Integrand size = 38, antiderivative size = 190 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 (-1)^{3/4} a^{5/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (2 i A+B) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \] Output:
2*(-1)^(3/4)*a^(5/2)*B*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*t an(d*x+c))^(1/2))/d+(-4+4*I)*a^(5/2)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x +c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d-2*a^2*(2*I*A+B)*(a+I*a*tan(d*x+c))^( 1/2)/d/tan(d*x+c)^(1/2)-2/3*a*A*(a+I*a*tan(d*x+c))^(3/2)/d/tan(d*x+c)^(3/2 )
Time = 6.74 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.63 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {i a^2 \left (-3 (-1)^{3/4} a (5 A-3 i B) \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \tan ^{\frac {3}{2}}(c+d x) (-i+\tan (c+d x))-\frac {15 i (i A+B) \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) (i a \tan (c+d x))^{3/2} (-i+\tan (c+d x))}{\sqrt {a}}+2 \sqrt {1+i \tan (c+d x)} \left (\frac {6 \sqrt {2} (i A+B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (i a \tan (c+d x))^{3/2} \sqrt {a+i a \tan (c+d x)}}{a}+a (-i+\tan (c+d x)) (A+(7 i A+3 B) \tan (c+d x))\right )\right )}{3 d \sqrt {1+i \tan (c+d x)} \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \] Input:
Integrate[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x] ^(5/2),x]
Output:
((-1/3*I)*a^2*(-3*(-1)^(3/4)*a*(5*A - (3*I)*B)*ArcSinh[(-1)^(1/4)*Sqrt[Tan [c + d*x]]]*Tan[c + d*x]^(3/2)*(-I + Tan[c + d*x]) - ((15*I)*(I*A + B)*Arc Sinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*(I*a*Tan[c + d*x])^(3/2)*(-I + Tan[c + d*x]))/Sqrt[a] + 2*Sqrt[1 + I*Tan[c + d*x]]*((6*Sqrt[2]*(I*A + B)*ArcTan h[(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)*Sqrt[a + I*a*Tan[c + d*x]])/a + a*(-I + Tan[c + d*x])*(A + ( (7*I)*A + 3*B)*Tan[c + d*x]))))/(d*Sqrt[1 + I*Tan[c + d*x]]*Tan[c + d*x]^( 3/2)*Sqrt[a + I*a*Tan[c + d*x]])
Time = 1.19 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 4076, 27, 3042, 4076, 27, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}
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.
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan (c+d x)^{5/2}}dx\) |
\(\Big \downarrow \) 4076 |
\(\displaystyle \frac {2}{3} \int \frac {3 (i \tan (c+d x) a+a)^{3/2} (a (2 i A+B)+i a B \tan (c+d x))}{2 \tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \int \frac {(i \tan (c+d x) a+a)^{3/2} (a (2 i A+B)+i a B \tan (c+d x))}{\tan ^{\frac {3}{2}}(c+d x)}dx-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(i \tan (c+d x) a+a)^{3/2} (a (2 i A+B)+i a B \tan (c+d x))}{\tan (c+d x)^{3/2}}dx-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4076 |
\(\displaystyle 2 \int -\frac {\sqrt {i \tan (c+d x) a+a} \left ((4 A-3 i B) a^2+B \tan (c+d x) a^2\right )}{2 \sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((4 A-3 i B) a^2+B \tan (c+d x) a^2\right )}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\int \frac {\sqrt {i \tan (c+d x) a+a} \left ((4 A-3 i B) a^2+B \tan (c+d x) a^2\right )}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4084 |
\(\displaystyle -4 a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -4 a^2 (A-i B) \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4027 |
\(\displaystyle \frac {8 i a^4 (A-i B) \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle -i a B \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 4082 |
\(\displaystyle -\frac {i a^3 B \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}-\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 65 |
\(\displaystyle -\frac {2 i a^3 B \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle -\frac {(4-4 i) a^{5/2} (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {2 (-1)^{3/4} a^{5/2} B \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 (B+2 i A) \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {2 a A (a+i a \tan (c+d x))^{3/2}}{3 d \tan ^{\frac {3}{2}}(c+d x)}\) |
Input:
Int[((a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/Tan[c + d*x]^(5/2) ,x]
Output:
(2*(-1)^(3/4)*a^(5/2)*B*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqr t[a + I*a*Tan[c + d*x]]])/d - ((4 - 4*I)*a^(5/2)*(A - I*B)*ArcTanh[((1 + I )*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d - (2*a^2*((2* I)*A + B)*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]) - (2*a*A*(a + I*a*Tan[c + d*x])^(3/2))/(3*d*Tan[c + d*x]^(3/2))
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2 Sub st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d }, x] && !GtQ[c, 0]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[(-a^2)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x] - Simp[a/(d*(b*c + a*d)*(n + 1)) Int[ (a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*b*d*(m - n - 2) - B*(b*c*(m - 1) + a*d*(n + 1)) + (a*A*d*(m + n) - B*(a*c*(m - 1) + b *d*(n + 1)))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]
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]
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*b + a*B)/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] - Simp[B/b Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ e + f*x]), 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] && NeQ[A*b + a*B, 0]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 617 vs. \(2 (154 ) = 308\).
