Integrand size = 19, antiderivative size = 185 \[ \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx=-\frac {8 (b c-a d)^2 \sqrt {a+b x} \sqrt [4]{c+d x}}{77 b d^2}+\frac {4 (b c-a d) (a+b x)^{3/2} \sqrt [4]{c+d x}}{77 b d}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}+\frac {16 (b c-a d)^{13/4} \sqrt {-\frac {d (a+b x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{77 b^{5/4} d^3 \sqrt {a+b x}} \]
4/77*(-a*d+b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/4)/b/d+4/11*(b*x+a)^(5/2)*(d*x+c) ^(1/4)/b-8/77*(-a*d+b*c)^2*(d*x+c)^(1/4)*(b*x+a)^(1/2)/b/d^2+16/77*(-a*d+b *c)^(13/4)*EllipticF(b^(1/4)*(d*x+c)^(1/4)/(-a*d+b*c)^(1/4),I)*(-d*(b*x+a) /(-a*d+b*c))^(1/2)/b^(5/4)/d^3/(b*x+a)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.39 \[ \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx=\frac {2 (a+b x)^{5/2} \sqrt [4]{c+d x} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {5}{2},\frac {7}{2},\frac {d (a+b x)}{-b c+a d}\right )}{5 b \sqrt [4]{\frac {b (c+d x)}{b c-a d}}} \]
(2*(a + b*x)^(5/2)*(c + d*x)^(1/4)*Hypergeometric2F1[-1/4, 5/2, 7/2, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*((b*(c + d*x))/(b*c - a*d))^(1/4))
Time = 0.29 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {60, 60, 60, 73, 765, 762}
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.
\(\displaystyle \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \int \frac {(a+b x)^{3/2}}{(c+d x)^{3/4}}dx}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \int \frac {\sqrt {a+b x}}{(c+d x)^{3/4}}dx}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\) |
\(\Big \downarrow \) 60 |
\(\displaystyle \frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {2 (b c-a d) \int \frac {1}{\sqrt {a+b x} (c+d x)^{3/4}}dx}{3 d}\right )}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \int \frac {1}{\sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}d\sqrt [4]{c+d x}}{3 d^2}\right )}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\) |
\(\Big \downarrow \) 765 |
\(\displaystyle \frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d) \sqrt {1-\frac {b (c+d x)}{b c-a d}} \int \frac {1}{\sqrt {1-\frac {b (c+d x)}{b c-a d}}}d\sqrt [4]{c+d x}}{3 d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\) |
\(\Big \downarrow \) 762 |
\(\displaystyle \frac {(b c-a d) \left (\frac {4 (a+b x)^{3/2} \sqrt [4]{c+d x}}{7 d}-\frac {6 (b c-a d) \left (\frac {4 \sqrt {a+b x} \sqrt [4]{c+d x}}{3 d}-\frac {8 (b c-a d)^{5/4} \sqrt {1-\frac {b (c+d x)}{b c-a d}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt [4]{b} \sqrt [4]{c+d x}}{\sqrt [4]{b c-a d}}\right ),-1\right )}{3 \sqrt [4]{b} d^2 \sqrt {a+\frac {b (c+d x)}{d}-\frac {b c}{d}}}\right )}{7 d}\right )}{11 b}+\frac {4 (a+b x)^{5/2} \sqrt [4]{c+d x}}{11 b}\) |
(4*(a + b*x)^(5/2)*(c + d*x)^(1/4))/(11*b) + ((b*c - a*d)*((4*(a + b*x)^(3 /2)*(c + d*x)^(1/4))/(7*d) - (6*(b*c - a*d)*((4*Sqrt[a + b*x]*(c + d*x)^(1 /4))/(3*d) - (8*(b*c - a*d)^(5/4)*Sqrt[1 - (b*(c + d*x))/(b*c - a*d)]*Elli pticF[ArcSin[(b^(1/4)*(c + d*x)^(1/4))/(b*c - a*d)^(1/4)], -1])/(3*b^(1/4) *d^2*Sqrt[a - (b*c)/d + (b*(c + d*x))/d])))/(7*d)))/(11*b)
3.17.29.3.1 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 + n + 1))), x] + Simp[n*((b*c - a*d)/( b*(m + n + 1))) Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] && !(IGtQ[m, 0] && ( !Integer Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) && !ILtQ[m + n + 2, 0] && IntLinear Q[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_)^4], x_Symbol] :> Simp[(1/(Sqrt[a]*Rt[-b/a, 4]) )*EllipticF[ArcSin[Rt[-b/a, 4]*x], -1], x] /; FreeQ[{a, b}, x] && NegQ[b/a] && GtQ[a, 0]
Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> Simp[Sqrt[1 + b*(x^4/a)]/Sqrt [a + b*x^4] Int[1/Sqrt[1 + b*(x^4/a)], x], x] /; FreeQ[{a, b}, x] && NegQ [b/a] && !GtQ[a, 0]
\[\int \left (b x +a \right )^{\frac {3}{2}} \left (d x +c \right )^{\frac {1}{4}}d x\]
\[ \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx=\int { {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {1}{4}} \,d x } \]
\[ \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx=\int \left (a + b x\right )^{\frac {3}{2}} \sqrt [4]{c + d x}\, dx \]
\[ \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx=\int { {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {1}{4}} \,d x } \]
\[ \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx=\int { {\left (b x + a\right )}^{\frac {3}{2}} {\left (d x + c\right )}^{\frac {1}{4}} \,d x } \]
Timed out. \[ \int (a+b x)^{3/2} \sqrt [4]{c+d x} \, dx=\int {\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{1/4} \,d x \]