\(\int \frac {(c x)^{19/2}}{(a+b x^3)^{3/2}} \, dx\) [336]

Optimal result

Integrand size = 19, antiderivative size = 133 \[ \int \frac {(c x)^{19/2}}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {2 c^2 (c x)^{15/2}}{3 b \sqrt {a+b x^3}}-\frac {5 a c^8 (c x)^{3/2} \sqrt {a+b x^3}}{4 b^3}+\frac {5 c^5 (c x)^{9/2} \sqrt {a+b x^3}}{6 b^2}+\frac {5 a^2 c^{19/2} \text {arctanh}\left (\frac {\sqrt {b} (c x)^{3/2}}{c^{3/2} \sqrt {a+b x^3}}\right )}{4 b^{7/2}} \] Output:

-2/3*c^2*(c*x)^(15/2)/b/(b*x^3+a)^(1/2)-5/4*a*c^8*(c*x)^(3/2)*(b*x^3+a)^(1 
/2)/b^3+5/6*c^5*(c*x)^(9/2)*(b*x^3+a)^(1/2)/b^2+5/4*a^2*c^(19/2)*arctanh(b 
^(1/2)*(c*x)^(3/2)/c^(3/2)/(b*x^3+a)^(1/2))/b^(7/2)
 

Mathematica [A] (verified)

Time = 1.69 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.73 \[ \int \frac {(c x)^{19/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {c^9 \sqrt {c x} \left (\frac {\sqrt {b} x^{3/2} \left (-15 a^2-5 a b x^3+2 b^2 x^6\right )}{\sqrt {a+b x^3}}+15 a^2 \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )\right )}{12 b^{7/2} \sqrt {x}} \] Input:

Integrate[(c*x)^(19/2)/(a + b*x^3)^(3/2),x]
 

Output:

(c^9*Sqrt[c*x]*((Sqrt[b]*x^(3/2)*(-15*a^2 - 5*a*b*x^3 + 2*b^2*x^6))/Sqrt[a 
 + b*x^3] + 15*a^2*Log[Sqrt[b]*x^(3/2) + Sqrt[a + b*x^3]]))/(12*b^(7/2)*Sq 
rt[x])
 

Rubi [A] (warning: unable to verify)

Time = 0.51 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.14, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {817, 843, 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.

\(\displaystyle \int \frac {(c x)^{19/2}}{\left (a+b x^3\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 817

\(\displaystyle \frac {5 c^3 \int \frac {(c x)^{13/2}}{\sqrt {b x^3+a}}dx}{b}-\frac {2 c^2 (c x)^{15/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {5 c^3 \left (\frac {c^2 (c x)^{9/2} \sqrt {a+b x^3}}{6 b}-\frac {3 a c^3 \int \frac {(c x)^{7/2}}{\sqrt {b x^3+a}}dx}{4 b}\right )}{b}-\frac {2 c^2 (c x)^{15/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {5 c^3 \left (\frac {c^2 (c x)^{9/2} \sqrt {a+b x^3}}{6 b}-\frac {3 a c^3 \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^3 \int \frac {\sqrt {c x}}{\sqrt {b x^3+a}}dx}{2 b}\right )}{4 b}\right )}{b}-\frac {2 c^2 (c x)^{15/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 851

\(\displaystyle \frac {5 c^3 \left (\frac {c^2 (c x)^{9/2} \sqrt {a+b x^3}}{6 b}-\frac {3 a c^3 \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^2 \int \frac {c x}{\sqrt {b x^3+a}}d\sqrt {c x}}{b}\right )}{4 b}\right )}{b}-\frac {2 c^2 (c x)^{15/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 807

\(\displaystyle \frac {5 c^3 \left (\frac {c^2 (c x)^{9/2} \sqrt {a+b x^3}}{6 b}-\frac {3 a c^3 \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^2 \int \frac {1}{\sqrt {a+\frac {b x}{c^2}}}d(c x)^{3/2}}{3 b}\right )}{4 b}\right )}{b}-\frac {2 c^2 (c x)^{15/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {5 c^3 \left (\frac {c^2 (c x)^{9/2} \sqrt {a+b x^3}}{6 b}-\frac {3 a c^3 \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^2 \int \frac {1}{1-\frac {b x}{c^2}}d\frac {(c x)^{3/2}}{\sqrt {a+\frac {b x}{c^2}}}}{3 b}\right )}{4 b}\right )}{b}-\frac {2 c^2 (c x)^{15/2}}{3 b \sqrt {a+b x^3}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {5 c^3 \left (\frac {c^2 (c x)^{9/2} \sqrt {a+b x^3}}{6 b}-\frac {3 a c^3 \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (c x)^{3/2}}{c^{3/2} \sqrt {a+\frac {b x}{c^2}}}\right )}{3 b^{3/2}}\right )}{4 b}\right )}{b}-\frac {2 c^2 (c x)^{15/2}}{3 b \sqrt {a+b x^3}}\)

