Integrand size = 28, antiderivative size = 83 \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=-\frac {6 i 2^{5/6} \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{6},\frac {5}{6},\frac {1}{2} (1-i \tan (e+f x))\right ) \left (a^2+i a^2 \tan (e+f x)\right )}{f \sqrt [3]{d \sec (e+f x)} (1+i \tan (e+f x))^{5/6}} \] Output:
-6*I*2^(5/6)*hypergeom([-5/6, -1/6],[5/6],1/2-1/2*I*tan(f*x+e))*(a^2+I*a^2 *tan(f*x+e))/f/(d*sec(f*x+e))^(1/3)/(1+I*tan(f*x+e))^(5/6)
Time = 0.32 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\frac {3 a^2 \left (\operatorname {Hypergeometric2F1}\left (-\frac {1}{2},-\frac {1}{6},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)+\operatorname {Hypergeometric2F1}\left (-\frac {1}{6},\frac {1}{2},\frac {5}{6},\sec ^2(e+f x)\right ) \tan (e+f x)-2 i \sqrt {-\tan ^2(e+f x)}\right )}{f \sqrt [3]{d \sec (e+f x)} \sqrt {-\tan ^2(e+f x)}} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^2/(d*Sec[e + f*x])^(1/3),x]
Output:
(3*a^2*(Hypergeometric2F1[-1/2, -1/6, 5/6, Sec[e + f*x]^2]*Tan[e + f*x] + Hypergeometric2F1[-1/6, 1/2, 5/6, Sec[e + f*x]^2]*Tan[e + f*x] - (2*I)*Sqr t[-Tan[e + f*x]^2]))/(f*(d*Sec[e + f*x])^(1/3)*Sqrt[-Tan[e + f*x]^2])
Time = 0.45 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3986, 3042, 4006, 80, 27, 79}
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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}}dx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle \frac {\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)} \int \frac {(i \tan (e+f x) a+a)^{11/6}}{\sqrt [6]{a-i a \tan (e+f x)}}dx}{\sqrt [3]{d \sec (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)} \int \frac {(i \tan (e+f x) a+a)^{11/6}}{\sqrt [6]{a-i a \tan (e+f x)}}dx}{\sqrt [3]{d \sec (e+f x)}}\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 \sqrt [6]{a-i a \tan (e+f x)} \sqrt [6]{a+i a \tan (e+f x)} \int \frac {(i \tan (e+f x) a+a)^{5/6}}{(a-i a \tan (e+f x))^{7/6}}d\tan (e+f x)}{f \sqrt [3]{d \sec (e+f x)}}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {2^{5/6} a^2 \sqrt [6]{a-i a \tan (e+f x)} (a+i a \tan (e+f x)) \int \frac {(i \tan (e+f x)+1)^{5/6}}{2^{5/6} (a-i a \tan (e+f x))^{7/6}}d\tan (e+f x)}{f (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a^2 \sqrt [6]{a-i a \tan (e+f x)} (a+i a \tan (e+f x)) \int \frac {(i \tan (e+f x)+1)^{5/6}}{(a-i a \tan (e+f x))^{7/6}}d\tan (e+f x)}{f (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)}}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {6 i 2^{5/6} a (a+i a \tan (e+f x)) \operatorname {Hypergeometric2F1}\left (-\frac {5}{6},-\frac {1}{6},\frac {5}{6},\frac {1}{2} (1-i \tan (e+f x))\right )}{f (1+i \tan (e+f x))^{5/6} \sqrt [3]{d \sec (e+f x)}}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^2/(d*Sec[e + f*x])^(1/3),x]
Output:
