Integrand size = 26, antiderivative size = 84 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {8 i}{a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i \sqrt {a+i a \tan (c+d x)}}{a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d} \] Output:
8*I/a^3/d/(a+I*a*tan(d*x+c))^(1/2)+8*I*(a+I*a*tan(d*x+c))^(1/2)/a^4/d-2/3* I*(a+I*a*tan(d*x+c))^(3/2)/a^5/d
Time = 0.13 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.58 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 i \left (23+10 i \tan (c+d x)+\tan ^2(c+d x)\right )}{3 a^3 d \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^(7/2),x]
Output:
(((2*I)/3)*(23 + (10*I)*Tan[c + d*x] + Tan[c + d*x]^2))/(a^3*d*Sqrt[a + I* a*Tan[c + d*x]])
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.88, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3042, 3968, 53, 2009}
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.
\(\displaystyle \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (c+d x)^6}{(a+i a \tan (c+d x))^{7/2}}dx\) |
\(\Big \downarrow \) 3968 |
\(\displaystyle -\frac {i \int \frac {(a-i a \tan (c+d x))^2}{(i \tan (c+d x) a+a)^{3/2}}d(i a \tan (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle -\frac {i \int \left (\frac {4 a^2}{(i \tan (c+d x) a+a)^{3/2}}-\frac {4 a}{\sqrt {i \tan (c+d x) a+a}}+\sqrt {i \tan (c+d x) a+a}\right )d(i a \tan (c+d x))}{a^5 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {i \left (-\frac {8 a^2}{\sqrt {a+i a \tan (c+d x)}}-8 a \sqrt {a+i a \tan (c+d x)}+\frac {2}{3} (a+i a \tan (c+d x))^{3/2}\right )}{a^5 d}\) |
Input:
Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^(7/2),x]
Output:
((-I)*((-8*a^2)/Sqrt[a + I*a*Tan[c + d*x]] - 8*a*Sqrt[a + I*a*Tan[c + d*x] ] + (2*(a + I*a*Tan[c + d*x])^(3/2))/3))/(a^5*d)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_ ), x_Symbol] :> Simp[1/(a^(m - 2)*b*f) Subst[Int[(a - x)^(m/2 - 1)*(a + x )^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]
Time = 0.98 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.75
method | result | size |
derivativedivides | \(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a \sqrt {a +i a \tan \left (d x +c \right )}+\frac {4 a^{2}}{\sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{5}}\) | \(63\) |
default | \(\frac {2 i \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+4 a \sqrt {a +i a \tan \left (d x +c \right )}+\frac {4 a^{2}}{\sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{5}}\) | \(63\) |
Input:
int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)
Output:
2*I/d/a^5*(-1/3*(a+I*a*tan(d*x+c))^(3/2)+4*a*(a+I*a*tan(d*x+c))^(1/2)+4*a^ 2/(a+I*a*tan(d*x+c))^(1/2))
Time = 0.08 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.92 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=-\frac {4 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-8 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 12 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\right )}}{3 \, {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )}} \] Input:
integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")
Output:
-4/3*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-8*I*e^(4*I*d*x + 4*I*c) - 12*I*e^(2*I*d*x + 2*I*c) - 3*I)/(a^4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d *x + I*c))
\[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\int \frac {\sec ^{6}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {7}{2}}}\, dx \] Input:
integrate(sec(d*x+c)**6/(a+I*a*tan(d*x+c))**(7/2),x)
Output:
Integral(sec(c + d*x)**6/(I*a*(tan(c + d*x) - I))**(7/2), x)
Time = 0.05 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.74 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2 i \, {\left (\frac {12}{\sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}} - \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 12 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a}{a^{4}}\right )}}{3 \, a d} \] Input:
integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")
Output:
2/3*I*(12/(sqrt(I*a*tan(d*x + c) + a)*a^2) - ((I*a*tan(d*x + c) + a)^(3/2) - 12*sqrt(I*a*tan(d*x + c) + a)*a)/a^4)/(a*d)
Exception generated. \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 0.81 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.31 \[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx=\frac {2\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,23{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,3{}\mathrm {i}+7\,\sin \left (2\,c+2\,d\,x\right )+3\,\sin \left (4\,c+4\,d\,x\right )+20{}\mathrm {i}\right )}{3\,a^4\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \] Input:
int(1/(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(7/2)),x)
Output:
(2*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1 ))^(1/2)*(cos(2*c + 2*d*x)*23i + cos(4*c + 4*d*x)*3i + 7*sin(2*c + 2*d*x) + 3*sin(4*c + 4*d*x) + 20i))/(3*a^4*d*(cos(2*c + 2*d*x) + 1))
\[ \int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx =\text {Too large to display} \] Input:
int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x)
Output:
(sqrt(a)*( - 4*sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**6*i - 18*int((sqrt(t an(c + d*x)*i + 1)*sec(c + d*x)**6*tan(c + d*x)**4)/(tan(c + d*x)**5*i + 3 *tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x )*i - 1),x)*tan(c + d*x)**2*d - 18*int((sqrt(tan(c + d*x)*i + 1)*sec(c + d *x)**6*tan(c + d*x)**4)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*d + 52*int((sqrt (tan(c + d*x)*i + 1)*sec(c + d*x)**6*tan(c + d*x)**3)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d *x)*i - 1),x)*tan(c + d*x)**2*d*i + 52*int((sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**6*tan(c + d*x)**3)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*tan (c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*d*i + 48*int ((sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**6*tan(c + d*x)**2)/(tan(c + d*x)* *5*i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan (c + d*x)*i - 1),x)*tan(c + d*x)**2*d + 48*int((sqrt(tan(c + d*x)*i + 1)*s ec(c + d*x)**6*tan(c + d*x)**2)/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2 *tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*d - 9*in t((sqrt(tan(c + d*x)*i + 1)*sec(c + d*x)**6*tan(c + d*x))/(tan(c + d*x)**5 *i + 3*tan(c + d*x)**4 - 2*tan(c + d*x)**3*i + 2*tan(c + d*x)**2 - 3*tan(c + d*x)*i - 1),x)*tan(c + d*x)**2*d*i - 9*int((sqrt(tan(c + d*x)*i + 1)*se c(c + d*x)**6*tan(c + d*x))/(tan(c + d*x)**5*i + 3*tan(c + d*x)**4 - 2*...