\(\int \frac {(e x)^{7/2} (A+B x^3)}{\sqrt {a+b x^3}} \, dx\) [265]

Optimal result

Integrand size = 26, antiderivative size = 121 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {(4 A b-3 a B) e^2 (e x)^{3/2} \sqrt {a+b x^3}}{12 b^2}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}-\frac {a (4 A b-3 a B) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{12 b^{5/2}} \] Output:

1/12*(4*A*b-3*B*a)*e^2*(e*x)^(3/2)*(b*x^3+a)^(1/2)/b^2+1/6*B*(e*x)^(9/2)*( 
b*x^3+a)^(1/2)/b/e-1/12*a*(4*A*b-3*B*a)*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 = 0.58 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.83 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {(e x)^{7/2} \sqrt {a+b x^3} \left (4 A b-3 a B+2 b B x^3\right )}{12 b^2 x^2}+\frac {a (-4 A b+3 a B) (e x)^{7/2} \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )}{12 b^{5/2} x^{7/2}} \] Input:

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

Output:

((e*x)^(7/2)*Sqrt[a + b*x^3]*(4*A*b - 3*a*B + 2*b*B*x^3))/(12*b^2*x^2) + ( 
a*(-4*A*b + 3*a*B)*(e*x)^(7/2)*Log[Sqrt[b]*x^(3/2) + Sqrt[a + b*x^3]])/(12 
*b^(5/2)*x^(7/2))
 

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {959, 843, 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.

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

\(\Big \downarrow \) 959

\(\displaystyle \frac {(4 A b-3 a B) \int \frac {(e x)^{7/2}}{\sqrt {b x^3+a}}dx}{4 b}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {(4 A b-3 a B) \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^3 \int \frac {\sqrt {e x}}{\sqrt {b x^3+a}}dx}{2 b}\right )}{4 b}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {(4 A b-3 a B) \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^2 \int \frac {e x}{\sqrt {b x^3+a}}d\sqrt {e x}}{b}\right )}{4 b}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {(4 A b-3 a B) \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^2 \int \frac {1}{\sqrt {a+\frac {b x}{e^2}}}d(e x)^{3/2}}{3 b}\right )}{4 b}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {(4 A b-3 a B) \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^2 \int \frac {1}{1-\frac {b x}{e^2}}d\frac {(e x)^{3/2}}{\sqrt {a+\frac {b x}{e^2}}}}{3 b}\right )}{4 b}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {(4 A b-3 a B) \left (\frac {e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+\frac {b x}{e^2}}}\right )}{3 b^{3/2}}\right )}{4 b}+\frac {B (e x)^{9/2} \sqrt {a+b x^3}}{6 b e}\)

Input:

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

Output:

(B*(e*x)^(9/2)*Sqrt[a + b*x^3])/(6*b*e) + ((4*A*b - 3*a*B)*((e^2*(e*x)^(3/ 
2)*Sqrt[a + b*x^3])/(3*b) - (a*e^(7/2)*ArcTanh[(Sqrt[b]*(e*x)^(3/2))/(e^(3 
/2)*Sqrt[a + (b*x)/e^2])])/(3*b^(3/2))))/(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 843

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*(m + n*p + 1))), x] - Simp[ 
a*c^n*((m - n + 1)/(b*(m + n*p + 1)))   Int[(c*x)^(m - n)*(a + b*x^n)^p, x] 
, x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && 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 = 1.41 (sec) , antiderivative size = 114, normalized size of antiderivative = 0.94

method	result	size
risch	\(\frac {x^{2} \left (2 b B \,x^{3}+4 A b -3 B a \right ) \sqrt {b \,x^{3}+a}\, e^{4}}{12 b^{2} \sqrt {e x}}-\frac {a \left (4 A b -3 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) e^{4} \sqrt {\left (b \,x^{3}+a \right ) e x}}{12 b^{2} \sqrt {b e}\, \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\)	\(114\)
default	\(-\frac {e^{3} \sqrt {e x}\, \sqrt {b \,x^{3}+a}\, \left (-2 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b \,x^{4}+4 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a b e -4 A \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, b x -3 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) a^{2} e +3 B \sqrt {\left (b \,x^{3}+a \right ) e x}\, \sqrt {b e}\, a x \right )}{12 \sqrt {\left (b \,x^{3}+a \right ) e x}\, b^{2} \sqrt {b e}}\)	\(166\)
elliptic	\(\text {Expression too large to display}\)	\(1093\)

Input:

int((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.42 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.02 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\left [-\frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} e^{3} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e + 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right ) - 4 \, {\left (2 \, B b e^{3} x^{4} - {\left (3 \, B a - 4 \, A b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{48 \, b^{2}}, -\frac {{\left (3 \, B a^{2} - 4 \, A a b\right )} e^{3} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right ) - 2 \, {\left (2 \, B b e^{3} x^{4} - {\left (3 \, B a - 4 \, A b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{24 \, b^{2}}\right ] \] Input:

integrate((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="fricas")
 

