\(\int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx\) [129]

Optimal result

Integrand size = 36, antiderivative size = 146 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {8 \sqrt [4]{-1} a^3 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {16 a^3 (5 A-6 i B) \sqrt {\tan (c+d x)}}{15 d}+\frac {2 i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}-\frac {2 (5 A-9 i B) \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d} \]

[Out]

-8*(-1)^(1/4)*a^3*(A-I*B)*arctan((-1)^(3/4)*tan(d*x+c)^(1/2))/d-16/15*a^3*(5*A-6*I*B)*tan(d*x+c)^(1/2)/d+2/5*I
*a*B*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^2/d-2/15*(5*A-9*I*B)*tan(d*x+c)^(1/2)*(a^3+I*a^3*tan(d*x+c))/d

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3675, 3673, 3614, 211} \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {8 \sqrt [4]{-1} a^3 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {16 a^3 (5 A-6 i B) \sqrt {\tan (c+d x)}}{15 d}-\frac {2 (5 A-9 i B) \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}+\frac {2 i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d} \]

[In]

Int[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Sqrt[Tan[c + d*x]],x]

[Out]

(-8*(-1)^(1/4)*a^3*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/d - (16*a^3*(5*A - (6*I)*B)*Sqrt[Tan[c + d
*x]])/(15*d) + (((2*I)/5)*a*B*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^2)/d - (2*(5*A - (9*I)*B)*Sqrt[Tan[c +
d*x]]*(a^3 + I*a^3*Tan[c + d*x]))/(15*d)

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 3614

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2*(c^2/f), S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3673

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

Rule 3675

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}+\frac {2}{5} \int \frac {(a+i a \tan (c+d x))^2 \left (\frac {1}{2} a (5 A-i B)+\frac {1}{2} a (5 i A+9 B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx \\ & = \frac {2 i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}-\frac {2 (5 A-9 i B) \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}+\frac {4}{15} \int \frac {(a+i a \tan (c+d x)) \left (a^2 (5 A-3 i B)+2 a^2 (5 i A+6 B) \tan (c+d x)\right )}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {16 a^3 (5 A-6 i B) \sqrt {\tan (c+d x)}}{15 d}+\frac {2 i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}-\frac {2 (5 A-9 i B) \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}+\frac {4}{15} \int \frac {15 a^3 (A-i B)+15 a^3 (i A+B) \tan (c+d x)}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {16 a^3 (5 A-6 i B) \sqrt {\tan (c+d x)}}{15 d}+\frac {2 i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}-\frac {2 (5 A-9 i B) \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d}+\frac {\left (120 a^6 (A-i B)^2\right ) \text {Subst}\left (\int \frac {1}{15 a^3 (A-i B)-15 a^3 (i A+B) x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d} \\ & = -\frac {8 \sqrt [4]{-1} a^3 (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{d}-\frac {16 a^3 (5 A-6 i B) \sqrt {\tan (c+d x)}}{15 d}+\frac {2 i a B \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^2}{5 d}-\frac {2 (5 A-9 i B) \sqrt {\tan (c+d x)} \left (a^3+i a^3 \tan (c+d x)\right )}{15 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.99 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.63 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {2 a^3 \left (60 \sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )+\sqrt {\tan (c+d x)} \left (45 A-60 i B+5 (i A+3 B) \tan (c+d x)+3 i B \tan ^2(c+d x)\right )\right )}{15 d} \]

[In]

Integrate[((a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]))/Sqrt[Tan[c + d*x]],x]

[Out]

(-2*a^3*(60*(-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]] + Sqrt[Tan[c + d*x]]*(45*A - (60*I)*B +
5*(I*A + 3*B)*Tan[c + d*x] + (3*I)*B*Tan[c + d*x]^2)))/(15*d)

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 252 vs. \(2 (122 ) = 244\).

Time = 0.03 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.73

method	result	size
derivativedivides	\(\frac {a^{3} \left (-\frac {2 i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-\frac {2 i A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+8 i B \left (\sqrt {\tan }\left (d x +c \right )\right )-6 A \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (4 i A +4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(253\)
default	\(\frac {a^{3} \left (-\frac {2 i B \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-\frac {2 i A \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-2 B \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )+8 i B \left (\sqrt {\tan }\left (d x +c \right )\right )-6 A \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\left (-4 i B +4 A \right ) \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}+\frac {\left (4 i A +4 B \right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(253\)
parts	\(\frac {\left (-i A \,a^{3}-3 B \,a^{3}\right ) \left (\frac {2 \left (\tan ^{\frac {3}{2}}\left (d x +c \right )\right )}{3}-\frac {\sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {\left (3 i A \,a^{3}+B \,a^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}+\frac {\left (3 i B \,a^{3}-3 A \,a^{3}\right ) \left (2 \left (\sqrt {\tan }\left (d x +c \right )\right )-\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}+\frac {A \,a^{3} \sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4 d}-\frac {i B \,a^{3} \left (\frac {2 \left (\tan ^{\frac {5}{2}}\left (d x +c \right )\right )}{5}-2 \left (\sqrt {\tan }\left (d x +c \right )\right )+\frac {\sqrt {2}\, \left (\ln \left (\frac {1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}{1-\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )+\tan \left (d x +c \right )}\right )+2 \arctan \left (1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )+2 \arctan \left (-1+\sqrt {2}\, \left (\sqrt {\tan }\left (d x +c \right )\right )\right )\right )}{4}\right )}{d}\)	\(539\)

