Integrand size = 28, antiderivative size = 82 \[ \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {3 a \operatorname {AppellF1}\left (\frac {5}{3},-\frac {1}{2},1,\frac {8}{3},-i \tan (c+d x),i \tan (c+d x)\right ) \tan ^{\frac {5}{3}}(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 d \sqrt {1+i \tan (c+d x)}} \] Output:
3/5*a*AppellF1(5/3,-1/2,1,8/3,-I*tan(d*x+c),I*tan(d*x+c))*tan(d*x+c)^(5/3) *(a+I*a*tan(d*x+c))^(1/2)/d/(1+I*tan(d*x+c))^(1/2)
\[ \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx \] Input:
Integrate[Tan[c + d*x]^(2/3)*(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
Integrate[Tan[c + d*x]^(2/3)*(a + I*a*Tan[c + d*x])^(3/2), x]
Time = 0.33 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.12, 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.
\(\displaystyle \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^{2/3} (a+i a \tan (c+d x))^{3/2}dx\) |
\(\Big \downarrow \) 4047 |
\(\displaystyle \frac {i a^2 \int -\frac {\tan ^{\frac {2}{3}}(c+d x) \sqrt {i \tan (c+d x) a+a}}{a (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i a^2 \int \frac {\tan ^{\frac {2}{3}}(c+d x) \sqrt {i \tan (c+d x) a+a}}{a (a-i a \tan (c+d x))}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {i a \int \frac {\tan ^{\frac {2}{3}}(c+d x) \sqrt {i \tan (c+d x) a+a}}{a-i a \tan (c+d x)}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {3 a^2 \int \frac {a^3 \tan ^4(c+d x) \sqrt {\tan ^3(c+d x) a^4+a}}{1-a^3 \tan ^3(c+d x)}d\sqrt [3]{\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 a \int \frac {a^4 \tan ^4(c+d x) \sqrt {\tan ^3(c+d x) a^4+a}}{1-a^3 \tan ^3(c+d x)}d\sqrt [3]{\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {3 a \sqrt {a^4 \tan ^3(c+d x)+a} \int \frac {a^4 \tan ^4(c+d x) \sqrt {a^3 \tan ^3(c+d x)+1}}{1-a^3 \tan ^3(c+d x)}d\sqrt [3]{\tan (c+d x)}}{d \sqrt {a^3 \tan ^3(c+d x)+1}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {3 i a^6 \tan ^5(c+d x) \sqrt {a^4 \tan ^3(c+d x)+a} \operatorname {AppellF1}\left (\frac {5}{3},1,-\frac {1}{2},\frac {8}{3},a^3 \tan ^3(c+d x),-a^3 \tan ^3(c+d x)\right )}{5 d \sqrt {a^3 \tan ^3(c+d x)+1}}\) |
Input:
Int[Tan[c + d*x]^(2/3)*(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(((3*I)/5)*a^6*AppellF1[5/3, 1, -1/2, 8/3, a^3*Tan[c + d*x]^3, -(a^3*Tan[c + d*x]^3)]*Tan[c + d*x]^5*Sqrt[a + a^4*Tan[c + d*x]^3])/(d*Sqrt[1 + a^3*T an[c + d*x]^3])
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]
\[\int \tan \left (d x +c \right )^{\frac {2}{3}} \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}d x\]
Input:
int(tan(d*x+c)^(2/3)*(a+I*a*tan(d*x+c))^(3/2),x)
Output:
int(tan(d*x+c)^(2/3)*(a+I*a*tan(d*x+c))^(3/2),x)
Timed out. \[ \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \] Input:
integrate(tan(d*x+c)^(2/3)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \tan ^{\frac {2}{3}}{\left (c + d x \right )}\, dx \] Input:
integrate(tan(d*x+c)**(2/3)*(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Integral((I*a*(tan(c + d*x) - I))**(3/2)*tan(c + d*x)**(2/3), x)
\[ \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \tan \left (d x + c\right )^{\frac {2}{3}} \,d x } \] Input:
integrate(tan(d*x+c)^(2/3)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((I*a*tan(d*x + c) + a)^(3/2)*tan(d*x + c)^(2/3), x)
Exception generated. \[ \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(d*x+c)^(2/3)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Degree mismatch inside factorisatio n over extensionUnable to transpose Error: Bad Argument ValueDone
Timed out. \[ \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int {\mathrm {tan}\left (c+d\,x\right )}^{2/3}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:
int(tan(c + d*x)^(2/3)*(a + a*tan(c + d*x)*1i)^(3/2),x)
Output:
int(tan(c + d*x)^(2/3)*(a + a*tan(c + d*x)*1i)^(3/2), x)
\[ \int \tan ^{\frac {2}{3}}(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {2 \sqrt {a}\, a i \left (-3 \tan \left (d x +c \right )^{\frac {2}{3}} \sqrt {\tan \left (d x +c \right ) i +1}+2 \left (\int \frac {\sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{\frac {1}{3}}}d x \right ) d +5 \left (\int \tan \left (d x +c \right )^{\frac {5}{3}} \sqrt {\tan \left (d x +c \right ) i +1}d x \right ) d \right )}{3 d} \] Input:
int(tan(d*x+c)^(2/3)*(a+I*a*tan(d*x+c))^(3/2),x)
Output:
(2*sqrt(a)*a*i*( - 3*tan(c + d*x)**(2/3)*sqrt(tan(c + d*x)*i + 1) + 2*int( (tan(c + d*x)**(2/3)*sqrt(tan(c + d*x)*i + 1))/tan(c + d*x),x)*d + 5*int(t an(c + d*x)**(2/3)*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*d))/(3*d)