\(\int \frac {(e x)^{7/2} (A+B x^3)}{(a+b x^3)^{3/2}} \, dx\) [273]

Optimal result

Integrand size = 26, antiderivative size = 119 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 (A b-a B) e^2 (e x)^{3/2}}{3 b^2 \sqrt {a+b x^3}}+\frac {B e^2 (e x)^{3/2} \sqrt {a+b x^3}}{3 b^2}+\frac {(2 A b-3 a B) e^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (e x)^{3/2}}{e^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{5/2}} \] Output:

-2/3*(A*b-B*a)*e^2*(e*x)^(3/2)/b^2/(b*x^3+a)^(1/2)+1/3*B*e^2*(e*x)^(3/2)*( 
b*x^3+a)^(1/2)/b^2+1/3*(2*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 = 0.85 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.78 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {(e x)^{7/2} \left (\frac {\sqrt {b} x^{3/2} \left (-2 A b+3 a B+b B x^3\right )}{\sqrt {a+b x^3}}+(2 A b-3 a B) \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )\right )}{3 b^{5/2} x^{7/2}} \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 0.48 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.01, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {959, 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.

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

\(\Big \downarrow \) 959

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

\(\Big \downarrow \) 817

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

\(\Big \downarrow \) 851

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

\(\Big \downarrow \) 807

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

(B*(e*x)^(9/2))/(3*b*e*Sqrt[a + b*x^3]) + ((2*A*b - 3*a*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))))/(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 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 = 1.35 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08

method	result	size
risch	\(\frac {B \,x^{2} \sqrt {b \,x^{3}+a}\, e^{4}}{3 b^{2} \sqrt {e x}}+\frac {\left (\frac {2 \left (2 A b -3 B a \right ) \operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right )}{3 \sqrt {b e}}-\frac {4 \left (A b -B a \right ) x^{2}}{3 \sqrt {\left (x^{3}+\frac {a}{b}\right ) b e x}}\right ) e^{4} \sqrt {\left (b \,x^{3}+a \right ) e x}}{2 b^{2} \sqrt {e x}\, \sqrt {b \,x^{3}+a}}\)	\(129\)
default	\(\frac {e^{3} \sqrt {e x}\, \left (B \sqrt {b e}\, b \,x^{5}-2 A \sqrt {b e}\, b \,x^{2}+3 B \sqrt {b e}\, a \,x^{2}+2 A \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) \sqrt {\left (b \,x^{3}+a \right ) e x}\, b -3 B \,\operatorname {arctanh}\left (\frac {\sqrt {\left (b \,x^{3}+a \right ) e x}}{x^{2} \sqrt {b e}}\right ) \sqrt {\left (b \,x^{3}+a \right ) e x}\, a \right )}{3 x \sqrt {b \,x^{3}+a}\, b^{2} \sqrt {b e}}\)	\(143\)
elliptic	\(\text {Expression too large to display}\)	\(1097\)

Input:

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

Output:

1/3*B*x^2/b^2*(b*x^3+a)^(1/2)*e^4/(e*x)^(1/2)+1/2/b^2*(2/3*(2*A*b-3*B*a)/( 
b*e)^(1/2)*arctanh(((b*x^3+a)*e*x)^(1/2)/x^2/(b*e)^(1/2))-4/3*(A*b-B*a)*x^ 
2/((x^3+a/b)*b*e*x)^(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 = 0.39 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.58 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\left [-\frac {{\left ({\left (3 \, B a b - 2 \, A b^{2}\right )} e^{3} x^{3} + {\left (3 \, B a^{2} - 2 \, A a b\right )} e^{3}\right )} \sqrt {\frac {e}{b}} \log \left (-8 \, b^{2} e x^{6} - 8 \, a b e x^{3} - a^{2} e - 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {e x} \sqrt {\frac {e}{b}}\right ) - 4 \, {\left (B b e^{3} x^{4} + {\left (3 \, B a - 2 \, A b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{12 \, {\left (b^{3} x^{3} + a b^{2}\right )}}, \frac {{\left ({\left (3 \, B a b - 2 \, A b^{2}\right )} e^{3} x^{3} + {\left (3 \, B a^{2} - 2 \, A a b\right )} e^{3}\right )} \sqrt {-\frac {e}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {e x} b x \sqrt {-\frac {e}{b}}}{2 \, b e x^{3} + a e}\right ) + 2 \, {\left (B b e^{3} x^{4} + {\left (3 \, B a - 2 \, A b\right )} e^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {e x}}{6 \, {\left (b^{3} x^{3} + a b^{2}\right )}}\right ] \] Input:

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

Output:

[-1/12*(((3*B*a*b - 2*A*b^2)*e^3*x^3 + (3*B*a^2 - 2*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*(B*b*e^3*x^4 + (3*B*a - 2*A*b)*e^3*x)*sqrt(b 
*x^3 + a)*sqrt(e*x))/(b^3*x^3 + a*b^2), 1/6*(((3*B*a*b - 2*A*b^2)*e^3*x^3 
+ (3*B*a^2 - 2*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*(B*b*e^3*x^4 + (3*B*a - 2*A*b)*e^3*x) 
*sqrt(b*x^3 + a)*sqrt(e*x))/(b^3*x^3 + a*b^2)]
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.11 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {{\left (\frac {B e^{4} x^{3}}{b} + \frac {3 \, B a b^{3} e^{4} - 2 \, A b^{4} e^{4}}{b^{5}}\right )} \sqrt {e x} e x}{3 \, \sqrt {b e^{4} x^{3} + a e^{4}}} + \frac {{\left (3 \, B a b^{3} e^{4} - 2 \, A b^{4} e^{4}\right )} e^{2} \log \left ({\left | -\sqrt {b e} \sqrt {e x} e x + \sqrt {b e^{4} x^{3} + a e^{4}} \right |}\right )}{3 \, \sqrt {b e} b^{5} {\left | e \right |}^{2}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.56 \[ \int \frac {(e x)^{7/2} \left (A+B x^3\right )}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {e}\, e^{3} \left (2 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b x +\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a -\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a \right )}{6 b^{2}} \] Input:

int((e*x)^(7/2)*(B*x^3+A)/(b*x^3+a)^(3/2),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)*a))/(6*b**2)