 b B \,x^{3}+4 A b +B a \right ) \sqrt {b \,x^{3}+a}\, e}{12 b \sqrt {e x}}+\frac {a \left (4 A b -B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) e \sqrt {\left (b \,x^{3}+a \right ) e x}}{12 b \sqrt {b e}\, \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\)	\(109\)
default	\(\frac {\sqrt {e x}\, \sqrt {b \,x^{3}+a}\, \left (2 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b \,x^{4}+4 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b x +4 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a b e +B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a x -B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{2} e \right )}{12 \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b}\)	\(162\)
elliptic	\(\text {Expression too large to display}\)	\(1096\)

Input:

int((e*x)^(1/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 221, normalized size of antiderivative = 1.83 \[ \int \sqrt {e x} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\left [-\frac {{\left (B a^{2} - 4 \, A a b\right )} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right ) - 4 \, {\left (2 \, B b x^{4} + {\left (B a + 4 \, A b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{48 \, b}, \frac {{\left (B a^{2} - 4 \, A a b\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right ) + 2 \, {\left (2 \, B b x^{4} + {\left (B a + 4 \, A b\right )} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{24 \, b}\right ] \] Input:

integrate((e*x)^(1/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="fricas")
 

Output:

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

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 253 vs. \(2 (104) = 208\).

Time = 1.62 (sec) , antiderivative size = 253, normalized size of antiderivative = 2.09 \[ \int \sqrt {e x} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\begin {cases} \frac {2 \left (\begin {cases} \text {NaN} & \text {for}\: e^{3} = 0 \\\frac {A e^{3} \left (\begin {cases} \frac {a \left (\begin {cases} \frac {\log {\left (\frac {2 b \left (e x\right )^{\frac {3}{2}}}{e^{3}} + 2 \sqrt {\frac {b}{e^{3}}} \sqrt {a + b x^{3}} \right )}}{\sqrt {\frac {b}{e^{3}}}} & \text {for}\: a \neq 0 \\\frac {\left (e x\right )^{\frac {3}{2}} \log {\left (\left (e x\right )^{\frac {3}{2}} \right )}}{\sqrt {b x^{3}}} & \text {otherwise} \end {cases}\right )}{2} + \frac {\left (e x\right )^{\frac {3}{2}} \sqrt {a + b x^{3}}}{2} & \text {for}\: \frac {b}{e^{3}} \neq 0 \\\sqrt {a} \left (e x\right )^{\frac {3}{2}} & \text {otherwise} \end {cases}\right ) + B \left (\begin {cases} - \frac {a^{2} e^{3} \left (\begin {cases} \frac {\log {\left (\frac {2 b \left (e x\right )^{\frac {3}{2}}}{e^{3}} + 2 \sqrt {\frac {b}{e^{3}}} \sqrt {a + b x^{3}} \right )}}{\sqrt {\frac {b}{e^{3}}}} & \text {for}\: a \neq 0 \\\frac {\left (e x\right )^{\frac {3}{2}} \log {\left (\left (e x\right )^{\frac {3}{2}} \right )}}{\sqrt {b x^{3}}} & \text {otherwise} \end {cases}\right )}{8 b} + \sqrt {a + b x^{3}} \left (\frac {a e^{3} \left (e x\right )^{\frac {3}{2}}}{8 b} + \frac {\left (e x\right )^{\frac {9}{2}}}{4}\right ) & \text {for}\: \frac {b}{e^{3}} \neq 0 \\\frac {\sqrt {a} \left (e x\right )^{\frac {9}{2}}}{3} & \text {otherwise} \end {cases}\right )}{3 e^{3}} & \text {otherwise} \end {cases}\right )}{e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

integrate((e*x)**(1/2)*(b*x**3+a)**(1/2)*(B*x**3+A),x)
 

Output:

