\(\int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [91]

Optimal result

Integrand size = 36, antiderivative size = 159 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(i A-B) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {4 (i A-B) \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {(3 i A-5 B) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d} \]

[Out]

1/2*(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)/a^(1/2)-4*(I*A-B)*(a+I*a*tan(d*x+c
))^(1/2)/a/d+(I*A-B)*tan(d*x+c)^2/d/(a+I*a*tan(d*x+c))^(1/2)+1/3*(3*I*A-5*B)*(a+I*a*tan(d*x+c))^(3/2)/a^2/d

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3676, 3673, 3608, 3561, 212} \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(-5 B+3 i A) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac {(B+i A) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(-B+i A) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {4 (-B+i A) \sqrt {a+i a \tan (c+d x)}}{a d} \]

[In]

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

[Out]

((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + ((I*A - B)*Tan[c + d*x
]^2)/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (4*(I*A - B)*Sqrt[a + I*a*Tan[c + d*x]])/(a*d) + (((3*I)*A - 5*B)*(a + I
*a*Tan[c + d*x])^(3/2))/(3*a^2*d)

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3561

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2*(b/d), Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3608

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*(
(a + b*Tan[e + f*x])^m/(f*m)), x] + Dist[(b*c + a*d)/b, Int[(a + b*Tan[e + f*x])^m, x], x] /; FreeQ[{a, b, c,
d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] &&  !LtQ[m, 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 3676

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*b - a*B))*(a + b*Tan[e + f*x])^m*((c + d*Tan[e + f*x])^n/(2*a*f
*m)), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d
*n) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*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] && LtQ[m, 0] && GtQ[n, 0]

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

Mathematica [A] (verified)

Time = 1.87 (sec) , antiderivative size = 111, normalized size of antiderivative = 0.70 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {(i A+B) \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {-9 i A+7 B+(6 A+2 i B) \tan (c+d x)+2 B \tan ^2(c+d x)}{3 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

((I*A + B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + ((-9*I)*A + 7*B + (6*A
+ (2*I)*B)*Tan[c + d*x] + 2*B*Tan[c + d*x]^2)/(3*d*Sqrt[a + I*a*Tan[c + d*x]])

Maple [A] (verified)

Time = 0.14 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.81

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-i a B \sqrt {a +i a \tan \left (d x +c \right )}-a A \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a^{2} \left (i B +A \right )}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{2}}\)	\(128\)
default	\(\frac {2 i \left (\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-i a B \sqrt {a +i a \tan \left (d x +c \right )}-a A \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a^{2} \left (i B +A \right )}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{2}}\)	\(128\)
parts	\(\frac {2 i A \left (-\sqrt {a +i a \tan \left (d x +c \right )}+\frac {\sqrt {a}\, \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d a}+\frac {2 B \left (-\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+a \sqrt {a +i a \tan \left (d x +c \right )}+\frac {a^{\frac {3}{2}} \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}+\frac {a^{2}}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{2}}\)	\(169\)

[In]

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

[Out]

2*I/d/a^2*(1/3*I*B*(a+I*a*tan(d*x+c))^(3/2)-I*a*B*(a+I*a*tan(d*x+c))^(1/2)-a*A*(a+I*a*tan(d*x+c))^(1/2)+1/4*a^
(3/2)*(A-I*B)*2^(1/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))-1/2*a^2*(A+I*B)/(a+I*a*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. 391 vs. \(2 (127) = 254\).

Time = 0.27 (sec) , antiderivative size = 391, normalized size of antiderivative = 2.46 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {3 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} + {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, \sqrt {2} {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}} \log \left (-\frac {4 \, {\left ({\left (-i \, A - B\right )} a e^{\left (i \, d x + i \, c\right )} - {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {A^{2} - 2 i \, A B - B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + 2 \, \sqrt {2} {\left ({\left (15 i \, A - 7 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 18 \, {\left (i \, A - B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i \, A - 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

[In]

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

[Out]

-1/12*(3*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a*d^2))*log(-4*(
(-I*A - B)*a*e^(I*d*x + I*c) + (a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A^2 -
2*I*A*B - B^2)/(a*d^2)))*e^(-I*d*x - I*c)/(I*A + B)) - 3*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c
))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a*d^2))*log(-4*((-I*A - B)*a*e^(I*d*x + I*c) - (a*d*e^(2*I*d*x + 2*I*c) + a*d)
*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(A^2 - 2*I*A*B - B^2)/(a*d^2)))*e^(-I*d*x - I*c)/(I*A + B)) + 2*sqrt(
2)*((15*I*A - 7*B)*e^(4*I*d*x + 4*I*c) + 18*(I*A - B)*e^(2*I*d*x + 2*I*c) + 3*I*A - 3*B)*sqrt(a/(e^(2*I*d*x +
2*I*c) + 1)))/(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.84 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {i \, {\left (3 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 8 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a + 24 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + i \, B\right )} a^{2} + \frac {12 \, {\left (A + i \, B\right )} a^{3}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}}{12 \, a^{3} d} \]

[In]

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

[Out]

-1/12*I*(3*sqrt(2)*(A - I*B)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sq
rt(I*a*tan(d*x + c) + a))) - 8*I*(I*a*tan(d*x + c) + a)^(3/2)*B*a + 24*sqrt(I*a*tan(d*x + c) + a)*(A + I*B)*a^
2 + 12*(A + I*B)*a^3/sqrt(I*a*tan(d*x + c) + a))/(a^3*d)

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 8.18 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.18 \[ \int \frac {\tan ^2(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {A\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {B}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a\,d}+\frac {2\,B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}-\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {-a}\,d}-\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {a}\,d} \]

[In]

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

[Out]

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