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

Optimal. Leaf size=159 \[ \frac {(-5 B+3 i A) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac {(-B+i A) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {4 (-B+i A) \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

[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

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Rubi [A]  time = 0.33, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.139, Rules used = {3595, 3592, 3527, 3480, 206} \[ \frac {(-5 B+3 i A) (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac {(-B+i A) \tan ^2(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {4 (-B+i A) \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {(B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3480

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 3527

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 3592

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 3595

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*} \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) \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) \operatorname {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) \tanh ^{-1}\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*}

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Mathematica [A]  time = 2.79, size = 147, normalized size = 0.92 \[ \frac {(A+B \tan (c+d x)) \left ((B+i A) \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+\frac {1}{3} \sec (c+d x) ((6 A+2 i B) \sin (2 (c+d x))+(5 B-9 i A) \cos (2 (c+d x))+9 (B-i A))\right )}{2 d \sqrt {a+i a \tan (c+d x)} (A \cos (c+d x)+B \sin (c+d x))} \]

Antiderivative was successfully verified.

[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)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))] + (Sec[c + d*x]*(9*((-I)*A + B) + ((-9*I)*A
+ 5*B)*Cos[2*(c + d*x)] + (6*A + (2*I)*B)*Sin[2*(c + d*x)]))/3)*(A + B*Tan[c + d*x]))/(2*d*(A*Cos[c + d*x] +
B*Sin[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.63, size = 396, normalized size = 2.49 \[ \frac {3 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} + 2 \, \sqrt {2} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 3 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {-\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}} \log \left (\frac {{\left ({\left (4 i \, A + 4 \, B\right )} a e^{\left (i \, d x + i \, c\right )} - 2 \, \sqrt {2} {\left (a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {2 \, A^{2} - 4 i \, A B - 2 \, B^{2}}{a d^{2}}}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) + \sqrt {2} {\left ({\left (-30 i \, A + 14 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (-36 i \, A + 36 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 6 i \, A + 6 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{12 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[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*(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-(2*A^2 - 4*I*A*B - 2*B^2)/(a*d^2))*log(((4*I*A +
4*B)*a*e^(I*d*x + I*c) + 2*sqrt(2)*(a*d*e^(2*I*d*x + 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(2
*A^2 - 4*I*A*B - 2*B^2)/(a*d^2)))*e^(-I*d*x - I*c)/(I*A + B)) - 3*(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*
c))*sqrt(-(2*A^2 - 4*I*A*B - 2*B^2)/(a*d^2))*log(((4*I*A + 4*B)*a*e^(I*d*x + I*c) - 2*sqrt(2)*(a*d*e^(2*I*d*x
+ 2*I*c) + a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-(2*A^2 - 4*I*A*B - 2*B^2)/(a*d^2)))*e^(-I*d*x - I*c)/(
I*A + B)) + sqrt(2)*((-30*I*A + 14*B)*e^(4*I*d*x + 4*I*c) + (-36*I*A + 36*B)*e^(2*I*d*x + 2*I*c) - 6*I*A + 6*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))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \tan \left (d x + c\right )^{2}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [A]  time = 0.26, size = 127, normalized size = 0.80 \[ -\frac {2 i \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+i B a \sqrt {a +i a \tan \left (d x +c \right )}+A \sqrt {a +i a \tan \left (d x +c \right )}\, a -\frac {a^{\frac {3}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}+\frac {a^{2} \left (i B +A \right )}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 1.16, size = 134, normalized size = 0.84 \[ -\frac {i \, {\left (3 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) - 8 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} B a + 24 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + i \, B\right )} a^{2} + \frac {12 \, {\left (A + i \, B\right )} a^{3}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}}{12 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [B]  time = 7.67, size = 188, normalized size = 1.18 \[ -\frac {A\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {B}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}-\frac {A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,2{}\mathrm {i}}{a\,d}+\frac {2\,B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}-\frac {2\,B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}-\frac {\sqrt {2}\,A\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\sqrt {-a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {-a}\,d}-\frac {\sqrt {2}\,B\,\mathrm {atan}\left (\frac {\sqrt {2}\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{2\,\sqrt {a}}\right )\,1{}\mathrm {i}}{2\,\sqrt {a}\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{2}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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