Integrand size = 19, antiderivative size = 74 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^3}} \, dx=\frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (c x)^{3/2}}{c^{3/2} \sqrt {a+b x^3}}\right )}{3 b^{3/2}} \] Output:
1/3*c^2*(c*x)^(3/2)*(b*x^3+a)^(1/2)/b-1/3*a*c^(7/2)*arctanh(b^(1/2)*(c*x)^ (3/2)/c^(3/2)/(b*x^3+a)^(1/2))/b^(3/2)
Time = 0.21 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^3}} \, dx=\frac {(c x)^{7/2} \sqrt {a+b x^3}}{3 b x^2}-\frac {a (c x)^{7/2} \log \left (\sqrt {b} x^{3/2}+\sqrt {a+b x^3}\right )}{3 b^{3/2} x^{7/2}} \] Input:
Integrate[(c*x)^(7/2)/Sqrt[a + b*x^3],x]
Output:
((c*x)^(7/2)*Sqrt[a + b*x^3])/(3*b*x^2) - (a*(c*x)^(7/2)*Log[Sqrt[b]*x^(3/ 2) + Sqrt[a + b*x^3]])/(3*b^(3/2)*x^(7/2))
Time = 0.37 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {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.
\(\displaystyle \int \frac {(c x)^{7/2}}{\sqrt {a+b x^3}} \, dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^3 \int \frac {\sqrt {c x}}{\sqrt {b x^3+a}}dx}{2 b}\) |
\(\Big \downarrow \) 851 |
\(\displaystyle \frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^2 \int \frac {c x}{\sqrt {b x^3+a}}d\sqrt {c x}}{b}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^2 \int \frac {1}{\sqrt {a+\frac {b x}{c^2}}}d(c x)^{3/2}}{3 b}\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^2 \int \frac {1}{1-\frac {b x}{c^2}}d\frac {(c x)^{3/2}}{\sqrt {a+\frac {b x}{c^2}}}}{3 b}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {c^2 (c x)^{3/2} \sqrt {a+b x^3}}{3 b}-\frac {a c^{7/2} \text {arctanh}\left (\frac {\sqrt {b} (c x)^{3/2}}{c^{3/2} \sqrt {a+\frac {b x}{c^2}}}\right )}{3 b^{3/2}}\) |
Input:
Int[(c*x)^(7/2)/Sqrt[a + b*x^3],x]
Output:
(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))
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^(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 = 0.65 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16
method | result | size |
default | \(-\frac {c^{3} \sqrt {c x}\, \sqrt {b \,x^{3}+a}\, \left (\operatorname {arctanh}\left (\frac {\sqrt {c x \left (b \,x^{3}+a \right )}}{x^{2} \sqrt {b c}}\right ) a c -\sqrt {c x \left (b \,x^{3}+a \right )}\, x \sqrt {b c}\right )}{3 \sqrt {c x \left (b \,x^{3}+a \right )}\, b \sqrt {b c}}\) | \(86\) |
risch | \(\frac {x^{2} \sqrt {b \,x^{3}+a}\, c^{4}}{3 b \sqrt {c x}}-\frac {a \,\operatorname {arctanh}\left (\frac {\sqrt {c x \left (b \,x^{3}+a \right )}}{x^{2} \sqrt {b c}}\right ) c^{4} \sqrt {c x \left (b \,x^{3}+a \right )}}{3 b \sqrt {b c}\, \sqrt {c x}\, \sqrt {b \,x^{3}+a}}\) | \(89\) |
elliptic | \(\text {Expression too large to display}\) | \(1039\) |
Input:
int((c*x)^(7/2)/(b*x^3+a)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/3*c^3*(c*x)^(1/2)*(b*x^3+a)^(1/2)*(arctanh((c*x*(b*x^3+a))^(1/2)/x^2/(b *c)^(1/2))*a*c-(c*x*(b*x^3+a))^(1/2)*x*(b*c)^(1/2))/(c*x*(b*x^3+a))^(1/2)/ b/(b*c)^(1/2)
