\(\int x^5 (a+c x^4)^{3/2} \, dx\) [789]

Optimal result

Integrand size = 15, antiderivative size = 95 \[ \int x^5 \left (a+c x^4\right )^{3/2} \, dx=\frac {a^2 x^2 \sqrt {a+c x^4}}{32 c}+\frac {1}{16} a x^6 \sqrt {a+c x^4}+\frac {1}{12} x^6 \left (a+c x^4\right )^{3/2}-\frac {a^3 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{32 c^{3/2}} \]

[Out]

1/12*x^6*(c*x^4+a)^(3/2)-1/32*a^3*arctanh(x^2*c^(1/2)/(c*x^4+a)^(1/2))/c^(3/2)+1/32*a^2*x^2*(c*x^4+a)^(1/2)/c+
1/16*a*x^6*(c*x^4+a)^(1/2)

Rubi [A] (verified)

Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {281, 285, 327, 223, 212} \[ \int x^5 \left (a+c x^4\right )^{3/2} \, dx=-\frac {a^3 \text {arctanh}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{32 c^{3/2}}+\frac {a^2 x^2 \sqrt {a+c x^4}}{32 c}+\frac {1}{12} x^6 \left (a+c x^4\right )^{3/2}+\frac {1}{16} a x^6 \sqrt {a+c x^4} \]

[In]

Int[x^5*(a + c*x^4)^(3/2),x]

[Out]

(a^2*x^2*Sqrt[a + c*x^4])/(32*c) + (a*x^6*Sqrt[a + c*x^4])/16 + (x^6*(a + c*x^4)^(3/2))/12 - (a^3*ArcTanh[(Sqr
t[c]*x^2)/Sqrt[a + c*x^4]])/(32*c^(3/2))

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 223

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 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[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 285

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] + Dist[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]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 327

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] - Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {Subst}\left (\int x^2 \left (a+c x^2\right )^{3/2} \, dx,x,x^2\right ) \\ & = \frac {1}{12} x^6 \left (a+c x^4\right )^{3/2}+\frac {1}{4} a \text {Subst}\left (\int x^2 \sqrt {a+c x^2} \, dx,x,x^2\right ) \\ & = \frac {1}{16} a x^6 \sqrt {a+c x^4}+\frac {1}{12} x^6 \left (a+c x^4\right )^{3/2}+\frac {1}{16} a^2 \text {Subst}\left (\int \frac {x^2}{\sqrt {a+c x^2}} \, dx,x,x^2\right ) \\ & = \frac {a^2 x^2 \sqrt {a+c x^4}}{32 c}+\frac {1}{16} a x^6 \sqrt {a+c x^4}+\frac {1}{12} x^6 \left (a+c x^4\right )^{3/2}-\frac {a^3 \text {Subst}\left (\int \frac {1}{\sqrt {a+c x^2}} \, dx,x,x^2\right )}{32 c} \\ & = \frac {a^2 x^2 \sqrt {a+c x^4}}{32 c}+\frac {1}{16} a x^6 \sqrt {a+c x^4}+\frac {1}{12} x^6 \left (a+c x^4\right )^{3/2}-\frac {a^3 \text {Subst}\left (\int \frac {1}{1-c x^2} \, dx,x,\frac {x^2}{\sqrt {a+c x^4}}\right )}{32 c} \\ & = \frac {a^2 x^2 \sqrt {a+c x^4}}{32 c}+\frac {1}{16} a x^6 \sqrt {a+c x^4}+\frac {1}{12} x^6 \left (a+c x^4\right )^{3/2}-\frac {a^3 \tanh ^{-1}\left (\frac {\sqrt {c} x^2}{\sqrt {a+c x^4}}\right )}{32 c^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.81 \[ \int x^5 \left (a+c x^4\right )^{3/2} \, dx=\frac {x^2 \sqrt {a+c x^4} \left (3 a^2+14 a c x^4+8 c^2 x^8\right )}{96 c}-\frac {a^3 \log \left (\sqrt {c} x^2+\sqrt {a+c x^4}\right )}{32 c^{3/2}} \]

[In]

Integrate[x^5*(a + c*x^4)^(3/2),x]

[Out]

(x^2*Sqrt[a + c*x^4]*(3*a^2 + 14*a*c*x^4 + 8*c^2*x^8))/(96*c) - (a^3*Log[Sqrt[c]*x^2 + Sqrt[a + c*x^4]])/(32*c
^(3/2))

