Integrand size = 28, antiderivative size = 84 \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\frac {3 \operatorname {AppellF1}\left (\frac {4}{3},\frac {5}{2},1,\frac {7}{3},-i \tan (c+d x),i \tan (c+d x)\right ) \sqrt {1+i \tan (c+d x)} \tan ^{\frac {4}{3}}(c+d x)}{4 a d \sqrt {a+i a \tan (c+d x)}} \] Output:
3/4*AppellF1(4/3,5/2,1,7/3,-I*tan(d*x+c),I*tan(d*x+c))*(1+I*tan(d*x+c))^(1 /2)*tan(d*x+c)^(4/3)/a/d/(a+I*a*tan(d*x+c))^(1/2)
\[ \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx \] Input:
Integrate[Tan[c + d*x]^(1/3)/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
Integrate[Tan[c + d*x]^(1/3)/(a + I*a*Tan[c + d*x])^(3/2), x]
Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.07, 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.
\(\displaystyle \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 4047 |
\(\displaystyle \frac {i a^2 \int -\frac {\sqrt [3]{\tan (c+d x)}}{a (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {i a^2 \int \frac {\sqrt [3]{\tan (c+d x)}}{a (a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {i a \int \frac {\sqrt [3]{\tan (c+d x)}}{(a-i a \tan (c+d x)) (i \tan (c+d x) a+a)^{5/2}}d(i a \tan (c+d x))}{d}\) |
\(\Big \downarrow \) 148 |
\(\displaystyle \frac {3 a^2 \int -\frac {i a^2 \tan ^3(c+d x)}{\left (1-a^3 \tan ^3(c+d x)\right ) \left (\tan ^3(c+d x) a^4+a\right )^{5/2}}d\sqrt [3]{\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 a \int -\frac {i a^3 \tan ^3(c+d x)}{\left (1-a^3 \tan ^3(c+d x)\right ) \left (\tan ^3(c+d x) a^4+a\right )^{5/2}}d\sqrt [3]{\tan (c+d x)}}{d}\) |
\(\Big \downarrow \) 1013 |
\(\displaystyle \frac {3 \sqrt {a^3 \tan ^3(c+d x)+1} \int -\frac {i a^3 \tan ^3(c+d x)}{\left (1-a^3 \tan ^3(c+d x)\right ) \left (a^3 \tan ^3(c+d x)+1\right )^{5/2}}d\sqrt [3]{\tan (c+d x)}}{a d \sqrt {a^4 \tan ^3(c+d x)+a}}\) |
\(\Big \downarrow \) 1012 |
\(\displaystyle \frac {3 a^3 \tan ^4(c+d x) \sqrt {a^3 \tan ^3(c+d x)+1} \operatorname {AppellF1}\left (\frac {4}{3},1,\frac {5}{2},\frac {7}{3},a^3 \tan ^3(c+d x),-a^3 \tan ^3(c+d x)\right )}{4 d \sqrt {a^4 \tan ^3(c+d x)+a}}\) |
Input:
Int[Tan[c + d*x]^(1/3)/(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(3*a^3*AppellF1[4/3, 1, 5/2, 7/3, a^3*Tan[c + d*x]^3, -(a^3*Tan[c + d*x]^3 )]*Tan[c + d*x]^4*Sqrt[1 + a^3*Tan[c + d*x]^3])/(4*d*Sqrt[a + a^4*Tan[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]
\[\int \frac {\tan \left (d x +c \right )^{\frac {1}{3}}}{\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}d x\]
Input:
int(tan(d*x+c)^(1/3)/(a+I*a*tan(d*x+c))^(3/2),x)
Output:
int(tan(d*x+c)^(1/3)/(a+I*a*tan(d*x+c))^(3/2),x)
Timed out. \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Timed out} \] Input:
integrate(tan(d*x+c)^(1/3)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
Timed out
\[ \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {\sqrt [3]{\tan {\left (c + d x \right )}}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate(tan(d*x+c)**(1/3)/(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Integral(tan(c + d*x)**(1/3)/(I*a*(tan(c + d*x) - I))**(3/2), x)
Exception generated. \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate(tan(d*x+c)^(1/3)/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
Exception generated. \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(tan(d*x+c)^(1/3)/(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. \[ \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx=\int \frac {{\mathrm {tan}\left (c+d\,x\right )}^{1/3}}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}} \,d x \] Input:
int(tan(c + d*x)^(1/3)/(a + a*tan(c + d*x)*1i)^(3/2),x)
Output:
int(tan(c + d*x)^(1/3)/(a + a*tan(c + d*x)*1i)^(3/2), x)
\[ \int \frac {\sqrt [3]{\tan (c+d x)}}{(a+i a \tan (c+d x))^{3/2}} \, dx =\text {Too large to display} \] Input:
int(tan(d*x+c)^(1/3)/(a+I*a*tan(d*x+c))^(3/2),x)
Output:
(sqrt(a)*( - 12*tan(c + d*x)**(1/3)*sqrt(tan(c + d*x)*i + 1)*i - 141854339 2358400000*int(( - tan(c + d*x)**(1/3)*sqrt(tan(c + d*x)*i + 1)*tan(c + d* x)**2)/(67549685350400000*tan(c + d*x)**3*a**2*i + 67549685350400000*tan(c + d*x)**2*a**2 + 67549685350400000*tan(c + d*x)*a**2*i + 6754968535040000 0*a**2),x)*tan(c + d*x)**2*a**2*d - 1418543392358400000*int(( - tan(c + d* x)**(1/3)*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)**2)/(67549685350400000*tan (c + d*x)**3*a**2*i + 67549685350400000*tan(c + d*x)**2*a**2 + 67549685350 400000*tan(c + d*x)*a**2*i + 67549685350400000*a**2),x)*a**2*d + 202649056 0512000000*int(( - tan(c + d*x)**(1/3)*sqrt(tan(c + d*x)*i + 1)*tan(c + d* x))/(67549685350400000*tan(c + d*x)**3*a**2*i + 67549685350400000*tan(c + d*x)**2*a**2 + 67549685350400000*tan(c + d*x)*a**2*i + 67549685350400000*a **2),x)*tan(c + d*x)**2*a**2*d*i + 2026490560512000000*int(( - tan(c + d*x )**(1/3)*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x))/(67549685350400000*tan(c + d*x)**3*a**2*i + 67549685350400000*tan(c + d*x)**2*a**2 + 675496853504000 00*tan(c + d*x)*a**2*i + 67549685350400000*a**2),x)*a**2*d*i - 40529811210 2400000*int(( - tan(c + d*x)**(1/3)*sqrt(tan(c + d*x)*i + 1))/(67549685350 400000*tan(c + d*x)**4*a**2*i + 67549685350400000*tan(c + d*x)**3*a**2 + 6 7549685350400000*tan(c + d*x)**2*a**2*i + 67549685350400000*tan(c + d*x)*a **2),x)*tan(c + d*x)**2*a**2*d*i - 405298112102400000*int(( - tan(c + d*x) **(1/3)*sqrt(tan(c + d*x)*i + 1))/(67549685350400000*tan(c + d*x)**4*a*...