Integrand size = 28, antiderivative size = 81 \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 \operatorname {AppellF1}\left (\frac {2}{3},\frac {3}{2},1,\frac {5}{3},-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \tan ^{\frac {2}{3}}(c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}} \]
3/2*AppellF1(2/3,3/2,1,5/3,-I*tan(d*x+c),I*tan(d*x+c))*(1+I*tan(d*x+c))^(1 /2)*tan(d*x+c)^(2/3)/d/(a+I*a*tan(d*x+c))^(1/2)
\[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx \]
Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.11, 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.
\(\displaystyle \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}dx\) |
\(\Big \downarrow \) 4047 |
\(\displaystyle \frac {i a^2 \int -\frac {1}{a \sqrt [3]{\tan (c+d x)} (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i a^2 \int \frac {1}{a \sqrt [3]{\tan (c+d x)} (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {i a \int \frac {1}{\sqrt [3]{\tan (c+d x)} (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {3 a^2 \int \frac {i \tan (c+d x)}{\left (1-a^3 \tan ^3(c+d x)\right ) \left (\tan ^3(c+d x) a^4+a\right )^{3/2}}d\sqrt [3]{\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 a \int \frac {\sqrt [3]{\tan (c+d x)}}{\left (1-a^3 \tan ^3(c+d x)\right ) \left (\tan ^3(c+d x) a^4+a\right )^{3/2}}d\sqrt [3]{\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {3 \sqrt {a^3 \tan ^3(c+d x)+1} \int \frac {\sqrt [3]{\tan (c+d x)}}{\left (1-a^3 \tan ^3(c+d x)\right ) \left (a^3 \tan ^3(c+d x)+1\right )^{3/2}}d\sqrt [3]{\tan (c+d x)}}{d \sqrt {a^4 \tan ^3(c+d x)+a}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle -\frac {3 a^2 \tan ^2(c+d x) \sqrt {a^3 \tan ^3(c+d x)+1} \operatorname {AppellF1}\left (\frac {2}{3},1,\frac {3}{2},\frac {5}{3},a^3 \tan ^3(c+d x),-a^3 \tan ^3(c+d x)\right )}{2 d \sqrt {a^4 \tan ^3(c+d x)+a}}\) |
(-3*a^2*AppellF1[2/3, 1, 3/2, 5/3, a^3*Tan[c + d*x]^3, -(a^3*Tan[c + d*x]^ 3)]*Tan[c + d*x]^2*Sqrt[1 + a^3*Tan[c + d*x]^3])/(2*d*Sqrt[a + a^4*Tan[c + d*x]^3])
3.3.63.3.1 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]
\[\int \frac {1}{\sqrt {a +i a \tan \left (d x +c \right )}\, \tan \left (d x +c \right )^{\frac {1}{3}}}d x\]
\[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\sqrt {i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {1}{3}}} \,d x } \]
(2*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*((-I*e^(2*I*d*x + 2*I*c) + I) /(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(e^(4*I*d*x + 4*I*c) + 2*e^(2*I*d*x + 2* I*c) + 1) + (a*d*e^(3*I*d*x + 3*I*c) - 4*a*d*e^(2*I*d*x + 2*I*c) + 4*a*d*e ^(I*d*x + I*c))*integral(1/6*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*((- I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^(2/3)*(3*e^(5*I*d*x + 5*I*c) + 30*e^(4*I*d*x + 4*I*c) + 46*e^(3*I*d*x + 3*I*c) - 20*e^(2*I*d*x + 2*I*c) + 43*e^(I*d*x + I*c) - 50)/(a*d*e^(5*I*d*x + 5*I*c) - 6*a*d*e^(4 *I*d*x + 4*I*c) + 11*a*d*e^(3*I*d*x + 3*I*c) - 2*a*d*e^(2*I*d*x + 2*I*c) - 12*a*d*e^(I*d*x + I*c) + 8*a*d), x))/(a*d*e^(3*I*d*x + 3*I*c) - 4*a*d*e^( 2*I*d*x + 2*I*c) + 4*a*d*e^(I*d*x + I*c))
\[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sqrt [3]{\tan {\left (c + d x \right )}}}\, dx \]
\[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {1}{\sqrt {i \, a \tan \left (d x + c\right ) + a} \tan \left (d x + c\right )^{\frac {1}{3}}} \,d x } \]
Exception generated. \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:The choice was done assuming [sageV ARa,t_nostep]=[63,-86]Precision problem choosing root in common_EXT, curre nt precis
Timed out. \[ \int \frac {1}{\sqrt [3]{\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {1}{{\mathrm {tan}\left (c+d\,x\right )}^{1/3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]