Input:

Int[(c*x)^(19/2)/(a + b*x^3)^(3/2),x]
 

Output:

(-2*c^2*(c*x)^(15/2))/(3*b*Sqrt[a + b*x^3]) + (5*c^3*((c^2*(c*x)^(9/2)*Sqr 
t[a + b*x^3])/(6*b) - (3*a*c^3*((c^2*(c*x)^(3/2)*Sqrt[a + b*x^3])/(3*b) - 
(a*c^(7/2)*ArcTanh[(Sqrt[b]*(c*x)^(3/2))/(c^(3/2)*Sqrt[a + (b*x)/c^2])])/( 
3*b^(3/2))))/(4*b)))/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 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]
 

Maple [A] (verified)

Time = 0.87 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {c^{9} \sqrt {c x}\, \left (2 \sqrt {b c}\, b^{2} x^{8}-5 \sqrt {b c}\, a b \,x^{5}-15 a^{2} x^{2} \sqrt {b c}+15 \,\operatorname {arctanh}\left (\frac {\sqrt {c x \left (b \,x^{3}+a \right )}}{x^{2} \sqrt {b c}}\right ) a^{2} \sqrt {c x \left (b \,x^{3}+a \right )}\right )}{12 x \sqrt {b \,x^{3}+a}\, b^{3} \sqrt {b c}}\)	\(109\)
risch	\(-\frac {x^{2} \left (-2 b \,x^{3}+7 a \right ) \sqrt {b \,x^{3}+a}\, c^{10}}{12 b^{3} \sqrt {c x}}+\frac {a^{2} \left (\frac {10 \,\operatorname {arctanh}\left (\frac {\sqrt {c x \left (b \,x^{3}+a \right )}}{x^{2} \sqrt {b c}}\right )}{\sqrt {b c}}-\frac {16 x^{2}}{3 \sqrt {\left (x^{3}+\frac {a}{b}\right ) b c x}}\right ) c^{10} \sqrt {c x \left (b \,x^{3}+a \right )}}{8 b^{3} \sqrt {c x}\, \sqrt {b \,x^{3}+a}}\)	\(124\)
elliptic	\(\text {Expression too large to display}\)	\(1098\)

Input:

int((c*x)^(19/2)/(b*x^3+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/12*c^9/x*(c*x)^(1/2)/(b*x^3+a)^(1/2)*(2*(b*c)^(1/2)*b^2*x^8-5*(b*c)^(1/2 
)*a*b*x^5-15*a^2*x^2*(b*c)^(1/2)+15*arctanh((c*x*(b*x^3+a))^(1/2)/x^2/(b*c 
)^(1/2))*a^2*(c*x*(b*x^3+a))^(1/2))/b^3/(b*c)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.17 \[ \int \frac {(c x)^{19/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\left [\frac {15 \, {\left (a^{2} b c^{9} x^{3} + a^{3} c^{9}\right )} \sqrt {\frac {c}{b}} \log \left (-8 \, b^{2} c x^{6} - 8 \, a b c x^{3} - a^{2} c - 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {c x} \sqrt {\frac {c}{b}}\right ) + 4 \, {\left (2 \, b^{2} c^{9} x^{7} - 5 \, a b c^{9} x^{4} - 15 \, a^{2} c^{9} x\right )} \sqrt {b x^{3} + a} \sqrt {c x}}{48 \, {\left (b^{4} x^{3} + a b^{3}\right )}}, -\frac {15 \, {\left (a^{2} b c^{9} x^{3} + a^{3} c^{9}\right )} \sqrt {-\frac {c}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {c x} b x \sqrt {-\frac {c}{b}}}{2 \, b c x^{3} + a c}\right ) - 2 \, {\left (2 \, b^{2} c^{9} x^{7} - 5 \, a b c^{9} x^{4} - 15 \, a^{2} c^{9} x\right )} \sqrt {b x^{3} + a} \sqrt {c x}}{24 \, {\left (b^{4} x^{3} + a b^{3}\right )}}\right ] \] Input:

integrate((c*x)^(19/2)/(b*x^3+a)^(3/2),x, algorithm="fricas")
 

Output:

