\(\int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\) [184]

Optimal result

Integrand size = 38, antiderivative size = 237 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {(23 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}} \]

[Out]

(-1/2-1/2*I)*(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/a^(1/2)+1/15*(61*A+35*
I*B)*(a+I*a*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(1/2)+(A+I*B)/d/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2)-1/5*(7*
A+5*I*B)*(a+I*a*tan(d*x+c))^(1/2)/a/d/tan(d*x+c)^(5/2)+1/15*(23*I*A-25*B)*(a+I*a*tan(d*x+c))^(1/2)/a/d/tan(d*x
+c)^(3/2)

Rubi [A] (verified)

Time = 0.80 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.132, Rules used = {3677, 3679, 12, 3625, 211} \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {(-25 B+23 i A) \sqrt {a+i a \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}-\frac {(7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {(61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}} \]

[In]

Int[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((-1/2 - I/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(Sqrt[a]*d)
+ (A + I*B)/(d*Tan[c + d*x]^(5/2)*Sqrt[a + I*a*Tan[c + d*x]]) - ((7*A + (5*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(
5*a*d*Tan[c + d*x]^(5/2)) + (((23*I)*A - 25*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*a*d*Tan[c + d*x]^(3/2)) + ((61*
A + (35*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(15*a*d*Sqrt[Tan[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3625

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
-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] && NeQ[c^2 + d^2, 0]

Rule 3677

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*A + b*B)*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(2*
f*m*(b*c - a*d))), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] &&  !GtQ[n,
0]

Rule 3679

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(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] - Dist[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] /; FreeQ[{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]

Rubi steps \begin{align*} \text {integral}& = \frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (7 A+5 i B)-3 a (i A-B) \tan (c+d x)\right )}{\tan ^{\frac {7}{2}}(c+d x)} \, dx}{a^2} \\ & = \frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (23 i A-25 B)-a^2 (7 A+5 i B) \tan (c+d x)\right )}{\tan ^{\frac {5}{2}}(c+d x)} \, dx}{5 a^3} \\ & = \frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {(23 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{8} a^3 (61 A+35 i B)+\frac {1}{4} a^3 (23 i A-25 B) \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^4} \\ & = \frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {(23 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}}+\frac {8 \int -\frac {15 a^4 (i A+B) \sqrt {a+i a \tan (c+d x)}}{16 \sqrt {\tan (c+d x)}} \, dx}{15 a^5} \\ & = \frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {(23 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}}-\frac {(i A+B) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{2 a} \\ & = \frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {(23 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}}-\frac {(a (A-i B)) \text {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \\ & = -\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (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 )}{\sqrt {a} d}+\frac {A+i B}{d \tan ^{\frac {5}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}}-\frac {(7 A+5 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d \tan ^{\frac {5}{2}}(c+d x)}+\frac {(23 i A-25 B) \sqrt {a+i a \tan (c+d x)}}{15 a d \tan ^{\frac {3}{2}}(c+d x)}+\frac {(61 A+35 i B) \sqrt {a+i a \tan (c+d x)}}{15 a d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.82 (sec) , antiderivative size = 172, normalized size of antiderivative = 0.73 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\frac {15 \sqrt {2} a (A-i B) \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \tan ^4(c+d x)}{(i a \tan (c+d x))^{3/2}}+\frac {2 \left (-6 A+2 i (A+5 i B) \tan (c+d x)+2 (19 A+5 i B) \tan ^2(c+d x)+(61 i A-35 B) \tan ^3(c+d x)\right )}{\sqrt {a+i a \tan (c+d x)}}}{30 d \tan ^{\frac {5}{2}}(c+d x)} \]

[In]

Integrate[(A + B*Tan[c + d*x])/(Tan[c + d*x]^(7/2)*Sqrt[a + I*a*Tan[c + d*x]]),x]

[Out]

((15*Sqrt[2]*a*(A - I*B)*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Tan[c + d*x]^4)/
(I*a*Tan[c + d*x])^(3/2) + (2*(-6*A + (2*I)*(A + (5*I)*B)*Tan[c + d*x] + 2*(19*A + (5*I)*B)*Tan[c + d*x]^2 + (
(61*I)*A - 35*B)*Tan[c + d*x]^3))/Sqrt[a + I*a*Tan[c + d*x]])/(30*d*Tan[c + d*x]^(5/2))

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 814 vs. \(2 (193 ) = 386\).

Time = 0.18 (sec) , antiderivative size = 815, normalized size of antiderivative = 3.44

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-396 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+140 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{4}\left (d x +c \right )\right )-15 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 ) a \left (\tan ^{5}\left (d x +c \right )\right )+244 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{4}\left (d x +c \right )\right )-15 i B \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 ^{3}\left (d x +c \right )\right )+30 B \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 ) a \left (\tan ^{4}\left (d x +c \right )\right )+180 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+16 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 )+15 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 ) \sqrt {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )-144 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+30 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 ) a \left (\tan ^{4}\left (d x +c \right )\right )+15 i B \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 ) a \left (\tan ^{5}\left (d x +c \right )\right )+40 B \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 )+24 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{60 d a \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\)	\(815\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (-396 i A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+140 i B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{4}\left (d x +c \right )\right )-15 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 ) a \left (\tan ^{5}\left (d x +c \right )\right )+244 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{4}\left (d x +c \right )\right )-15 i B \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 ^{3}\left (d x +c \right )\right )+30 B \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 ) a \left (\tan ^{4}\left (d x +c \right )\right )+180 B \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )+16 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 )+15 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 ) \sqrt {2}\, a \left (\tan ^{3}\left (d x +c \right )\right )-144 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+30 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 ) a \left (\tan ^{4}\left (d x +c \right )\right )+15 i B \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 ) a \left (\tan ^{5}\left (d x +c \right )\right )+40 B \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 )+24 A \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right )}{60 d a \tan \left (d x +c \right )^{\frac {5}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\)	\(815\)
parts	\(\text {Expression too large to display}\)	\(870\)

