Integrand size = 28, antiderivative size = 120 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {(1-i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \]
(-1+I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*a^ (1/2)/d-2/3*I*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(1/2)-2/3*(a+I*a*tan(d *x+c))^(1/2)/d/tan(d*x+c)^(3/2)
Time = 0.58 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (i a \tan (c+d x))^{3/2}}{a}-2 i (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \]
((3*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*(I*a*Tan[c + d*x])^(3/2))/a - (2*I)*(-I + Tan[c + d*x])*Sqrt[a + I* a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2))
Time = 0.64 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4044, 27, 3042, 4081, 27, 3042, 4027, 218}
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.
\(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan (c+d x)^{5/2}}dx\) |
\(\Big \downarrow \) 4044 |
\(\displaystyle \frac {2 \int \frac {(i a-2 a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{2 \tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {(i a-2 a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan ^{\frac {3}{2}}(c+d x)}dx}{3 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {(i a-2 a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\tan (c+d x)^{3/2}}dx}{3 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4081 |
\(\displaystyle \frac {\frac {2 \int -\frac {3 a^2 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a}-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-3 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-3 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {\frac {6 i a^3 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {-\frac {(3-3 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 i a \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}}{3 a}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
(-2*Sqrt[a + I*a*Tan[c + d*x]])/(3*d*Tan[c + d*x]^(3/2)) + (((-3 + 3*I)*a^ (3/2)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d* x]]])/d - ((2*I)*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(3* a)
3.2.91.3.1 Defintions of rubi rules used
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_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f) Subst[Int[1/(a*c - b*d - 2* a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N eQ[c^2 + d^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(c^2 + d^2)*(n + 1)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - a*c*(n + 1) + a*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d , e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[n, -1] && (IntegerQ[n] || IntegersQ[2*m, 2*n])
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[(A*d - B*c)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(f*(n + 1)*(c^2 + d^2))), x] - Simp[1/(a*(n + 1)*(c^2 + d^2)) Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c* m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x], x] /; Fr eeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 195 vs. \(2 (96 ) = 192\).
Time = 1.15 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.63
method | result | size |
derivativedivides | \(\frac {\left (3 i \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \left (\tan ^{2}\left (d x +c \right )\right )-4 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}}\) | \(196\) |
default | \(\frac {\left (3 i \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {2}\, a \left (\tan ^{2}\left (d x +c \right )\right )-4 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}}\) | \(196\) |
1/6/d/tan(d*x+c)^(3/2)*(3*I*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I* tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*2^(1/2)*a*tan(d*x+c )^2-4*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-4*(a *tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))*(a*(1+I*tan(d*x+c)))^(1/ 2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-I*a)^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 355 vs. \(2 (88) = 176\).
Time = 0.25 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.96 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {8 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (5 i \, d x + 5 i \, c\right )} + e^{\left (3 i \, d x + 3 i \, c\right )}\right )} - 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + d \sqrt {-\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - d \sqrt {-\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{6 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]
1/6*(8*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I *c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(5*I*d*x + 5*I*c) + e^(3*I*d*x + 3* I*c)) - 3*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-2*I* a/d^2)*log((sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1) + d*sqr t(-2*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 3*(d*e^(4*I*d*x + 4*I*c ) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-2*I*a/d^2)*log((sqrt(2)*sqrt(a/(e^( 2*I*d*x + 2*I*c) + 1))*sqrt((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I *c) + 1))*(e^(2*I*d*x + 2*I*c) + 1) - d*sqrt(-2*I*a/d^2)*e^(I*d*x + I*c))* e^(-I*d*x - I*c)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 974 vs. \(2 (88) = 176\).
Time = 0.56 (sec) , antiderivative size = 974, normalized size of antiderivative = 8.12 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \]
1/6*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((3*I + 3)*cos(3*d*x + 3*c) + (I + 1)*cos(d*x + c) + (3*I - 3)*sin(3* d*x + 3*c) + (I - 1)*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c), -cos( 2*d*x + 2*c) + 1)) + ((3*I - 3)*cos(3*d*x + 3*c) + (I - 1)*cos(d*x + c) - (3*I + 3)*sin(3*d*x + 3*c) - (I + 1)*sin(d*x + c))*sin(3/2*arctan2(sin(2*d *x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a) + 3*(2*(-(I + 1)*cos(2*d*x + 2 *c)^2 - (I + 1)*sin(2*d*x + 2*c)^2 + (2*I + 2)*cos(2*d*x + 2*c) - I - 1)*a rctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^ (1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x + c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^( 1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + c)) + (-(I - 1)*cos(2*d*x + 2*c)^2 - (I - 1)*sin(2*d*x + 2*c)^2 + (2*I - 2)*cos(2*d*x + 2*c) - I + 1)*log(cos(d*x + c)^2 + sin(d*x + c)^2 + sqrt(co s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*a rctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2 *d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d* x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2* c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(si n(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 2*(((-(I + 1)*cos(d...
Exception generated. \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Non regular value [0] was discarded and replaced randomly by 0=[-96]Warning, replacing -96 by 32, a substitut ion varia
Timed out. \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \]