Integrand size = 28, antiderivative size = 82 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx=\frac {3 a \operatorname {AppellF1}\left (\frac {2}{3},-\frac {1}{2},1,\frac {5}{3},-i \tan (c+d x),i \tan (c+d x)\right ) \tan ^{\frac {2}{3}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d \sqrt {1+i \tan (c+d x)}} \]
3/2*a*AppellF1(2/3,-1/2,1,5/3,-I*tan(d*x+c),I*tan(d*x+c))*(a+I*a*tan(d*x+c ))^(1/2)*tan(d*x+c)^(2/3)/d/(1+I*tan(d*x+c))^(1/2)
\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx=\int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx \]
Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.10, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4047, 25, 27, 148, 27, 1013, 1012}
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.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4047 |
\(\displaystyle \frac {i a^2 \int -\frac {\sqrt {i \tan (c+d x) a+a}}{a \sqrt [3]{\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i a^2 \int \frac {\sqrt {i \tan (c+d x) a+a}}{a \sqrt [3]{\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {i a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt [3]{\tan (c+d x)} (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {3 a^2 \int \frac {i \tan (c+d x) \sqrt {\tan ^3(c+d x) a^4+a}}{1-a^3 \tan ^3(c+d x)}d\sqrt [3]{\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 a \int \frac {\sqrt [3]{\tan (c+d x)} \sqrt {\tan ^3(c+d x) a^4+a}}{1-a^3 \tan ^3(c+d x)}d\sqrt [3]{\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {3 a \sqrt {a^4 \tan ^3(c+d x)+a} \int \frac {\sqrt [3]{\tan (c+d x)} \sqrt {a^3 \tan ^3(c+d x)+1}}{1-a^3 \tan ^3(c+d x)}d\sqrt [3]{\tan (c+d x)}}{d \sqrt {a^3 \tan ^3(c+d x)+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {3 a^3 \tan ^2(c+d x) \sqrt {a^4 \tan ^3(c+d x)+a} \operatorname {AppellF1}\left (\frac {2}{3},1,-\frac {1}{2},\frac {5}{3},a^3 \tan ^3(c+d x),-a^3 \tan ^3(c+d x)\right )}{2 d \sqrt {a^3 \tan ^3(c+d x)+1}}\) |
(-3*a^3*AppellF1[2/3, 1, -1/2, 5/3, a^3*Tan[c + d*x]^3, -(a^3*Tan[c + d*x] ^3)]*Tan[c + d*x]^2*Sqrt[a + a^4*Tan[c + d*x]^3])/(2*d*Sqrt[1 + a^3*Tan[c + d*x]^3])
3.3.57.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))^(p_.), x_] :> With[{k = Denominator[m]}, Simp[k/b Subst[Int[x^(k*(m + 1) - 1)*(c + d*(x^k/b))^n*(e + f*(x^k/b))^p, x], x, (b*x)^(1/k)], x]] /; FreeQ[{b, c, d, e, f, n, p}, x] && FractionQ[m] && IntegerQ[p]
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^p*c^q*((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])
Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ ))^(q_), x_Symbol] :> Simp[a^IntPart[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^ n/a))^FracPart[p]) Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] & & NeQ[m, n - 1] && !(IntegerQ[p] || GtQ[a, 0])
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(b/f) Subst[Int[(a + x)^(m - 1)*(( c + (d/b)*x)^n/(b^2 + a*x)), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d ^2, 0]
\[\int \frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{\tan \left (d x +c \right )^{\frac {1}{3}}}d x\]
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}{\sqrt [3]{\tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}{\tan \left (d x + c\right )^{\frac {1}{3}}} \,d x } \]
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt [3]{\tan (c+d x)}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{1/3}} \,d x \]