3.1.90 \(\int \frac {\tan ^3(c+d x) (A+B \tan (c+d x))}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [90]

Optimal. Leaf size=205 \[ \frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d}+\frac {(i A-B) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}-\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d} \]

[Out]

1/2*(A-I*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)+4/5*(5*A+7*I*B)*(a+I*a*tan
(d*x+c))^(1/2)/a/d-1/5*(5*A+7*I*B)*(a+I*a*tan(d*x+c))^(1/2)*tan(d*x+c)^2/a/d+(I*A-B)*tan(d*x+c)^3/d/(a+I*a*tan
(d*x+c))^(1/2)-1/15*(25*A+23*I*B)*(a+I*a*tan(d*x+c))^(3/2)/a^2/d

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Rubi [A]
time = 0.34, antiderivative size = 205, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3676, 3678, 3673, 3608, 3561, 212} \begin {gather*} -\frac {(25 A+23 i B) (a+i a \tan (c+d x))^{3/2}}{15 a^2 d}+\frac {(-B+i A) \tan ^3(c+d x)}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(5 A+7 i B) \tan ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{5 a d}+\frac {4 (5 A+7 i B) \sqrt {a+i a \tan (c+d x)}}{5 a d}+\frac {(A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{\sqrt {2} \sqrt {a} d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((A - I*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
]^3)/(d*Sqrt[a + I*a*Tan[c + d*x]]) + (4*(5*A + (7*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/(5*a*d) - ((5*A + (7*I)*B
)*Tan[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(5*a*d) - ((25*A + (23*I)*B)*(a + I*a*Tan[c + d*x])^(3/2))/(15*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]

Rule 3678

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

Rubi steps

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

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Mathematica [A]
time = 3.17, size = 176, normalized size = 0.86 \begin {gather*} \frac {\left ((A-i B) \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )+\frac {1}{30} \sec ^2(c+d x) (5 (23 A+37 i B) \cos (c+d x)+(25 A+59 i B) \cos (3 (c+d x))+4 i (5 A+16 i B+(5 A+22 i B) \cos (2 (c+d x))) \sin (c+d x))\right ) (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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((A - I*B)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))] + (Sec[c + d*x]^2*(5*(23*A + (37*I)*B)*Cos[
c + d*x] + (25*A + (59*I)*B)*Cos[3*(c + d*x)] + (4*I)*(5*A + (16*I)*B + (5*A + (22*I)*B)*Cos[2*(c + d*x)])*Sin
[c + d*x]))/30)*(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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Maple [A]
time = 0.11, size = 168, normalized size = 0.82

method	result	size
derivativedivides	\(-\frac {2 \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}-a^{2} A \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a^{3} \left (i B +A \right )}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{3}}\)	\(168\)
default	\(-\frac {2 \left (-\frac {i B \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}+\frac {2 i B a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}+\frac {A a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {3}{2}}}{3}-2 i a^{2} B \sqrt {a +i a \tan \left (d x +c \right )}-a^{2} A \sqrt {a +i a \tan \left (d x +c \right )}-\frac {a^{\frac {5}{2}} \left (-i B +A \right ) \sqrt {2}\, \arctanh \left (\frac {\sqrt {a +i a \tan \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right )}{4}-\frac {a^{3} \left (i B +A \right )}{2 \sqrt {a +i a \tan \left (d x +c \right )}}\right )}{d \,a^{3}}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 157, normalized size = 0.77 \begin {gather*} -\frac {15 \, \sqrt {2} {\left (A - i \, B\right )} a^{\frac {7}{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 ) - 24 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} B a + 40 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (A + 2 i \, B\right )} a^{2} - 120 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (A + 2 i \, B\right )} a^{3} - \frac {60 \, {\left (A + i \, B\right )} a^{4}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}}{60 \, a^{4} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*sqrt(2)*(A - I*B)*a^(7/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqr
t(I*a*tan(d*x + c) + a))) - 24*I*(I*a*tan(d*x + c) + a)^(5/2)*B*a + 40*(I*a*tan(d*x + c) + a)^(3/2)*(A + 2*I*B
)*a^2 - 120*sqrt(I*a*tan(d*x + c) + a)*(A + 2*I*B)*a^3 - 60*(A + I*B)*a^4/sqrt(I*a*tan(d*x + c) + a))/(a^4*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 447 vs. \(2 (163) = 326\).
time = 2.27, size = 447, normalized size = 2.18 \begin {gather*} -\frac {15 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, 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 (i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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 ) - 15 \, \sqrt {2} {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, 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 (-i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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 (35 \, A + 103 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 5 \, {\left (25 \, A + 41 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 15 \, {\left (7 \, A + 11 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \, A + 15 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{60 \, {\left (a d e^{\left (5 i \, d x + 5 i \, c\right )} + 2 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*sqrt(2)*(a*d*e^(5*I*d*x + 5*I*c) + 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) + (I*a*d*e^(2*I*d*x + 2*I*c) + I*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)) - 15*sqrt(2)*(a*d*e^(5*I*d
*x + 5*I*c) + 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) + (-I*a*d*e^(2*I*d*x + 2*I*c) - I*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)*((35*A + 103*I*B)*e^(6*I*d*x + 6*I*c) + 5*(2
5*A + 41*I*B)*e^(4*I*d*x + 4*I*c) + 15*(7*A + 11*I*B)*e^(2*I*d*x + 2*I*c) + 15*A + 15*I*B)*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1)))/(a*d*e^(5*I*d*x + 5*I*c) + 2*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \tan ^{3}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(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)**3/sqrt(I*a*(tan(c + d*x) - I)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(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)^3/sqrt(I*a*tan(d*x + c) + a), x)

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Mupad [B]
time = 7.95, size = 236, normalized size = 1.15 \begin {gather*} \frac {A}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {B\,1{}\mathrm {i}}{d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}+\frac {2\,A\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{a\,d}-\frac {2\,A\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}}{3\,a^2\,d}+\frac {B\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a\,d}-\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,4{}\mathrm {i}}{3\,a^2\,d}+\frac {B\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,2{}\mathrm {i}}{5\,a^3\,d}+\frac {\sqrt {2}\,B\,\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}\,A\,\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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

A/(d*(a + a*tan(c + d*x)*1i)^(1/2)) + (B*1i)/(d*(a + a*tan(c + d*x)*1i)^(1/2)) + (2*A*(a + a*tan(c + d*x)*1i)^
(1/2))/(a*d) - (2*A*(a + a*tan(c + d*x)*1i)^(3/2))/(3*a^2*d) + (B*(a + a*tan(c + d*x)*1i)^(1/2)*4i)/(a*d) - (B
*(a + a*tan(c + d*x)*1i)^(3/2)*4i)/(3*a^2*d) + (B*(a + a*tan(c + d*x)*1i)^(5/2)*2i)/(5*a^3*d) + (2^(1/2)*B*ata
n((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)*A*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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