3.210 \(\int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=308 \[ -\frac{(1-m) \left (-A \left (2 m^2-7 m+3\right )+i B \left (-2 m^2+m+3\right )\right ) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+1}{2},\frac{m+3}{2},-\tan ^2(c+d x)\right )}{24 a^3 d (m+1)}+\frac{(2-m) m (i A (5-2 m)+2 B m+B) \tan ^{m+2}(c+d x) \text{Hypergeometric2F1}\left (1,\frac{m+2}{2},\frac{m+4}{2},-\tan ^2(c+d x)\right )}{24 a^3 d (m+2)}+\frac{(2-m) (A (5-2 m)-i (2 B m+B)) \tan ^{m+1}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(A (7-2 m)+i B (1-2 m)) \tan ^{m+1}(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{6 d (a+i a \tan (c+d x))^3} \]

[Out]

-((1 - m)*(I*B*(3 + m - 2*m^2) - A*(3 - 7*m + 2*m^2))*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]
^2]*Tan[c + d*x]^(1 + m))/(24*a^3*d*(1 + m)) + ((2 - m)*m*(B + I*A*(5 - 2*m) + 2*B*m)*Hypergeometric2F1[1, (2
+ m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/(24*a^3*d*(2 + m)) + ((A + I*B)*Tan[c + d*x]^(1 + m)
)/(6*d*(a + I*a*Tan[c + d*x])^3) + ((I*B*(1 - 2*m) + A*(7 - 2*m))*Tan[c + d*x]^(1 + m))/(24*a*d*(a + I*a*Tan[c
+ d*x])^2) + ((2 - m)*(A*(5 - 2*m) - I*(B + 2*B*m))*Tan[c + d*x]^(1 + m))/(24*d*(a^3 + I*a^3*Tan[c + d*x]))

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Rubi [A]  time = 0.832992, antiderivative size = 308, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used = {3596, 3538, 3476, 364} \[ -\frac{(1-m) \left (-A \left (2 m^2-7 m+3\right )+i B \left (-2 m^2+m+3\right )\right ) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac{m+1}{2};\frac{m+3}{2};-\tan ^2(c+d x)\right )}{24 a^3 d (m+1)}+\frac{(2-m) m (i A (5-2 m)+2 B m+B) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac{m+2}{2};\frac{m+4}{2};-\tan ^2(c+d x)\right )}{24 a^3 d (m+2)}+\frac{(2-m) (A (5-2 m)-i (2 B m+B)) \tan ^{m+1}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{(A (7-2 m)+i B (1-2 m)) \tan ^{m+1}(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{(A+i B) \tan ^{m+1}(c+d x)}{6 d (a+i a \tan (c+d x))^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((1 - m)*(I*B*(3 + m - 2*m^2) - A*(3 - 7*m + 2*m^2))*Hypergeometric2F1[1, (1 + m)/2, (3 + m)/2, -Tan[c + d*x]
^2]*Tan[c + d*x]^(1 + m))/(24*a^3*d*(1 + m)) + ((2 - m)*m*(B + I*A*(5 - 2*m) + 2*B*m)*Hypergeometric2F1[1, (2
+ m)/2, (4 + m)/2, -Tan[c + d*x]^2]*Tan[c + d*x]^(2 + m))/(24*a^3*d*(2 + m)) + ((A + I*B)*Tan[c + d*x]^(1 + m)
)/(6*d*(a + I*a*Tan[c + d*x])^3) + ((I*B*(1 - 2*m) + A*(7 - 2*m))*Tan[c + d*x]^(1 + m))/(24*a*d*(a + I*a*Tan[c
+ d*x])^2) + ((2 - m)*(A*(5 - 2*m) - I*(B + 2*B*m))*Tan[c + d*x]^(1 + m))/(24*d*(a^3 + I*a^3*Tan[c + d*x]))