[In]

int((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/d*a^3*(-2/5*I*B*tan(d*x+c)^(5/2)-2/3*I*A*tan(d*x+c)^(3/2)-2*B*tan(d*x+c)^(3/2)+8*I*B*tan(d*x+c)^(1/2)-6*A*ta
n(d*x+c)^(1/2)+1/4*(-4*I*B+4*A)*2^(1/2)*(ln((1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1-2^(1/2)*tan(d*x+c)^(1/2
)+tan(d*x+c)))+2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)))+1/4*(4*I*A+4*B)*2^(
1/2)*(ln((1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))+2*arctan(1+2^(1/2)*t
an(d*x+c)^(1/2))+2*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))))

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. 447 vs. \(2 (116) = 232\).

Time = 0.27 (sec) , antiderivative size = 447, normalized size of antiderivative = 3.06 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 \, {\left (15 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 15 \, \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{6}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (-i \, A - B\right )} a^{3}}\right ) - 2 \, {\left ({\left (25 \, A - 39 i \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (15 \, A - 19 i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 4 \, {\left (5 \, A - 6 i \, B\right )} a^{3}\right )} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )}}{15 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

2/15*(15*sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*(
(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqr
t((-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)/((-I*A - B)*a^3)) - 15*sqrt(-(
I*A^2 + 2*A*B - I*B^2)*a^6/d^2)*(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-2*((A - I*B)*a^3*e^
(2*I*d*x + 2*I*c) + sqrt(-(I*A^2 + 2*A*B - I*B^2)*a^6/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*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)/((-I*A - B)*a^3)) - 2*((25*A - 39*I*B)*a^3*e^
(4*I*d*x + 4*I*c) + 3*(15*A - 19*I*B)*a^3*e^(2*I*d*x + 2*I*c) + 4*(5*A - 6*I*B)*a^3)*sqrt((-I*e^(2*I*d*x + 2*I
*c) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(4*I*d*x + 4*I*c) + 2*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=- i a^{3} \left (\int \left (- 3 A \sqrt {\tan {\left (c + d x \right )}}\right )\, dx + \int A \tan ^{\frac {5}{2}}{\left (c + d x \right )}\, dx + \int \left (- 3 B \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{\frac {7}{2}}{\left (c + d x \right )}\, dx + \int \frac {i A}{\sqrt {\tan {\left (c + d x \right )}}}\, dx + \int \left (- 3 i A \tan ^{\frac {3}{2}}{\left (c + d x \right )}\right )\, dx + \int i B \sqrt {\tan {\left (c + d x \right )}}\, dx + \int \left (- 3 i B \tan ^{\frac {5}{2}}{\left (c + d x \right )}\right )\, dx\right ) \]

[In]

integrate((a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c))/tan(d*x+c)**(1/2),x)

[Out]

-I*a**3*(Integral(-3*A*sqrt(tan(c + d*x)), x) + Integral(A*tan(c + d*x)**(5/2), x) + Integral(-3*B*tan(c + d*x
)**(3/2), x) + Integral(B*tan(c + d*x)**(7/2), x) + Integral(I*A/sqrt(tan(c + d*x)), x) + Integral(-3*I*A*tan(
c + d*x)**(3/2), x) + Integral(I*B*sqrt(tan(c + d*x)), x) + Integral(-3*I*B*tan(c + d*x)**(5/2), x))

Maxima [A] (verification not implemented)

none

Time = 0.45 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.34 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {6 i \, B a^{3} \tan \left (d x + c\right )^{\frac {5}{2}} + 10 \, {\left (i \, A + 3 \, B\right )} a^{3} \tan \left (d x + c\right )^{\frac {3}{2}} + 30 \, {\left (3 \, A - 4 i \, B\right )} a^{3} \sqrt {\tan \left (d x + c\right )} - 15 \, {\left (2 \, \sqrt {2} {\left (\left (i + 1\right ) \, A - \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} {\left (\left (i + 1\right ) \, A - \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{3}}{15 \, d} \]

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

-1/15*(6*I*B*a^3*tan(d*x + c)^(5/2) + 10*(I*A + 3*B)*a^3*tan(d*x + c)^(3/2) + 30*(3*A - 4*I*B)*a^3*sqrt(tan(d*
x + c)) - 15*(2*sqrt(2)*((I + 1)*A - (I - 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2*sqrt(tan(d*x + c)))) + 2*sqrt(
2)*((I + 1)*A - (I - 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2*sqrt(tan(d*x + c)))) - sqrt(2)*((I - 1)*A + (I + 1
)*B)*log(sqrt(2)*sqrt(tan(d*x + c)) + tan(d*x + c) + 1) + sqrt(2)*((I - 1)*A + (I + 1)*B)*log(-sqrt(2)*sqrt(ta
n(d*x + c)) + tan(d*x + c) + 1))*a^3)/d