((-6*I)*2^(5/6)*a*Hypergeometric2F1[-5/6, -1/6, 5/6, (1 - I*Tan[e + f*x])/ 2]*(a + I*a*Tan[e + f*x]))/(f*(d*Sec[e + f*x])^(1/3)*(1 + I*Tan[e + f*x])^ (5/6))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_.), x_Symbol] :> Simp[(d*Sec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/ 2)*(a - b*Tan[e + f*x])^(m/2)) Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a - b* Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*( c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
\[\int \frac {\left (a +i a \tan \left (f x +e \right )\right )^{2}}{\left (d \sec \left (f x +e \right )\right )^{\frac {1}{3}}}d x\]
Input:
int((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x)
Output:
int((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x)
\[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="fricas")
Output:
-1/2*(3*2^(2/3)*(4*I*a^2*e^(2*I*f*x + 2*I*e) + I*a^2*e^(I*f*x + I*e) + 5*I *a^2)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*e^(2/3*I*f*x + 2/3*I*e) - 2*(d*f *e^(I*f*x + I*e) - d*f)*integral(-5*2^(2/3)*(I*a^2*e^(2*I*f*x + 2*I*e) + I *a^2*e^(I*f*x + I*e) + I*a^2)*(d/(e^(2*I*f*x + 2*I*e) + 1))^(2/3)*e^(2/3*I *f*x + 2/3*I*e)/(d*f*e^(3*I*f*x + 3*I*e) - 2*d*f*e^(2*I*f*x + 2*I*e) + d*f *e^(I*f*x + I*e)), x))/(d*f*e^(I*f*x + I*e) - d*f)
\[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=- a^{2} \left (\int \left (- \frac {1}{\sqrt [3]{d \sec {\left (e + f x \right )}}}\right )\, dx + \int \frac {\tan ^{2}{\left (e + f x \right )}}{\sqrt [3]{d \sec {\left (e + f x \right )}}}\, dx + \int \left (- \frac {2 i \tan {\left (e + f x \right )}}{\sqrt [3]{d \sec {\left (e + f x \right )}}}\right )\, dx\right ) \] Input:
integrate((a+I*a*tan(f*x+e))**2/(d*sec(f*x+e))**(1/3),x)
Output:
-a**2*(Integral(-1/(d*sec(e + f*x))**(1/3), x) + Integral(tan(e + f*x)**2/ (d*sec(e + f*x))**(1/3), x) + Integral(-2*I*tan(e + f*x)/(d*sec(e + f*x))* *(1/3), x))
\[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="maxima")
Output:
integrate((I*a*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(1/3), x)
\[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{2}}{\left (d \sec \left (f x + e\right )\right )^{\frac {1}{3}}} \,d x } \] Input:
integrate((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x, algorithm="giac")
Output:
integrate((I*a*tan(f*x + e) + a)^2/(d*sec(f*x + e))^(1/3), x)
Timed out. \[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^2}{{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^{1/3}} \,d x \] Input:
int((a + a*tan(e + f*x)*1i)^2/(d/cos(e + f*x))^(1/3),x)
Output:
int((a + a*tan(e + f*x)*1i)^2/(d/cos(e + f*x))^(1/3), x)
\[ \int \frac {(a+i a \tan (e+f x))^2}{\sqrt [3]{d \sec (e+f x)}} \, dx=\frac {a^{2} \left (-\left (\int \frac {\tan \left (f x +e \right )^{2}}{\sec \left (f x +e \right )^{\frac {1}{3}}}d x \right )+2 \left (\int \frac {\tan \left (f x +e \right )}{\sec \left (f x +e \right )^{\frac {1}{3}}}d x \right ) i +\int \frac {1}{\sec \left (f x +e \right )^{\frac {1}{3}}}d x \right )}{d^{\frac {1}{3}}} \] Input:
int((a+I*a*tan(f*x+e))^2/(d*sec(f*x+e))^(1/3),x)
Output:
(a**2*( - int(tan(e + f*x)**2/sec(e + f*x)**(1/3),x) + 2*int(tan(e + f*x)/ sec(e + f*x)**(1/3),x)*i + int(1/sec(e + f*x)**(1/3),x)))/d**(1/3)