Integrand size = 19, antiderivative size = 457 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx=-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}-\frac {4 d \sqrt [3]{c+d x}}{45 b (b c-a d) (a+b x)^{3/2}}+\frac {28 d^2 \sqrt [3]{c+d x}}{135 b (b c-a d)^2 \sqrt {a+b x}}-\frac {28 \sqrt {2-\sqrt {3}} d^2 \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {(b c-a d)^{2/3}+\sqrt [3]{b} \sqrt [3]{b c-a d} \sqrt [3]{c+d x}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{135 \sqrt [4]{3} b^{4/3} (b c-a d)^2 \sqrt {a+b x} \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}}} \]
-2/5*(d*x+c)^(1/3)/b/(b*x+a)^(5/2)-4/45*d*(d*x+c)^(1/3)/b/(-a*d+b*c)/(b*x+ a)^(3/2)+28/135*d^2*(d*x+c)^(1/3)/b/(-a*d+b*c)^2/(b*x+a)^(1/2)-28/405*d^2* ((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3))*EllipticF((-b^(1/3)*(d*x+c)^(1/3) +(-a*d+b*c)^(1/3)*(1+3^(1/2)))/(-b^(1/3)*(d*x+c)^(1/3)+(-a*d+b*c)^(1/3)*(1 -3^(1/2))),2*I-I*3^(1/2))*(((-a*d+b*c)^(2/3)+b^(1/3)*(-a*d+b*c)^(1/3)*(d*x +c)^(1/3)+b^(2/3)*(d*x+c)^(2/3))/(-b^(1/3)*(d*x+c)^(1/3)+(-a*d+b*c)^(1/3)* (1-3^(1/2)))^2)^(1/2)*(1/2*6^(1/2)-1/2*2^(1/2))*3^(3/4)/b^(4/3)/(-a*d+b*c) ^2/(b*x+a)^(1/2)/(-(-a*d+b*c)^(1/3)*((-a*d+b*c)^(1/3)-b^(1/3)*(d*x+c)^(1/3 ))/(-b^(1/3)*(d*x+c)^(1/3)+(-a*d+b*c)^(1/3)*(1-3^(1/2)))^2)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.02 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.16 \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx=-\frac {2 \sqrt [3]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {1}{3},-\frac {3}{2},\frac {d (a+b x)}{-b c+a d}\right )}{5 b (a+b x)^{5/2} \sqrt [3]{\frac {b (c+d x)}{b c-a d}}} \]
(-2*(c + d*x)^(1/3)*Hypergeometric2F1[-5/2, -1/3, -3/2, (d*(a + b*x))/(-(b *c) + a*d)])/(5*b*(a + b*x)^(5/2)*((b*(c + d*x))/(b*c - a*d))^(1/3))
Time = 0.36 (sec) , antiderivative size = 479, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {57, 61, 61, 73, 760}
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.
\(\displaystyle \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx\) |
\(\Big \downarrow \) 57 |
\(\displaystyle \frac {2 d \int \frac {1}{(a+b x)^{5/2} (c+d x)^{2/3}}dx}{15 b}-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {2 d \left (-\frac {7 d \int \frac {1}{(a+b x)^{3/2} (c+d x)^{2/3}}dx}{9 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{15 b}-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}\) |
\(\Big \downarrow \) 61 |
\(\displaystyle \frac {2 d \left (-\frac {7 d \left (-\frac {d \int \frac {1}{\sqrt {a+b x} (c+d x)^{2/3}}dx}{3 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{15 b}-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 d \left (-\frac {7 d \left (-\frac {\int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [3]{c+d x}}{b c-a d}-\frac {2 \sqrt [3]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{15 b}-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}\) |
\(\Big \downarrow \) 760 |