Maple [A] (verified)

Time = 4.53 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.69

method	result	size
risch	\(\frac {x^{2} \left (8 c^{2} x^{8}+14 a \,x^{4} c +3 a^{2}\right ) \sqrt {x^{4} c +a}}{96 c}-\frac {a^{3} \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{32 c^{\frac {3}{2}}}\)	\(66\)
default	\(\frac {c \,x^{10} \sqrt {x^{4} c +a}}{12}+\frac {7 a \,x^{6} \sqrt {x^{4} c +a}}{48}+\frac {a^{2} x^{2} \sqrt {x^{4} c +a}}{32 c}-\frac {a^{3} \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{32 c^{\frac {3}{2}}}\)	\(78\)
elliptic	\(\frac {c \,x^{10} \sqrt {x^{4} c +a}}{12}+\frac {7 a \,x^{6} \sqrt {x^{4} c +a}}{48}+\frac {a^{2} x^{2} \sqrt {x^{4} c +a}}{32 c}-\frac {a^{3} \ln \left (x^{2} \sqrt {c}+\sqrt {x^{4} c +a}\right )}{32 c^{\frac {3}{2}}}\)	\(78\)
pseudoelliptic	\(\frac {8 c^{\frac {5}{2}} \sqrt {x^{4} c +a}\, x^{10}+14 a \,c^{\frac {3}{2}} x^{6} \sqrt {x^{4} c +a}+3 a^{2} x^{2} \sqrt {c}\, \sqrt {x^{4} c +a}-3 \,\operatorname {arctanh}\left (\frac {\sqrt {x^{4} c +a}}{x^{2} \sqrt {c}}\right ) a^{3}}{96 c^{\frac {3}{2}}}\)	\(84\)

[In]

int(x^5*(c*x^4+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/96*x^2*(8*c^2*x^8+14*a*c*x^4+3*a^2)*(c*x^4+a)^(1/2)/c-1/32/c^(3/2)*a^3*ln(x^2*c^(1/2)+(c*x^4+a)^(1/2))

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.61 \[ \int x^5 \left (a+c x^4\right )^{3/2} \, dx=\left [\frac {3 \, a^{3} \sqrt {c} \log \left (-2 \, c x^{4} + 2 \, \sqrt {c x^{4} + a} \sqrt {c} x^{2} - a\right ) + 2 \, {\left (8 \, c^{3} x^{10} + 14 \, a c^{2} x^{6} + 3 \, a^{2} c x^{2}\right )} \sqrt {c x^{4} + a}}{192 \, c^{2}}, \frac {3 \, a^{3} \sqrt {-c} \arctan \left (\frac {\sqrt {-c} x^{2}}{\sqrt {c x^{4} + a}}\right ) + {\left (8 \, c^{3} x^{10} + 14 \, a c^{2} x^{6} + 3 \, a^{2} c x^{2}\right )} \sqrt {c x^{4} + a}}{96 \, c^{2}}\right ] \]

[In]

integrate(x^5*(c*x^4+a)^(3/2),x, algorithm="fricas")

[Out]

[1/192*(3*a^3*sqrt(c)*log(-2*c*x^4 + 2*sqrt(c*x^4 + a)*sqrt(c)*x^2 - a) + 2*(8*c^3*x^10 + 14*a*c^2*x^6 + 3*a^2
*c*x^2)*sqrt(c*x^4 + a))/c^2, 1/96*(3*a^3*sqrt(-c)*arctan(sqrt(-c)*x^2/sqrt(c*x^4 + a)) + (8*c^3*x^10 + 14*a*c
^2*x^6 + 3*a^2*c*x^2)*sqrt(c*x^4 + a))/c^2]

Sympy [A] (verification not implemented)

Time = 3.37 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.28 \[ \int x^5 \left (a+c x^4\right )^{3/2} \, dx=\frac {a^{\frac {5}{2}} x^{2}}{32 c \sqrt {1 + \frac {c x^{4}}{a}}} + \frac {17 a^{\frac {3}{2}} x^{6}}{96 \sqrt {1 + \frac {c x^{4}}{a}}} + \frac {11 \sqrt {a} c x^{10}}{48 \sqrt {1 + \frac {c x^{4}}{a}}} - \frac {a^{3} \operatorname {asinh}{\left (\frac {\sqrt {c} x^{2}}{\sqrt {a}} \right )}}{32 c^{\frac {3}{2}}} + \frac {c^{2} x^{14}}{12 \sqrt {a} \sqrt {1 + \frac {c x^{4}}{a}}} \]