Output:

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

Sympy [A] (verification not implemented)

Time = 14.92 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.60 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\begin {cases} \frac {2 \left (\begin {cases} \text {NaN} & \text {for}\: e^{3} = 0 \\\frac {\begin {cases} - \frac {a e^{3} \left (A e^{3} - \frac {3 B a e^{3}}{4 b}\right ) \left (\begin {cases} \frac {\log {\left (\frac {2 b \left (e x\right )^{\frac {3}{2}}}{e^{3}} + 2 \sqrt {\frac {b}{e^{3}}} \sqrt {a + b x^{3}} \right )}}{\sqrt {\frac {b}{e^{3}}}} & \text {for}\: a \neq 0 \\\frac {\left (e x\right )^{\frac {3}{2}} \log {\left (\left (e x\right )^{\frac {3}{2}} \right )}}{\sqrt {b x^{3}}} & \text {otherwise} \end {cases}\right )}{2 b} + \sqrt {a + b x^{3}} \left (\frac {B e^{3} \left (e x\right )^{\frac {9}{2}}}{4 b} + \frac {e^{3} \left (e x\right )^{\frac {3}{2}} \left (A e^{3} - \frac {3 B a e^{3}}{4 b}\right )}{2 b}\right ) & \text {for}\: \frac {b}{e^{3}} \neq 0 \\\frac {\frac {A e^{3} \left (e x\right )^{\frac {9}{2}}}{3} + \frac {B \left (e x\right )^{\frac {15}{2}}}{5}}{\sqrt {a}} & \text {otherwise} \end {cases}}{3 e^{3}} & \text {otherwise} \end {cases}\right )}{e} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \] Input:

integrate((e*x)**(7/2)*(B*x**3+A)/(b*x**3+a)**(1/2),x)
 

Output:

Piecewise((2*Piecewise((nan, Eq(e**3, 0)), (Piecewise((-a*e**3*(A*e**3 - 3 
*B*a*e**3/(4*b))*Piecewise((log(2*b*(e*x)**(3/2)/e**3 + 2*sqrt(b/e**3)*sqr 
t(a + b*x**3))/sqrt(b/e**3), Ne(a, 0)), ((e*x)**(3/2)*log((e*x)**(3/2))/sq 
rt(b*x**3), True))/(2*b) + sqrt(a + b*x**3)*(B*e**3*(e*x)**(9/2)/(4*b) + e 
**3*(e*x)**(3/2)*(A*e**3 - 3*B*a*e**3/(4*b))/(2*b)), Ne(b/e**3, 0)), ((A*e 
**3*(e*x)**(9/2)/3 + B*(e*x)**(15/2)/5)/sqrt(a), True))/(3*e**3), True))/e 
, Ne(e, 0)), (0, True))
 

Maxima [F]

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

integrate((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="maxima")
 

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.21 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {\sqrt {b e^{4} x^{3} + a e^{4}} \sqrt {e x} e^{5} x {\left (\frac {2 \, B x^{3}}{b e^{2}} - \frac {3 \, B a b^{3} e^{5} - 4 \, A b^{4} e^{5}}{b^{5} e^{7}}\right )}}{12 \, {\left | e \right |}^{2}} - \frac {{\left (3 \, B a^{2} b^{3} e^{9} - 4 \, A a b^{4} e^{9}\right )} \log \left ({\left | -\sqrt {b e} \sqrt {e x} e x + \sqrt {b e^{4} x^{3} + a e^{4}} \right |}\right )}{12 \, \sqrt {b e} b^{5} e {\left | e \right |}^{4}} \] Input:

integrate((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(1/2),x, algorithm="giac")
 

Output:

1/12*sqrt(b*e^4*x^3 + a*e^4)*sqrt(e*x)*e^5*x*(2*B*x^3/(b*e^2) - (3*B*a*b^3 
*e^5 - 4*A*b^4*e^5)/(b^5*e^7))/abs(e)^2 - 1/12*(3*B*a^2*b^3*e^9 - 4*A*a*b^ 
4*e^9)*log(abs(-sqrt(b*e)*sqrt(e*x)*e*x + sqrt(b*e^4*x^3 + a*e^4)))/(sqrt( 
b*e)*b^5*e*abs(e)^4)
 

Mupad [F(-1)]

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

int(((A + B*x^3)*(e*x)^(7/2))/(a + b*x^3)^(1/2),x)
 

Output:

int(((A + B*x^3)*(e*x)^(7/2))/(a + b*x^3)^(1/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.74 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\sqrt {a+b x^3}} \, dx=\frac {\sqrt {e}\, e^{3} \left (2 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a b x +4 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b^{2} x^{4}+\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a^{2}-\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a^{2}\right )}{24 b^{2}} \] Input:

int((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(1/2),x)
 

Output:

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