\(\displaystyle \frac {2 d \left (-\frac {7 d \left (\frac {2 \sqrt {2-\sqrt {3}} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right ) \sqrt {\frac {\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{b c-a d}+(b c-a d)^{2/3}+b^{2/3} (c+d x)^{2/3}}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\left (1+\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}{\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}}\right ),-7+4 \sqrt {3}\right )}{\sqrt [4]{3} \sqrt [3]{b} (b c-a d) \sqrt {-\frac {\sqrt [3]{b c-a d} \left (\sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{\left (\left (1-\sqrt {3}\right ) \sqrt [3]{b c-a d}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )^2}} \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}-\frac {2 \sqrt [3]{c+d x}}{\sqrt {a+b x} (b c-a d)}\right )}{9 (b c-a d)}-\frac {2 \sqrt [3]{c+d x}}{3 (a+b x)^{3/2} (b c-a d)}\right )}{15 b}-\frac {2 \sqrt [3]{c+d x}}{5 b (a+b x)^{5/2}}\) |
(-2*(c + d*x)^(1/3))/(5*b*(a + b*x)^(5/2)) + (2*d*((-2*(c + d*x)^(1/3))/(3 *(b*c - a*d)*(a + b*x)^(3/2)) - (7*d*((-2*(c + d*x)^(1/3))/((b*c - a*d)*Sq rt[a + b*x]) + (2*Sqrt[2 - Sqrt[3]]*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x) ^(1/3))*Sqrt[((b*c - a*d)^(2/3) + b^(1/3)*(b*c - a*d)^(1/3)*(c + d*x)^(1/3 ) + b^(2/3)*(c + d*x)^(2/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2]*EllipticF[ArcSin[((1 + Sqrt[3])*(b*c - a*d)^(1/3) - b^(1 /3)*(c + d*x)^(1/3))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^ (1/3))], -7 + 4*Sqrt[3]])/(3^(1/4)*b^(1/3)*(b*c - a*d)*Sqrt[-(((b*c - a*d) ^(1/3)*((b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3)))/((1 - Sqrt[3])*(b*c - a*d)^(1/3) - b^(1/3)*(c + d*x)^(1/3))^2)]*Sqrt[a - (b*c)/d + (b*(c + d*x ))/d])))/(9*(b*c - a*d))))/(15*b)
3.16.64.3.1 Defintions of rubi rules used
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & & GtQ[n, 0] && LtQ[m, -1] && !(IntegerQ[n] && !IntegerQ[m]) && !(ILeQ[m + n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c , d, m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ (a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(( m + n + 2)/((b*c - a*d)*(m + 1))) Int[(a + b*x)^(m + 1)*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && LtQ[m, -1] && !(LtQ[n, -1] && (EqQ[a, 0 ] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d , m, n, x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[1/Sqrt[(a_) + (b_.)*(x_)^3], x_Symbol] :> With[{r = Numer[Rt[b/a, 3]], s = Denom[Rt[b/a, 3]]}, Simp[2*Sqrt[2 - Sqrt[3]]*(s + r*x)*(Sqrt[(s^2 - r*s *x + r^2*x^2)/((1 - Sqrt[3])*s + r*x)^2]/(3^(1/4)*r*Sqrt[a + b*x^3]*Sqrt[(- s)*((s + r*x)/((1 - Sqrt[3])*s + r*x)^2)]))*EllipticF[ArcSin[((1 + Sqrt[3]) *s + r*x)/((1 - Sqrt[3])*s + r*x)], -7 + 4*Sqrt[3]], x]] /; FreeQ[{a, b}, x ] && NegQ[a]
\[\int \frac {\left (d x +c \right )^{\frac {1}{3}}}{\left (b x +a \right )^{\frac {7}{2}}}d x\]
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {7}{2}}} \,d x } \]
integral(sqrt(b*x + a)*(d*x + c)^(1/3)/(b^4*x^4 + 4*a*b^3*x^3 + 6*a^2*b^2* x^2 + 4*a^3*b*x + a^4), x)
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx=\int \frac {\sqrt [3]{c + d x}}{\left (a + b x\right )^{\frac {7}{2}}}\, dx \]
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {7}{2}}} \,d x } \]
\[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx=\int { \frac {{\left (d x + c\right )}^{\frac {1}{3}}}{{\left (b x + a\right )}^{\frac {7}{2}}} \,d x } \]
Timed out. \[ \int \frac {\sqrt [3]{c+d x}}{(a+b x)^{7/2}} \, dx=\int \frac {{\left (c+d\,x\right )}^{1/3}}{{\left (a+b\,x\right )}^{7/2}} \,d x \]