3.114 \(\int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=80 \[ \frac{2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{2 a (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.120946, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3592, 3528, 3533, 205} \[ \frac{2 \sqrt [4]{-1} a (B+i A) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{2 a (B+i A) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3528

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] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]

Rule 3533

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 205

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

Rubi steps

\begin{align*} \int \sqrt{\tan (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx &=\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\int \sqrt{\tan (c+d x)} (a (A-i B)+a (i A+B) \tan (c+d x)) \, dx\\ &=\frac{2 a (i A+B) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\int \frac{-a (i A+B)+a (A-i B) \tan (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{2 a (i A+B) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{\left (2 a^2 (i A+B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{-a (i A+B)-a (A-i B) x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{2 \sqrt [4]{-1} a (i A+B) \tan ^{-1}\left ((-1)^{3/4} \sqrt{\tan (c+d x)}\right )}{d}+\frac{2 a (i A+B) \sqrt{\tan (c+d x)}}{d}+\frac{2 i a B \tan ^{\frac{3}{2}}(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.80183, size = 112, normalized size = 1.4 \[ \frac{2 a \sqrt{\tan (c+d x)} \left (\sqrt{i \tan (c+d x)} (3 i A+i B \tan (c+d x)+3 B)+(-3 B-3 i A) \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{3 d \sqrt{i \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sqrt[Tan[c + d*x]]*(((-3*I)*A - 3*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]
+ Sqrt[I*Tan[c + d*x]]*((3*I)*A + 3*B + I*B*Tan[c + d*x])))/(3*d*Sqrt[I*Tan[c + d*x]])

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Maple [B]  time = 0.011, size = 475, normalized size = 5.9 \begin{align*}{\frac{{\frac{2\,i}{3}}aB}{d} \left ( \tan \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}}}+{\frac{2\,iaA}{d}\sqrt{\tan \left ( dx+c \right ) }}+2\,{\frac{aB\sqrt{\tan \left ( dx+c \right ) }}{d}}-{\frac{{\frac{i}{2}}aA\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{\frac{i}{4}}aA\sqrt{2}}{d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{{\frac{i}{2}}aA\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{aB\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{aB\sqrt{2}}{4\,d}\ln \left ({ \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{aB\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{\frac{i}{4}}aB\sqrt{2}}{d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }-{\frac{{\frac{i}{2}}aB\sqrt{2}}{d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }-{\frac{{\frac{i}{2}}aB\sqrt{2}}{d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{Aa\sqrt{2}}{4\,d}\ln \left ({ \left ( 1-\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) }+\tan \left ( dx+c \right ) \right ) ^{-1}} \right ) }+{\frac{Aa\sqrt{2}}{2\,d}\arctan \left ( -1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) }+{\frac{Aa\sqrt{2}}{2\,d}\arctan \left ( 1+\sqrt{2}\sqrt{\tan \left ( dx+c \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.86613, size = 230, normalized size = 2.88 \begin{align*} -\frac{-8 i \, B a \tan \left (d x + c\right )^{\frac{3}{2}} + 24 \,{\left (-i \, A - B\right )} a \sqrt{\tan \left (d x + c\right )} + 3 \,{\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}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 1.83584, size = 977, normalized size = 12.21 \begin{align*} -\frac{3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac{{\left (2 \,{\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \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}\right ) - 3 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (\frac{{\left (2 \,{\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} -{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt{\frac{{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} \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}\right ) -{\left ({\left (24 i \, A + 32 \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} +{\left (24 i \, A + 16 \, B\right )} a\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}}}{12 \,{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((4*I*A^2 + 8*A*B - 4*I*B^2)*a^2/d^2)*log((2*(A - I*B)*a*e^(2*I*d*x +
2*I*c) + (d*e^(2*I*d*x + 2*I*c) + d)*sqrt((4*I*A^2 + 8*A*B - 4*I*B^2)*a^2/d^2)*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*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt((4*I*
A^2 + 8*A*B - 4*I*B^2)*a^2/d^2)*log((2*(A - I*B)*a*e^(2*I*d*x + 2*I*c) - (d*e^(2*I*d*x + 2*I*c) + d)*sqrt((4*I
*A^2 + 8*A*B - 4*I*B^2)*a^2/d^2)*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)) - ((24*I*A + 32*B)*a*e^(2*I*d*x + 2*I*c) + (24*I*A + 16*B)*a)*sqrt((-I*e^(2*I*d*x + 2*I*c
) + I)/(e^(2*I*d*x + 2*I*c) + 1)))/(d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int A \sqrt{\tan{\left (c + d x \right )}}\, dx + \int B \tan ^{\frac{3}{2}}{\left (c + d x \right )}\, dx + \int i A \tan ^{\frac{3}{2}}{\left (c + d x \right )}\, dx + \int i B \tan ^{\frac{5}{2}}{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.20468, size = 112, normalized size = 1.4 \begin{align*} -\frac{\left (i - 1\right ) \, \sqrt{2}{\left (4 \, A a - 4 i \, B a\right )} \arctan \left (-\left (\frac{1}{2} i - \frac{1}{2}\right ) \, \sqrt{2} \sqrt{\tan \left (d x + c\right )}\right )}{4 \, d} - \frac{-2 i \, B a d^{2} \tan \left (d x + c\right )^{\frac{3}{2}} - 6 i \, A a d^{2} \sqrt{\tan \left (d x + c\right )} - 6 \, B a d^{2} \sqrt{\tan \left (d x + c\right )}}{3 \, d^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(1/4*I - 1/4)*sqrt(2)*(4*A*a - 4*I*B*a)*arctan(-(1/2*I - 1/2)*sqrt(2)*sqrt(tan(d*x + c)))/d - 1/3*(-2*I*B*a*d
^2*tan(d*x + c)^(3/2) - 6*I*A*a*d^2*sqrt(tan(d*x + c)) - 6*B*a*d^2*sqrt(tan(d*x + c)))/d^3