Giac [A] (verification not implemented)

none

Time = 0.87 (sec) , antiderivative size = 127, normalized size of antiderivative = 0.87 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=\frac {\left (4 i - 4\right ) \, \sqrt {2} {\left (-i \, A a^{3} - B a^{3}\right )} \arctan \left (-\left (\frac {1}{2} i - \frac {1}{2}\right ) \, \sqrt {2} \sqrt {\tan \left (d x + c\right )}\right )}{d} - \frac {2 \, {\left (3 i \, B a^{3} d^{4} \tan \left (d x + c\right )^{\frac {5}{2}} + 5 i \, A a^{3} d^{4} \tan \left (d x + c\right )^{\frac {3}{2}} + 15 \, B a^{3} d^{4} \tan \left (d x + c\right )^{\frac {3}{2}} + 45 \, A a^{3} d^{4} \sqrt {\tan \left (d x + c\right )} - 60 i \, B a^{3} d^{4} \sqrt {\tan \left (d x + c\right )}\right )}}{15 \, d^{5}} \]

[In]

integrate((a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c))/tan(d*x+c)^(1/2),x, algorithm="giac")

[Out]

(4*I - 4)*sqrt(2)*(-I*A*a^3 - B*a^3)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 2/15*(3*I*B*a^3*d^4
*tan(d*x + c)^(5/2) + 5*I*A*a^3*d^4*tan(d*x + c)^(3/2) + 15*B*a^3*d^4*tan(d*x + c)^(3/2) + 45*A*a^3*d^4*sqrt(t
an(d*x + c)) - 60*I*B*a^3*d^4*sqrt(tan(d*x + c)))/d^5

Mupad [B] (verification not implemented)

Time = 8.59 (sec) , antiderivative size = 257, normalized size of antiderivative = 1.76 \[ \int \frac {(a+i a \tan (c+d x))^3 (A+B \tan (c+d x))}{\sqrt {\tan (c+d x)}} \, dx=-\frac {6\,A\,a^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}}{d}-\frac {A\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}\,2{}\mathrm {i}}{3\,d}+\frac {B\,a^3\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,8{}\mathrm {i}}{d}-\frac {2\,B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{3/2}}{d}-\frac {B\,a^3\,{\mathrm {tan}\left (c+d\,x\right )}^{5/2}\,2{}\mathrm {i}}{5\,d}+\frac {\sqrt {2}\,A\,a^3\,\ln \left (A\,a^3\,d\,8{}\mathrm {i}+\sqrt {2}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4+4{}\mathrm {i}\right )\right )\,\left (2-2{}\mathrm {i}\right )}{d}-\frac {\sqrt {-16{}\mathrm {i}}\,A\,a^3\,\ln \left (A\,a^3\,d\,8{}\mathrm {i}+2\,\sqrt {-16{}\mathrm {i}}\,A\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d}+\frac {\sqrt {2}\,B\,a^3\,\ln \left (8\,B\,a^3\,d+\sqrt {2}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\,\left (-4-4{}\mathrm {i}\right )\right )\,\left (2+2{}\mathrm {i}\right )}{d}-\frac {\sqrt {16{}\mathrm {i}}\,B\,a^3\,\ln \left (8\,B\,a^3\,d+2\,\sqrt {16{}\mathrm {i}}\,B\,a^3\,d\,\sqrt {\mathrm {tan}\left (c+d\,x\right )}\right )}{d} \]

[In]

int(((A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3)/tan(c + d*x)^(1/2),x)

[Out]

(B*a^3*tan(c + d*x)^(1/2)*8i)/d - (A*a^3*tan(c + d*x)^(3/2)*2i)/(3*d) - (6*A*a^3*tan(c + d*x)^(1/2))/d - (2*B*
a^3*tan(c + d*x)^(3/2))/d - (B*a^3*tan(c + d*x)^(5/2)*2i)/(5*d) + (2^(1/2)*A*a^3*log(A*a^3*d*8i - 2^(1/2)*A*a^
3*d*tan(c + d*x)^(1/2)*(4 - 4i))*(2 - 2i))/d - ((-16i)^(1/2)*A*a^3*log(A*a^3*d*8i + 2*(-16i)^(1/2)*A*a^3*d*tan
(c + d*x)^(1/2)))/d + (2^(1/2)*B*a^3*log(8*B*a^3*d - 2^(1/2)*B*a^3*d*tan(c + d*x)^(1/2)*(4 + 4i))*(2 + 2i))/d
- (16i^(1/2)*B*a^3*log(8*B*a^3*d + 2*16i^(1/2)*B*a^3*d*tan(c + d*x)^(1/2)))/d