Piecewise((2*Piecewise((nan, Eq(e**3, 0)), ((A*e**3*Piecewise((a*Piecewise 
((log(2*b*(e*x)**(3/2)/e**3 + 2*sqrt(b/e**3)*sqrt(a + b*x**3))/sqrt(b/e**3 
), Ne(a, 0)), ((e*x)**(3/2)*log((e*x)**(3/2))/sqrt(b*x**3), True))/2 + (e* 
x)**(3/2)*sqrt(a + b*x**3)/2, Ne(b/e**3, 0)), (sqrt(a)*(e*x)**(3/2), True) 
) + B*Piecewise((-a**2*e**3*Piecewise((log(2*b*(e*x)**(3/2)/e**3 + 2*sqrt( 
b/e**3)*sqrt(a + b*x**3))/sqrt(b/e**3), Ne(a, 0)), ((e*x)**(3/2)*log((e*x) 
**(3/2))/sqrt(b*x**3), True))/(8*b) + sqrt(a + b*x**3)*(a*e**3*(e*x)**(3/2 
)/(8*b) + (e*x)**(9/2)/4), Ne(b/e**3, 0)), (sqrt(a)*(e*x)**(9/2)/3, True)) 
)/(3*e**3), True))/e, Ne(e, 0)), (0, True))
 

Maxima [F]

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

integrate((e*x)^(1/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="maxima")
 

Output:

integrate((B*x^3 + A)*sqrt(b*x^3 + a)*sqrt(e*x), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.46 \[ \int \sqrt {e x} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {B a^{2} e \log \left ({\left | -\sqrt {b e} \sqrt {e x} e x + \sqrt {b e^{4} x^{3} + a e^{4}} \right |}\right )}{12 \, \sqrt {b e} b} - \frac {{\left (\frac {a e^{4} \log \left ({\left | -\sqrt {b e} \sqrt {e x} e x + \sqrt {b e^{4} x^{3} + a e^{4}} \right |}\right )}{\sqrt {b e}} - \sqrt {b e^{4} x^{3} + a e^{4}} \sqrt {e x} e x\right )} A {\left | e \right |}^{2}}{3 \, e^{5}} + \frac {\sqrt {b e^{4} x^{3} + a e^{4}} {\left (2 \, e^{3} x^{3} + \frac {a e^{3}}{b}\right )} \sqrt {e x} B x {\left | e \right |}^{2}}{12 \, e^{7}} \] Input:

integrate((e*x)^(1/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x, algorithm="giac")
 

Output:

1/12*B*a^2*e*log(abs(-sqrt(b*e)*sqrt(e*x)*e*x + sqrt(b*e^4*x^3 + a*e^4)))/ 
(sqrt(b*e)*b) - 1/3*(a*e^4*log(abs(-sqrt(b*e)*sqrt(e*x)*e*x + sqrt(b*e^4*x 
^3 + a*e^4)))/sqrt(b*e) - sqrt(b*e^4*x^3 + a*e^4)*sqrt(e*x)*e*x)*A*abs(e)^ 
2/e^5 + 1/12*sqrt(b*e^4*x^3 + a*e^4)*(2*e^3*x^3 + a*e^3/b)*sqrt(e*x)*B*x*a 
bs(e)^2/e^7
 

Mupad [F(-1)]

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

int((A + B*x^3)*(e*x)^(1/2)*(a + b*x^3)^(1/2),x)
 

Output:

int((A + B*x^3)*(e*x)^(1/2)*(a + b*x^3)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.73 \[ \int \sqrt {e x} \sqrt {a+b x^3} \left (A+B x^3\right ) \, dx=\frac {\sqrt {e}\, \left (10 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a b x +4 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b^{2} x^{4}-3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a^{2}+3 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a^{2}\right )}{24 b} \] Input:

int((e*x)^(1/2)*(b*x^3+a)^(1/2)*(B*x^3+A),x)
 

Output:

(sqrt(e)*(10*sqrt(x)*sqrt(a + b*x**3)*a*b*x + 4*sqrt(x)*sqrt(a + b*x**3)*b 
**2*x**4 - 3*sqrt(b)*log(sqrt(a + b*x**3) - sqrt(x)*sqrt(b)*x)*a**2 + 3*sq 
rt(b)*log(sqrt(a + b*x**3) + sqrt(x)*sqrt(b)*x)*a**2))/(24*b)