[1/48*(15*(a^2*b*c^9*x^3 + a^3*c^9)*sqrt(c/b)*log(-8*b^2*c*x^6 - 8*a*b*c*x 
^3 - a^2*c - 4*(2*b^2*x^4 + a*b*x)*sqrt(b*x^3 + a)*sqrt(c*x)*sqrt(c/b)) + 
4*(2*b^2*c^9*x^7 - 5*a*b*c^9*x^4 - 15*a^2*c^9*x)*sqrt(b*x^3 + a)*sqrt(c*x) 
)/(b^4*x^3 + a*b^3), -1/24*(15*(a^2*b*c^9*x^3 + a^3*c^9)*sqrt(-c/b)*arctan 
(2*sqrt(b*x^3 + a)*sqrt(c*x)*b*x*sqrt(-c/b)/(2*b*c*x^3 + a*c)) - 2*(2*b^2* 
c^9*x^7 - 5*a*b*c^9*x^4 - 15*a^2*c^9*x)*sqrt(b*x^3 + a)*sqrt(c*x))/(b^4*x^ 
3 + a*b^3)]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(c x)^{19/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate((c*x)**(19/2)/(b*x**3+a)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(c x)^{19/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\int { \frac {\left (c x\right )^{\frac {19}{2}}}{{\left (b x^{3} + a\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate((c*x)^(19/2)/(b*x^3+a)^(3/2),x, algorithm="maxima")
 

Output:

integrate((c*x)^(19/2)/(b*x^3 + a)^(3/2), x)
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 119, normalized size of antiderivative = 0.89 \[ \int \frac {(c x)^{19/2}}{\left (a+b x^3\right )^{3/2}} \, dx=-\frac {5 \, a^{2} c^{12} \log \left ({\left | -\sqrt {b c} \sqrt {c x} c x + \sqrt {b c^{4} x^{3} + a c^{4}} \right |}\right )}{4 \, \sqrt {b c} b^{3} {\left | c \right |}^{2}} - \frac {{\left (\frac {15 \, a^{2} c^{11}}{b^{3}} - {\left (\frac {2 \, c^{8} x^{3}}{b} - \frac {5 \, a c^{8}}{b^{2}}\right )} c^{3} x^{3}\right )} \sqrt {c x} x}{12 \, \sqrt {b c^{4} x^{3} + a c^{4}}} \] Input:

integrate((c*x)^(19/2)/(b*x^3+a)^(3/2),x, algorithm="giac")
 

Output:

-5/4*a^2*c^12*log(abs(-sqrt(b*c)*sqrt(c*x)*c*x + sqrt(b*c^4*x^3 + a*c^4))) 
/(sqrt(b*c)*b^3*abs(c)^2) - 1/12*(15*a^2*c^11/b^3 - (2*c^8*x^3/b - 5*a*c^8 
/b^2)*c^3*x^3)*sqrt(c*x)*x/sqrt(b*c^4*x^3 + a*c^4)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c x)^{19/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\int \frac {{\left (c\,x\right )}^{19/2}}{{\left (b\,x^3+a\right )}^{3/2}} \,d x \] Input:

int((c*x)^(19/2)/(a + b*x^3)^(3/2),x)
 

Output:

int((c*x)^(19/2)/(a + b*x^3)^(3/2), x)
 

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.32 \[ \int \frac {(c x)^{19/2}}{\left (a+b x^3\right )^{3/2}} \, dx=\frac {\sqrt {c}\, c^{9} \left (-30 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a^{2} b x -10 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a \,b^{2} x^{4}+4 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b^{3} x^{7}-15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a^{3}-15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a^{2} b \,x^{3}+15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a^{3}+15 \sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a^{2} b \,x^{3}\right )}{24 b^{4} \left (b \,x^{3}+a \right )} \] Input:

int((c*x)^(19/2)/(b*x^3+a)^(3/2),x)
 

Output:

(sqrt(c)*c**9*( - 30*sqrt(x)*sqrt(a + b*x**3)*a**2*b*x - 10*sqrt(x)*sqrt(a 
 + b*x**3)*a*b**2*x**4 + 4*sqrt(x)*sqrt(a + b*x**3)*b**3*x**7 - 15*sqrt(b) 
*log(sqrt(a + b*x**3) - sqrt(x)*sqrt(b)*x)*a**3 - 15*sqrt(b)*log(sqrt(a + 
b*x**3) - sqrt(x)*sqrt(b)*x)*a**2*b*x**3 + 15*sqrt(b)*log(sqrt(a + b*x**3) 
 + sqrt(x)*sqrt(b)*x)*a**3 + 15*sqrt(b)*log(sqrt(a + b*x**3) + sqrt(x)*sqr 
t(b)*x)*a**2*b*x**3))/(24*b**4*(a + b*x**3))