Integrand size = 24, antiderivative size = 79 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}+\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d} \] Output:
-2*a^(3/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d+2*2^(1/2)*a^(3/2)*a rctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.90 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2 a^{3/2} \left (\text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )\right )}{d} \] Input:
Integrate[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(-2*a^(3/2)*(ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]] - Sqrt[2]*ArcTanh [Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]))/d
Time = 0.52 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4037, 3042, 3961, 219, 4082, 73, 221}
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.
\(\displaystyle \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{3/2}}{\tan (c+d x)}dx\) |
\(\Big \downarrow \) 4037 |
\(\displaystyle 2 i a \int \sqrt {i \tan (c+d x) a+a}dx+\int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle 2 i a \int \sqrt {i \tan (c+d x) a+a}dx+\int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx\) |
\(\Big \downarrow \) 3961 |
\(\displaystyle \frac {4 a^2 \int \frac {1}{a-i a \tan (c+d x)}d\sqrt {i \tan (c+d x) a+a}}{d}+\int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)}dx+\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\) |
\(\Big \downarrow \) 4082 |
\(\displaystyle \frac {a^2 \int \frac {\cot (c+d x)}{\sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}+\frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}\) |
\(\Big \downarrow \) 73 |
\(\displaystyle \frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 i a \int \frac {1}{i-\frac {i (i \tan (c+d x) a+a)}{a}}d\sqrt {i \tan (c+d x) a+a}}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {2 a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{d}\) |
Input:
Int[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^(3/2),x]
Output:
(-2*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + (2*Sqrt[2]*a^ (3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ {p = Denominator[m]}, Simp[p/b Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL inearQ[a, b, c, d, m, n, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(2*a - x^2), x], x, Sqrt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a , b, c, d}, x] && EqQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[2*(a^2/(a*c - b*d)) Int[Sqrt[a + b*Tan[e + f*x]], x], x] - Simp[(2*b*c*d + a*(c^2 - d^2))/(a*(c^2 + d^2)) Int[(a - b*Tan[e + f*x])*(Sqrt[a + b*Tan[e + f*x]]/(c + d*Tan[e + f*x])), x], x] / ; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(B/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
Time = 1.54 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.81
method | result | size |
derivativedivides | \(\frac {2 a^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}}\right )}{d}\) | \(64\) |
default | \(\frac {2 a^{2} \left (-\frac {\operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}}{\sqrt {a}}\right )}{\sqrt {a}}+\frac {\sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{\sqrt {a}}\right )}{d}\) | \(64\) |
Input:
int(cot(d*x+c)*(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
2/d*a^2*(-1/a^(1/2)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))+2^(1/2)/a^(1 /2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 357 vs. \(2 (60) = 120\).
Time = 0.10 (sec) , antiderivative size = 357, normalized size of antiderivative = 4.52 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\sqrt {2} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - \sqrt {2} \sqrt {\frac {a^{3}}{d^{2}}} \log \left (\frac {4 \, {\left (a^{2} e^{\left (i \, d x + i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - \frac {1}{2} \, \sqrt {\frac {a^{3}}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 \, \sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) + \frac {1}{2} \, \sqrt {\frac {a^{3}}{d^{2}}} \log \left (16 \, {\left (3 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - 2 \, \sqrt {2} {\left (d e^{\left (3 i \, d x + 3 i \, c\right )} + d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + a^{2}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}\right ) \] Input:
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
Output:
sqrt(2)*sqrt(a^3/d^2)*log(4*(a^2*e^(I*d*x + I*c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a) - sqrt(2)*sqrt(a^3/d^2)*log(4*(a^2*e^(I*d*x + I*c) - (d*e^(2*I*d*x + 2*I*c ) + d)*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/a ) - 1/2*sqrt(a^3/d^2)*log(16*(3*a^2*e^(2*I*d*x + 2*I*c) + 2*sqrt(2)*(d*e^( 3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c))*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2 *I*c) + 1)) + a^2)*e^(-2*I*d*x - 2*I*c)) + 1/2*sqrt(a^3/d^2)*log(16*(3*a^2 *e^(2*I*d*x + 2*I*c) - 2*sqrt(2)*(d*e^(3*I*d*x + 3*I*c) + d*e^(I*d*x + I*c ))*sqrt(a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + a^2)*e^(-2*I*d*x - 2* I*c))
\[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \cot {\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))**(3/2),x)
Output:
Integral((I*a*(tan(c + d*x) - I))**(3/2)*cot(c + d*x), x)
Time = 0.12 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.35 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {\sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - a^{\frac {3}{2}} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{d} \] Input:
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
Output:
-(sqrt(2)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqr t(2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a))) - a^(3/2)*log((sqrt(I*a*tan(d* x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a))))/d
Exception generated. \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Exception raised: TypeError} \] Input:
integrate(cot(d*x+c)*(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:Error: Bad Argument TypeError: Bad Argument TypeDone
Time = 0.98 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.96 \[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a^2}\right )\,\sqrt {a^3}}{d}+\frac {2\,\sqrt {2}\,\mathrm {atanh}\left (\frac {\sqrt {2}\,\sqrt {a^3}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,a^2}\right )\,\sqrt {a^3}}{d} \] Input:
int(cot(c + d*x)*(a + a*tan(c + d*x)*1i)^(3/2),x)
Output:
(2*2^(1/2)*atanh((2^(1/2)*(a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/(2*a^ 2))*(a^3)^(1/2))/d - (2*atanh(((a^3)^(1/2)*(a + a*tan(c + d*x)*1i)^(1/2))/ a^2)*(a^3)^(1/2))/d
\[ \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\left (\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right ) \tan \left (d x +c \right )d x \right ) i +\int \sqrt {\tan \left (d x +c \right ) i +1}\, \cot \left (d x +c \right )d x \right ) \] Input:
int(cot(d*x+c)*(a+I*a*tan(d*x+c))^(3/2),x)
Output:
sqrt(a)*a*(int(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x)*tan(c + d*x),x)*i + i nt(sqrt(tan(c + d*x)*i + 1)*cot(c + d*x),x))