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

Optimal. Leaf size=226 \[ \frac {(1-m) (A (1-m)-i B (m+1)) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{4 a^2 d (m+1)}+\frac {m (B m+i A (2-m)) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )}{4 a^2 d (m+2)}+\frac {(A (2-m)-i B m) \tan ^{m+1}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

[Out]

1/4*(1-m)*(A*(1-m)-I*B*(1+m))*hypergeom([1, 1/2+1/2*m],[3/2+1/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(1+m)/a^2/d/(1+m)
+1/4*(A*(2-m)-I*B*m)*tan(d*x+c)^(1+m)/a^2/d/(1+I*tan(d*x+c))+1/4*m*(I*A*(2-m)+B*m)*hypergeom([1, 1+1/2*m],[2+1
/2*m],-tan(d*x+c)^2)*tan(d*x+c)^(2+m)/a^2/d/(2+m)+1/4*(A+I*B)*tan(d*x+c)^(1+m)/d/(a+I*a*tan(d*x+c))^2

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Rubi [A]  time = 0.48, antiderivative size = 226, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3596, 3538, 3476, 364} \[ \frac {(1-m) (A (1-m)-i B (m+1)) \tan ^{m+1}(c+d x) \, _2F_1\left (1,\frac {m+1}{2};\frac {m+3}{2};-\tan ^2(c+d x)\right )}{4 a^2 d (m+1)}+\frac {m (B m+i A (2-m)) \tan ^{m+2}(c+d x) \, _2F_1\left (1,\frac {m+2}{2};\frac {m+4}{2};-\tan ^2(c+d x)\right )}{4 a^2 d (m+2)}+\frac {(A (2-m)-i B m) \tan ^{m+1}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \tan ^{m+1}(c+d x)}{4 d (a+i a \tan (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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 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 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]

Rubi steps

\begin {align*} \int \frac {\tan ^m(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^2} \, dx &=\frac {(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \frac {\tan ^m(c+d x) (a (A (3-m)-i B (1+m))-a (i A-B) (1-m) \tan (c+d x))}{a+i a \tan (c+d x)} \, dx}{4 a^2}\\ &=\frac {(A (2-m)-i B m) \tan ^{1+m}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {\int \tan ^m(c+d x) \left (2 a^2 (1-m) (A (1-m)-i B (1+m))+2 a^2 m (i A (2-m)+B m) \tan (c+d x)\right ) \, dx}{8 a^4}\\ &=\frac {(A (2-m)-i B m) \tan ^{1+m}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {(m (i A (2-m)+B m)) \int \tan ^{1+m}(c+d x) \, dx}{4 a^2}+\frac {((1-m) (A (1-m)-i B (1+m))) \int \tan ^m(c+d x) \, dx}{4 a^2}\\ &=\frac {(A (2-m)-i B m) \tan ^{1+m}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}+\frac {(m (i A (2-m)+B m)) \operatorname {Subst}\left (\int \frac {x^{1+m}}{1+x^2} \, dx,x,\tan (c+d x)\right )}{4 a^2 d}+\frac {((1-m) (A (1-m)-i B (1+m))) \operatorname {Subst}\left (\int \frac {x^m}{1+x^2} \, dx,x,\tan (c+d x)\right )}{4 a^2 d}\\ &=\frac {(1-m) (A (1-m)-i B (1+m)) \, _2F_1\left (1,\frac {1+m}{2};\frac {3+m}{2};-\tan ^2(c+d x)\right ) \tan ^{1+m}(c+d x)}{4 a^2 d (1+m)}+\frac {(A (2-m)-i B m) \tan ^{1+m}(c+d x)}{4 a^2 d (1+i \tan (c+d x))}+\frac {m (i A (2-m)+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)}{4 a^2 d (2+m)}+\frac {(A+i B) \tan ^{1+m}(c+d x)}{4 d (a+i a \tan (c+d x))^2}\\ \end {align*}