Time = 0.23 (sec) , antiderivative size = 618, normalized size of antiderivative = 3.25
method | result | size |
derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-9 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{2}+3 i \sqrt {i a}\, \sqrt {2}\, \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 ) \tan \left (d x +c \right )^{2} a +14 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+12 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{2}-3 \sqrt {i a}\, \sqrt {2}\, \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 ) \tan \left (d x +c \right )^{2} a +6 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{2}+6 B \sqrt {i a}\, \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 )+6 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{2}+2 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{3 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\) | \(618\) |
default | \(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (-9 i B \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{2}+3 i \sqrt {i a}\, \sqrt {2}\, \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 ) \tan \left (d x +c \right )^{2} a +14 i A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+12 A \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{2}-3 \sqrt {i a}\, \sqrt {2}\, \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 ) \tan \left (d x +c \right )^{2} a +6 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{2}+6 B \sqrt {i a}\, \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 )+6 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{2}+2 A \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{3 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {i a}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}}\) | \(618\) |
parts | \(-\frac {A \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2} \left (3 i \sqrt {i a}\, \sqrt {2}\, \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 ) \tan \left (d x +c \right )^{2} a -3 \sqrt {i a}\, \sqrt {2}\, \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 ) \tan \left (d x +c \right )^{2} a +12 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )^{2}+14 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\, \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right )}{3 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}\, \sqrt {-i a}}+\frac {B \left (-i \sqrt {i a}\, \sqrt {2}\, \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 ) \tan \left (d x +c \right ) a -i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\, \tan \left (d x +c \right )-\sqrt {2}\, \sqrt {i a}\, \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 ) \tan \left (d x +c \right ) a -4 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}\, a \tan \left (d x +c \right )-2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}{d \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\) | \(758\) |
Input:
int((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x,method=_R ETURNVERBOSE)
Output:
-1/3/d*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(3/2)*(-9*I*B*ln(1/2*(2*I *a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a) ^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^2+3*I*(I*a)^(1/2)*2^(1/2)*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))/(ta n(d*x+c)+I))*tan(d*x+c)^2*a+14*I*A*(I*a)^(1/2)*(-I*a)^(1/2)*tan(d*x+c)*(a* tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+12*A*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan( d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a* tan(d*x+c)^2-3*(I*a)^(1/2)*2^(1/2)*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))*tan(d*x+c)^2 *a+6*I*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I *a)^(1/2)+a)/(I*a)^(1/2))*(-I*a)^(1/2)*a*tan(d*x+c)^2+6*B*(I*a)^(1/2)*(-I* a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+6*ln(1/2*(2*I*a* tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1 /2))*(-I*a)^(1/2)*a*tan(d*x+c)^2+2*A*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2) *(I*a)^(1/2)*(-I*a)^(1/2))/(I*a)^(1/2)/(-I*a)^(1/2)/(a*tan(d*x+c)*(1+I*tan (d*x+c)))^(1/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 846 vs. \(2 (144) = 288\).
Time = 0.13 (sec) , antiderivative size = 846, normalized size of antiderivative = 4.45 \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, al gorithm="fricas")
Output:
1/6*(12*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I *c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((I*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I *B^2)*a^5/d^2)*d*e^(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a^2*e^(2*I*d*x + 2* I*c) + (-I*A - B)*a^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^(-I*d*x - I*c)/((-I*A - B)* a^2)) - 12*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((-I*sqrt(2)*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^5/d^2)*d*e^(I*d*x + I*c) + sqrt(2)*((-I*A - B)*a^2*e^(2*I*d*x + 2*I*c) + (-I*A - B)*a^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^(-I*d*x - I*c)/((-I*A - B)*a^2)) + 4*sqrt(2)*((8*A - 3*I*B)*a^2*e^(5*I*d*x + 5*I*c) + 2*A*a^2*e^( 3*I*d*x + 3*I*c) - 3*(2*A - I*B)*a^2*e^(I*d*x + I*c))*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)) + 3*sqrt(4*I*B^2*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c ) + d)*log((sqrt(2)*(B*a^2*e^(2*I*d*x + 2*I*c) + B*a^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 )) + I*sqrt(4*I*B^2*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(B*a^2)) - 3*sqrt(4*I*B^2*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*(B*a^2*e^(2*I*d*x + 2*I*c) + B*a^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) +...
\[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \left (A + B \tan {\left (c + d x \right )}\right )}{\tan ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx \] Input:
integrate((a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)**(5/2),x)
Output:
Integral((I*a*(tan(c + d*x) - I))**(5/2)*(A + B*tan(c + d*x))/tan(c + d*x) **(5/2), x)
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, al gorithm="maxima")
Output:
Timed out
Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x, al gorithm="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 TypeWarning, need to choose a branch for the root of a polynomial with par
Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \] Input:
int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/tan(c + d*x)^(5/2 ),x)
Output:
int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(5/2))/tan(c + d*x)^(5/2 ), x)
\[ \int \frac {(a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x))}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {2 \sqrt {a}\, a^{2} \left (3 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2} b i +5 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) a i +3 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) b -\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, a +6 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \tan \left (d x +c \right )^{2} a d i +3 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{2}}d x \right ) \tan \left (d x +c \right )^{2} b d -3 \left (\int \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )d x \right ) \tan \left (d x +c \right )^{2} b d i \right )}{3 \tan \left (d x +c \right )^{2} d} \] Input:
int((a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c))/tan(d*x+c)^(5/2),x)
Output:
(2*sqrt(a)*a**2*(3*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x )**2*b*i + 5*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*a*i + 3*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*b - sqrt(tan( c + d*x))*sqrt(tan(c + d*x)*i + 1)*a + 6*int((sqrt(tan(c + d*x))*sqrt(tan( c + d*x)*i + 1))/tan(c + d*x)**2,x)*tan(c + d*x)**2*a*d*i + 3*int((sqrt(ta n(c + d*x))*sqrt(tan(c + d*x)*i + 1))/tan(c + d*x)**2,x)*tan(c + d*x)**2*b *d - 3*int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)*tan (c + d*x)**2*b*d*i))/(3*tan(c + d*x)**2*d)