Time = 0.18 (sec) , antiderivative size = 179, normalized size of antiderivative = 2.42 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^3}} \, dx=\left [\frac {a c^{3} \sqrt {\frac {c}{b}} \log \left (-8 \, b^{2} c x^{6} - 8 \, a b c x^{3} - a^{2} c + 4 \, {\left (2 \, b^{2} x^{4} + a b x\right )} \sqrt {b x^{3} + a} \sqrt {c x} \sqrt {\frac {c}{b}}\right ) + 4 \, \sqrt {b x^{3} + a} \sqrt {c x} c^{3} x}{12 \, b}, \frac {a c^{3} \sqrt {-\frac {c}{b}} \arctan \left (\frac {2 \, \sqrt {b x^{3} + a} \sqrt {c x} b x \sqrt {-\frac {c}{b}}}{2 \, b c x^{3} + a c}\right ) + 2 \, \sqrt {b x^{3} + a} \sqrt {c x} c^{3} x}{6 \, b}\right ] \] Input:
integrate((c*x)^(7/2)/(b*x^3+a)^(1/2),x, algorithm="fricas")
Output:
[1/12*(a*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*sqrt(b*x^3 + a)*sqrt( c*x)*c^3*x)/b, 1/6*(a*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*sqrt(b*x^3 + a)*sqrt(c*x)*c^3*x)/b]
Time = 13.01 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.81 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^3}} \, dx=\frac {\sqrt {a} c^{\frac {7}{2}} x^{\frac {3}{2}} \sqrt {1 + \frac {b x^{3}}{a}}}{3 b} - \frac {a c^{\frac {7}{2}} \operatorname {asinh}{\left (\frac {\sqrt {b} x^{\frac {3}{2}}}{\sqrt {a}} \right )}}{3 b^{\frac {3}{2}}} \] Input:
integrate((c*x)**(7/2)/(b*x**3+a)**(1/2),x)
Output:
sqrt(a)*c**(7/2)*x**(3/2)*sqrt(1 + b*x**3/a)/(3*b) - a*c**(7/2)*asinh(sqrt (b)*x**(3/2)/sqrt(a))/(3*b**(3/2))
\[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^3}} \, dx=\int { \frac {\left (c x\right )^{\frac {7}{2}}}{\sqrt {b x^{3} + a}} \,d x } \] Input:
integrate((c*x)^(7/2)/(b*x^3+a)^(1/2),x, algorithm="maxima")
Output:
integrate((c*x)^(7/2)/sqrt(b*x^3 + a), x)
Time = 0.16 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.16 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^3}} \, dx=\frac {a c^{8} \log \left ({\left | -\sqrt {b c} \sqrt {c x} c x + \sqrt {b c^{4} x^{3} + a c^{4}} \right |}\right )}{3 \, \sqrt {b c} b {\left | c \right |}^{4}} + \frac {\sqrt {b c^{4} x^{3} + a c^{4}} \sqrt {c x} c^{3} x}{3 \, b {\left | c \right |}^{2}} \] Input:
integrate((c*x)^(7/2)/(b*x^3+a)^(1/2),x, algorithm="giac")
Output:
1/3*a*c^8*log(abs(-sqrt(b*c)*sqrt(c*x)*c*x + sqrt(b*c^4*x^3 + a*c^4)))/(sq rt(b*c)*b*abs(c)^4) + 1/3*sqrt(b*c^4*x^3 + a*c^4)*sqrt(c*x)*c^3*x/(b*abs(c )^2)
Timed out. \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^3}} \, dx=\int \frac {{\left (c\,x\right )}^{7/2}}{\sqrt {b\,x^3+a}} \,d x \] 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 = 0.24 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.91 \[ \int \frac {(c x)^{7/2}}{\sqrt {a+b x^3}} \, dx=\frac {\sqrt {c}\, c^{3} \left (2 \sqrt {x}\, \sqrt {b \,x^{3}+a}\, b x +\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}-\sqrt {x}\, \sqrt {b}\, x \right ) a -\sqrt {b}\, \mathrm {log}\left (\sqrt {b \,x^{3}+a}+\sqrt {x}\, \sqrt {b}\, x \right ) a \right )}{6 b^{2}} \] Input:
int((c*x)^(7/2)/(b*x^3+a)^(1/2),x)
Output:
(sqrt(c)*c**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)