Rule 3596

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

Rule 3538

Int[((b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*T
an[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Tan[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x] && NeQ
[c^2 + d^2, 0] &&  !IntegerQ[2*m]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin{align*} \int \frac{\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^3} \, dx &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{\int \frac{\tan ^m(c+d x) (a (A (5-m)-i B (1+m))-a (i A-B) (2-m) \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx}{6 a^2}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(i B (1-2 m)+A (7-2 m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{\int \frac{\tan ^m(c+d x) \left (-a^2 \left (i B \left (5+3 m-2 m^2\right )-A \left (13-9 m+2 m^2\right )\right )+a^2 (B (1-2 m)-i A (7-2 m)) (1-m) \tan (c+d x)\right )}{a+i a \tan (c+d x)} \, dx}{24 a^4}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(i B (1-2 m)+A (7-2 m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{(2-m) (A (5-2 m)-i (B+2 B m)) \tan ^{1+m}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{\int \tan ^m(c+d x) \left (-2 a^3 (1-m) \left (i B \left (3+m-2 m^2\right )-A \left (3-7 m+2 m^2\right )\right )+2 a^3 (2-m) m (B+i A (5-2 m)+2 B m) \tan (c+d x)\right ) \, dx}{48 a^6}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(i B (1-2 m)+A (7-2 m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{(2-m) (A (5-2 m)-i (B+2 B m)) \tan ^{1+m}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{((2-m) m (B+i A (5-2 m)+2 B m)) \int \tan ^{1+m}(c+d x) \, dx}{24 a^3}-\frac{\left ((1-m) \left (i B \left (3+m-2 m^2\right )-A \left (3-7 m+2 m^2\right )\right )\right ) \int \tan ^m(c+d x) \, dx}{24 a^3}\\ &=\frac{(A+i B) \tan ^{1+m}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(i B (1-2 m)+A (7-2 m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{(2-m) (A (5-2 m)-i (B+2 B m)) \tan ^{1+m}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}+\frac{((2-m) m (B+i A (5-2 m)+2 B m)) \operatorname{Subst}\left (\int \frac{x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{24 a^3 d}-\frac{\left ((1-m) \left (i B \left (3+m-2 m^2\right )-A \left (3-7 m+2 m^2\right )\right )\right ) \operatorname{Subst}\left (\int \frac{x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{24 a^3 d}\\ &=-\frac{(1-m) \left (i B \left (3+m-2 m^2\right )-A \left (3-7 m+2 m^2\right )\right ) \, _2F_1\left (1,\frac{1+m}{2};\frac{3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{24 a^3 d (1+m)}+\frac{(2-m) m (B+i A (5-2 m)+2 B m) \, _2F_1\left (1,\frac{2+m}{2};\frac{4+m}{2};-\tan ^2(c+d x)\right ) \tan ^{2+m}(c+d x)}{24 a^3 d (2+m)}+\frac{(A+i B) \tan ^{1+m}(c+d x)}{6 d (a+i a \tan (c+d x))^3}+\frac{(i B (1-2 m)+A (7-2 m)) \tan ^{1+m}(c+d x)}{24 a d (a+i a \tan (c+d x))^2}+\frac{(2-m) (A (5-2 m)-i (B+2 B m)) \tan ^{1+m}(c+d x)}{24 d \left (a^3+i a^3 \tan (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 113.738, size = 712, normalized size = 2.31 \[ -\frac{i e^{-3 i c} \left (-\frac{i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \left (\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-m} \sec ^2(c+d x) (\cos (d x)+i \sin (d x))^3 (A+B \tan (c+d x)) \left (\frac{3 e^{6 i c} 2^{3-m} \left (A \left (-2 m^2+7 m-4\right )+i B \left (-2 m^2+m+2\right )\right ) \left (-1+e^{2 i (c+d x)}\right )^{m+1} \text{Hypergeometric2F1}\left (m-2,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )}{m+1}+\frac{e^{6 i c} 2^{1-m} \left (A \left (4 m^3-18 m^2+20 m-3\right )+i B \left (4 m^3-6 m^2-4 m+3\right )\right ) \left (\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (2^m (m+1) \text{Hypergeometric2F1}\left (1,m,m+1,\frac{1-e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )-\left (1+e^{2 i (c+d x)}\right )^m \left (2 m \left (-1+e^{2 i (c+d x)}\right ) \text{Hypergeometric2F1}\left (m-1,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+(m+1) \text{Hypergeometric2F1}\left (m,m,m+1,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+m \left (-1+e^{2 i (c+d x)}\right ) \text{Hypergeometric2F1}\left (m,m+1,m+2,\frac{1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right )\right )}{m (m+1)}-2 \left (A \left (-2 m^2+7 m-4\right )+i B \left (-2 m^2+m+2\right )\right ) e^{-2 i (d x-2 c)} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{3-m}+2 (A+i B) e^{-6 i d x} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{3-m}+(A (5-2 m)-i (2 B m+B)) e^{2 i (c-2 d x)} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{3-m}\right )}{96 d (a+i a \tan (c+d x))^3 (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((-I/96)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*((2*(A + I*B)*(-1 + E^((2*I)*(c + d*x
)))^(1 + m)*(1 + E^((2*I)*(c + d*x)))^(3 - m))/E^((6*I)*d*x) + E^((2*I)*(c - 2*d*x))*(-1 + E^((2*I)*(c + d*x))
)^(1 + m)*(1 + E^((2*I)*(c + d*x)))^(3 - m)*(A*(5 - 2*m) - I*(B + 2*B*m)) - (2*(-1 + E^((2*I)*(c + d*x)))^(1 +
m)*(1 + E^((2*I)*(c + d*x)))^(3 - m)*(I*B*(2 + m - 2*m^2) + A*(-4 + 7*m - 2*m^2)))/E^((2*I)*(-2*c + d*x)) + (
3*2^(3 - m)*E^((6*I)*c)*(-1 + E^((2*I)*(c + d*x)))^(1 + m)*(I*B*(2 + m - 2*m^2) + A*(-4 + 7*m - 2*m^2))*Hyperg
eometric2F1[-2 + m, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x)))/2])/(1 + m) + (2^(1 - m)*E^((6*I)*c)*((-1 + E^((2*
I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^m*(A*(-3 + 20*m - 18*m^2 + 4*m^3) + I*B*(3 - 4*m - 6*m^2 + 4*m^3))*(
2^m*(1 + m)*Hypergeometric2F1[1, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))] - (1 + E^((2*I
)*(c + d*x)))^m*(2*(-1 + E^((2*I)*(c + d*x)))*m*Hypergeometric2F1[-1 + m, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x
)))/2] + (1 + m)*Hypergeometric2F1[m, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/2] + (-1 + E^((2*I)*(c + d*x)))*m*Hy
pergeometric2F1[m, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x)))/2])))/(m*(1 + m)))*Sec[c + d*x]^2*(Cos[d*x] + I*Sin
[d*x])^3*(A + B*Tan[c + d*x]))/(d*E^((3*I)*c)*((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^m*(A*Cos[
c + d*x] + B*Sin[c + d*x])*(a + I*a*Tan[c + d*x])^3)

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Maple [F]  time = 1.496, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( A+B\tan \left ( dx+c \right ) \right ) }{ \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x)

[Out]

int(tan(d*x+c)^m*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^3,x)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left ({\left (A - i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} +{\left (3 \, A - i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} +{\left (3 \, A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\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 )^{m} e^{\left (-6 i \, d x - 6 i \, c\right )}}{8 \, a^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(1/8*((A - I*B)*e^(6*I*d*x + 6*I*c) + (3*A - I*B)*e^(4*I*d*x + 4*I*c) + (3*A + I*B)*e^(2*I*d*x + 2*I*c
) + A + I*B)*((-I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) + 1))^m*e^(-6*I*d*x - 6*I*c)/a^3, x)

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Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: AttributeError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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