Integrand size = 19, antiderivative size = 102 \[ \int (c x)^{7/2} \sqrt {a+b x^3} \, dx=\frac {a c^2 (c x)^{3/2} \sqrt {a+b x^3}}{12 b}+\frac {(c x)^{9/2} \sqrt {a+b x^3}}{6 c}-\frac {a^2 c^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (c x)^{3/2}}{c^{3/2} \sqrt {a+b x^3}}\right )}{12 b^{3/2}} \] Output:
1/12*a*c^2*(c*x)^(3/2)*(b*x^3+a)^(1/2)/b+1/6*(c*x)^(9/2)*(b*x^3+a)^(1/2)/c -1/12*a^2*c^(7/2)*arctanh(b^(1/2)*(c*x)^(3/2)/c^(3/2)/(b*x^3+a)^(1/2))/b^( 3/2)
Time = 1.73 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.83 \[ \int (c x)^{7/2} \sqrt {a+b x^3} \, dx=\frac {(c x)^{7/2} \sqrt {a+b x^3} \left (a+2 b x^3\right )}{12 b x^2}-\frac {a^2 (c x)^{7/2} \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )}{12 b^{3/2} x^{7/2}} \] Input:
Integrate[(c*x)^(7/2)*Sqrt[a + b*x^3],x]
Output:
((c*x)^(7/2)*Sqrt[a + b*x^3]*(a + 2*b*x^3))/(12*b*x^2) - (a^2*(c*x)^(7/2)* Log[Sqrt[b]*x^(3/2) + Sqrt[a + b*x^3]])/(12*b^(3/2)*x^(7/2))
Time = 0.44 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {811, 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.
\(\displaystyle \int (c x)^{7/2} \sqrt {a+b x^3} \, dx\) |
\(\Big \downarrow \) 811 |
\(\displaystyle \frac {1}{4} a \int \frac {(c x)^{7/2}}{\sqrt {b x^3+a}}dx+\frac {(c x)^{9/2} \sqrt {a+b x^3}}{6 c}\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{4} a \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^3 \int \frac {\sqrt {c x}}{\sqrt {b x^3+a}}dx}{2 b}\right )+\frac {(c x)^{9/2} \sqrt {a+b x^3}}{6 c}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {1}{4} a \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^2 \int \frac {c x}{\sqrt {b x^3+a}}d\sqrt {c x}}{b}\right )+\frac {(c x)^{9/2} \sqrt {a+b x^3}}{6 c}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{4} a \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^2 \int \frac {1}{\sqrt {a+\frac {b x}{c^2}}}d(c x)^{3/2}}{3 b}\right )+\frac {(c x)^{9/2} \sqrt {a+b x^3}}{6 c}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{4} a \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^2 \int \frac {1}{1-\frac {b x}{c^2}}d\frac {(c x)^{3/2}}{\sqrt {a+\frac {b x}{c^2}}}}{3 b}\right )+\frac {(c x)^{9/2} \sqrt {a+b x^3}}{6 c}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} a \left (\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (c x)^{3/2}}{c^{3/2} \sqrt {a+\frac {b x}{c^2}}}\right )}{3 b^{3/2}}\right )+\frac {(c x)^{9/2} \sqrt {a+b x^3}}{6 c}\) |
Input:
Int[(c*x)^(7/2)*Sqrt[a + b*x^3],x]
Output:
((c*x)^(9/2)*Sqrt[a + b*x^3])/(6*c) + (a*((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
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])
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]
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]
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]
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]
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]
Time = 3.55 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97
method | result | size |
risch | \(\frac {x^{2} \left (2 b \,x^{3}+a \right ) \sqrt {b \,x^{3}+a}\, c^{4}}{12 b \sqrt {c x}}-\frac {a^{2} \operatorname {arctanh}\left (\frac {\sqrt {c x \left (b \,x^{3}+a \right )}}{x^{2} \sqrt {b c}}\right ) c^{4} \sqrt {c x \left (b \,x^{3}+a \right )}}{12 b \sqrt {b c}\, \sqrt {c x}\, \sqrt {b \,x^{3}+a}}\) | \(99\) |
default | \(-\frac {c^{3} \sqrt {c x}\, \sqrt {b \,x^{3}+a}\, \left (-2 \sqrt {c x \left (b \,x^{3}+a \right )}\, \sqrt {b c}\, b \,x^{4}+\operatorname {arctanh}\left (\frac {\sqrt {c x \left (b \,x^{3}+a \right )}}{x^{2} \sqrt {b c}}\right ) c \,a^{2}-a x \sqrt {c x \left (b \,x^{3}+a \right )}\, \sqrt {b c}\right )}{12 \sqrt {c x \left (b \,x^{3}+a \right )}\, \sqrt {b c}\, b}\) | \(112\) |