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Mathematica [B]  time = 9.19, size = 565, normalized size = 2.50 \[ -\frac {i e^{-2 i c} \left (-\frac {i \left (-1+e^{2 i (c+d x)}\right )}{1+e^{2 i (c+d x)}}\right )^m \left (\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{-m} \sec (c+d x) (\cos (d x)+i \sin (d x))^2 (A+B \tan (c+d x)) \left (\frac {e^{4 i c} 2^{1-m} \left (A \left (2 m^2-4 m+1\right )+i B \left (2 m^2-1\right )\right ) \left (\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (\left (1+e^{2 i (c+d x)}\right )^m \left ((m+1) \, _2F_1\left (m,m;m+1;\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )+m \left (-1+e^{2 i (c+d x)}\right ) \, _2F_1\left (m,m+1;m+2;\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )\right )-2^m (m+1) \, _2F_1\left (1,m;m+1;\frac {1-e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}\right )\right )}{m (m+1)}+\frac {e^{4 i c} 2^{2-m} (A (2 m-3)+i (2 B m+B)) \left (-1+e^{2 i (c+d x)}\right )^{m+1} \, _2F_1\left (m-1,m+1;m+2;\frac {1}{2} \left (1-e^{2 i (c+d x)}\right )\right )}{m+1}+(A+i B) e^{-4 i d x} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{2-m}+(A (3-2 m)-i (2 B m+B)) e^{2 i (c-d x)} \left (-1+e^{2 i (c+d x)}\right )^{m+1} \left (1+e^{2 i (c+d x)}\right )^{2-m}\right )}{16 d (a+i a \tan (c+d x))^2 (A \cos (c+d x)+B \sin (c+d x))} \]

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])^2,x]

[Out]

((-1/16*I)*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*(((A + I*B)*(-1 + E^((2*I)*(c + d*x
)))^(1 + m)*(1 + E^((2*I)*(c + d*x)))^(2 - m))/E^((4*I)*d*x) + E^((2*I)*(c - d*x))*(-1 + E^((2*I)*(c + d*x)))^
(1 + m)*(1 + E^((2*I)*(c + d*x)))^(2 - m)*(A*(3 - 2*m) - I*(B + 2*B*m)) + (2^(2 - m)*E^((4*I)*c)*(-1 + E^((2*I
)*(c + d*x)))^(1 + m)*(A*(-3 + 2*m) + I*(B + 2*B*m))*Hypergeometric2F1[-1 + m, 1 + m, 2 + m, (1 - E^((2*I)*(c
+ d*x)))/2])/(1 + m) + (2^(1 - m)*E^((4*I)*c)*((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^m*(I*B*(-
1 + 2*m^2) + A*(1 - 4*m + 2*m^2))*(-(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*((1 + m)*Hypergeometric2F1[m, m, 1 + m, (1 - E^((2*I)*(c
+ d*x)))/2] + (-1 + E^((2*I)*(c + d*x)))*m*Hypergeometric2F1[m, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x)))/2])))
/(m*(1 + m)))*Sec[c + d*x]*(Cos[d*x] + I*Sin[d*x])^2*(A + B*Tan[c + d*x]))/(d*E^((2*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])^2)

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left ({\left (A - i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, A e^{\left (2 i \, d x + 2 i \, c\right )} + A + i \, B\right )} \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m} e^{\left (-4 i \, d x - 4 i \, c\right )}}{4 \, a^{2}}, x\right ) \]

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))^2,x, algorithm="fricas")

[Out]

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

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

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))^2,x, algorithm="giac")

[Out]

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

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maple [F]  time = 4.65, size = 0, normalized size = 0.00 \[ \int \frac {\left (\tan ^{m}\left (d x +c \right )\right ) \left (A +B \tan \left (d x +c \right )\right )}{\left (a +i a \tan \left (d x +c \right )\right )^{2}}\, dx \]

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))^2,x)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

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))^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \frac {\int \frac {A \tan ^{m}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx + \int \frac {B \tan {\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}}{\tan ^{2}{\left (c + d x \right )} - 2 i \tan {\left (c + d x \right )} - 1}\, dx}{a^{2}} \]

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))**2,x)

[Out]

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

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