[In]

integrate(x**5*(c*x**4+a)**(3/2),x)

[Out]

a**(5/2)*x**2/(32*c*sqrt(1 + c*x**4/a)) + 17*a**(3/2)*x**6/(96*sqrt(1 + c*x**4/a)) + 11*sqrt(a)*c*x**10/(48*sq
rt(1 + c*x**4/a)) - a**3*asinh(sqrt(c)*x**2/sqrt(a))/(32*c**(3/2)) + c**2*x**14/(12*sqrt(a)*sqrt(1 + c*x**4/a)
)

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 160 vs. \(2 (75) = 150\).

Time = 0.28 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.68 \[ \int x^5 \left (a+c x^4\right )^{3/2} \, dx=\frac {a^{3} \log \left (-\frac {\sqrt {c} - \frac {\sqrt {c x^{4} + a}}{x^{2}}}{\sqrt {c} + \frac {\sqrt {c x^{4} + a}}{x^{2}}}\right )}{64 \, c^{\frac {3}{2}}} + \frac {\frac {3 \, \sqrt {c x^{4} + a} a^{3} c^{2}}{x^{2}} - \frac {8 \, {\left (c x^{4} + a\right )}^{\frac {3}{2}} a^{3} c}{x^{6}} - \frac {3 \, {\left (c x^{4} + a\right )}^{\frac {5}{2}} a^{3}}{x^{10}}}{96 \, {\left (c^{4} - \frac {3 \, {\left (c x^{4} + a\right )} c^{3}}{x^{4}} + \frac {3 \, {\left (c x^{4} + a\right )}^{2} c^{2}}{x^{8}} - \frac {{\left (c x^{4} + a\right )}^{3} c}{x^{12}}\right )}} \]

[In]

integrate(x^5*(c*x^4+a)^(3/2),x, algorithm="maxima")

[Out]

1/64*a^3*log(-(sqrt(c) - sqrt(c*x^4 + a)/x^2)/(sqrt(c) + sqrt(c*x^4 + a)/x^2))/c^(3/2) + 1/96*(3*sqrt(c*x^4 +
a)*a^3*c^2/x^2 - 8*(c*x^4 + a)^(3/2)*a^3*c/x^6 - 3*(c*x^4 + a)^(5/2)*a^3/x^10)/(c^4 - 3*(c*x^4 + a)*c^3/x^4 +
3*(c*x^4 + a)^2*c^2/x^8 - (c*x^4 + a)^3*c/x^12)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.33 \[ \int x^5 \left (a+c x^4\right )^{3/2} \, dx=\frac {1}{16} \, {\left (\sqrt {c x^{4} + a} {\left (2 \, x^{4} + \frac {a}{c}\right )} x^{2} + \frac {a^{2} \log \left ({\left | -\sqrt {c} x^{2} + \sqrt {c x^{4} + a} \right |}\right )}{c^{\frac {3}{2}}}\right )} a + \frac {1}{96} \, {\left ({\left (2 \, {\left (4 \, x^{4} + \frac {a}{c}\right )} x^{4} - \frac {3 \, a^{2}}{c^{2}}\right )} \sqrt {c x^{4} + a} x^{2} - \frac {3 \, a^{3} \log \left ({\left | -\sqrt {c} x^{2} + \sqrt {c x^{4} + a} \right |}\right )}{c^{\frac {5}{2}}}\right )} c \]

[In]

integrate(x^5*(c*x^4+a)^(3/2),x, algorithm="giac")

[Out]

1/16*(sqrt(c*x^4 + a)*(2*x^4 + a/c)*x^2 + a^2*log(abs(-sqrt(c)*x^2 + sqrt(c*x^4 + a)))/c^(3/2))*a + 1/96*((2*(
4*x^4 + a/c)*x^4 - 3*a^2/c^2)*sqrt(c*x^4 + a)*x^2 - 3*a^3*log(abs(-sqrt(c)*x^2 + sqrt(c*x^4 + a)))/c^(5/2))*c

Mupad [F(-1)]

Timed out. \[ \int x^5 \left (a+c x^4\right )^{3/2} \, dx=\int x^5\,{\left (c\,x^4+a\right )}^{3/2} \,d x \]

[In]

int(x^5*(a + c*x^4)^(3/2),x)

[Out]

int(x^5*(a + c*x^4)^(3/2), x)