[In]

int((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

1/60/d*(a*(1+I*tan(d*x+c)))^(1/2)/a/tan(d*x+c)^(5/2)*(-396*I*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1
/2)*tan(d*x+c)^3+140*I*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^4-15*A*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))*a*tan(d*x+c)^5+24
4*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^4-15*I*B*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))*a*tan(d*x+c)^3+30*B*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))*a*tan(d*x+c)^4+18
0*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3+16*I*A*(-I*a)^(1/2)*tan(d*x+c)*(a*tan(d*x+
c)*(1+I*tan(d*x+c)))^(1/2)+15*A*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)^3-144*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x
+c)^2+30*I*A*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))*a*tan(d*x+c)^4+15*I*B*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))*a*tan(d*x+c)^5+40*B*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*
x+c)+24*A*(-I*a)^(1/2)*(a*tan(d*x+c)*(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)/(-tan(d*x+c)+I)^2

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 596 vs. \(2 (181) = 362\).

Time = 0.28 (sec) , antiderivative size = 596, normalized size of antiderivative = 2.51 \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {15 \, \sqrt {2} {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} - 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} \log \left (\frac {\sqrt {2} a d \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \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}}}{4 i \, A + 4 \, B}\right ) - 15 \, \sqrt {2} {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} - 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} \log \left (-\frac {\sqrt {2} a d \sqrt {-\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} {\left ({\left (i \, A + B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + i \, A + B\right )} \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}}}{4 i \, A + 4 \, B}\right ) + 2 \, \sqrt {2} {\left ({\left (-103 i \, A + 35 \, B\right )} e^{\left (8 i \, d x + 8 i \, c\right )} + 6 \, {\left (17 i \, A - 15 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 20 \, {\left (2 i \, A - B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 30 \, {\left (-5 i \, A + 3 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, A - 15 \, B\right )} \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}}}{60 \, {\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} - 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="fricas")

[Out]

-1/60*(15*sqrt(2)*(a*d*e^(7*I*d*x + 7*I*c) - 3*a*d*e^(5*I*d*x + 5*I*c) + 3*a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*
d*x + I*c))*sqrt(-(-I*A^2 - 2*A*B + I*B^2)/(a*d^2))*log((sqrt(2)*a*d*sqrt(-(-I*A^2 - 2*A*B + I*B^2)/(a*d^2))*e
^(I*d*x + I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) - 15*sqrt(2)*(a*d*e^(7*I*d*x + 7*I*c) - 3*
a*d*e^(5*I*d*x + 5*I*c) + 3*a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))*sqrt(-(-I*A^2 - 2*A*B + I*B^2)/(a*d
^2))*log(-(sqrt(2)*a*d*sqrt(-(-I*A^2 - 2*A*B + I*B^2)/(a*d^2))*e^(I*d*x + I*c) - sqrt(2)*((I*A + B)*e^(2*I*d*x
+ 2*I*c) + I*A + B)*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)))/(4*I*A + 4*B)) + 2*sqrt(2)*((-103*I*A + 35*B)*e^(8*I*d*x + 8*I*c) + 6*(17*I*A - 15*B)*e^(6*I*d*x + 6*I*
c) + 20*(2*I*A - B)*e^(4*I*d*x + 4*I*c) + 30*(-5*I*A + 3*B)*e^(2*I*d*x + 2*I*c) + 15*I*A - 15*B)*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)))/(a*d*e^(7*I*d*x + 7*I*c) -
3*a*d*e^(5*I*d*x + 5*I*c) + 3*a*d*e^(3*I*d*x + 3*I*c) - a*d*e^(I*d*x + I*c))

Sympy [F]

\[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {A + B \tan {\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \tan ^{\frac {7}{2}}{\left (c + d x \right )}}\, dx \]

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(1/2)/tan(d*x+c)**(7/2),x)

[Out]

Integral((A + B*tan(c + d*x))/(sqrt(I*a*(tan(c + d*x) - I))*tan(c + d*x)**(7/2)), x)

Maxima [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [F(-2)]

Exception generated. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\text {Exception raised: NotImplementedError} \]

[In]

integrate((A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(7/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> unable to parse Giac output: 2*((-(345*i*sageVARa^3*sageVARA*sageVARd
^2-150*sageVARa^3*sageVARd^2*sageVARB)/(225*i)/sageVARa^6/sageVARd^3*sqrt(i*sageVARa*tan(sageVARc+sageVARd*sag
eVARx)+sageVARa)*sqrt

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \tan (c+d x)}{\tan ^{\frac {7}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {tan}\left (c+d\,x\right )}^{7/2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

[In]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

int((A + B*tan(c + d*x))/(tan(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(1/2)), x)