elliptic | \(\text {Expression too large to display}\) | \(1063\) |
Input:
int((c*x)^(7/2)*(b*x^3+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
1/12*x^2*(2*b*x^3+a)*(b*x^3+a)^(1/2)/b*c^4/(c*x)^(1/2)-1/12/b*a^2/(b*c)^(1 /2)*arctanh((c*x*(b*x^3+a))^(1/2)/x^2/(b*c)^(1/2))*c^4*(c*x*(b*x^3+a))^(1/ 2)/(c*x)^(1/2)/(b*x^3+a)^(1/2)
Time = 0.17 (sec) , antiderivative size = 207, normalized size of antiderivative = 2.03 \[ \int (c x)^{7/2} \sqrt {a+b x^3} \, dx=\left [\frac {a^{2} c^{3} \sqrt {\frac {c}{b}} \log \left (-8 \, b^{2} c x^{6} - 8 \, a b c x^{3} - a^{2} c + 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {c x} \sqrt {\frac {c}{b}}\right ) + 4 \, {\left (2 \, b c^{3} x^{4} + a c^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {c x}}{48 \, b}, \frac {a^{2} c^{3} \sqrt {-\frac {c}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {c x} b x \sqrt {-\frac {c}{b}}}{2 \, b c x^{3} + a c}\right ) + 2 \, {\left (2 \, b c^{3} x^{4} + a c^{3} x\right )} \sqrt {b x^{3} + a} \sqrt {c x}}{24 \, b}\right ] \] Input:
integrate((c*x)^(7/2)*(b*x^3+a)^(1/2),x, algorithm="fricas")
Output:
[1/48*(a^2*c^3*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*c^3*x^4 + a*c^ 3*x)*sqrt(b*x^3 + a)*sqrt(c*x))/b, 1/24*(a^2*c^3*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*c^3*x^4 + a*c^3*x)*sqrt(b*x^3 + a)*sqrt(c*x))/b]
Time = 17.02 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.19 \[ \int (c x)^{7/2} \sqrt {a+b x^3} \, dx=\frac {a^{\frac {3}{2}} c^{\frac {7}{2}} x^{\frac {3}{2}}}{12 b \sqrt {1 + \frac {b x^{3}}{a}}} + \frac {\sqrt {a} c^{\frac {7}{2}} x^{\frac {9}{2}}}{4 \sqrt {1 + \frac {b x^{3}}{a}}} - \frac {a^{2} c^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{12 b^{\frac {3}{2}}} + \frac {b c^{\frac {7}{2}} x^{\frac {15}{2}}}{6 \sqrt {a} \sqrt {1 + \frac {b x^{3}}{a}}} \] Input:
integrate((c*x)**(7/2)*(b*x**3+a)**(1/2),x)
Output:
a**(3/2)*c**(7/2)*x**(3/2)/(12*b*sqrt(1 + b*x**3/a)) + sqrt(a)*c**(7/2)*x* *(9/2)/(4*sqrt(1 + b*x**3/a)) - a**2*c**(7/2)*asinh(sqrt(b)*x**(3/2)/sqrt( a))/(12*b**(3/2)) + b*c**(7/2)*x**(15/2)/(6*sqrt(a)*sqrt(1 + b*x**3/a))
\[ \int (c x)^{7/2} \sqrt {a+b x^3} \, dx=\int { \sqrt {b x^{3} + a} \left (c x\right )^{\frac {7}{2}} \,d x } \] Input:
integrate((c*x)^(7/2)*(b*x^3+a)^(1/2),x, algorithm="maxima")
Output:
integrate(sqrt(b*x^3 + a)*(c*x)^(7/2), x)
Time = 0.16 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.96 \[ \int (c x)^{7/2} \sqrt {a+b x^3} \, dx=\frac {a^{2} c^{4} \log \left ({\left | -\sqrt {b c} \sqrt {c x} c x + \sqrt {b c^{4} x^{3} + a c^{4}} \right |}\right )}{12 \, \sqrt {b c} b} + \frac {\sqrt {b c^{4} x^{3} + a c^{4}} {\left (2 \, c^{3} x^{3} + \frac {a c^{3}}{b}\right )} \sqrt {c x} x {\left | c \right |}^{2}}{12 \, c^{4}} \] Input:
integrate((c*x)^(7/2)*(b*x^3+a)^(1/2),x, algorithm="giac")
Output:
1/12*a^2*c^4*log(abs(-sqrt(b*c)*sqrt(c*x)*c*x + sqrt(b*c^4*x^3 + a*c^4)))/ (sqrt(b*c)*b) + 1/12*sqrt(b*c^4*x^3 + a*c^4)*(2*c^3*x^3 + a*c^3/b)*sqrt(c* x)*x*abs(c)^2/c^4
Timed out. \[ \int (c x)^{7/2} \sqrt {a+b x^3} \, dx=\int {\left (c\,x\right )}^{7/2}\,\sqrt {b\,x^3+a} \,d x \] Input:
int((c*x)^(7/2)*(a + b*x^3)^(1/2),x)
Output:
int((c*x)^(7/2)*(a + b*x^3)^(1/2), x)
Time = 0.22 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88 \[ \int (c x)^{7/2} \sqrt {a+b x^3} \, dx=\frac {\sqrt {c}\, c^{3} \left (2 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, a b x +4 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b^{2} x^{4}+\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a^{2}-\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a^{2}\right )}{24 b^{2}} \] Input:
int((c*x)^(7/2)*(b*x^3+a)^(1/2),x)
Output:
(sqrt(c)*c**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)