3.204 \(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=290 \[ \frac{8 a^4 (A-i B) \tan ^{m+1}(c+d x) \text{Hypergeometric2F1}(1,m+1,m+2,i \tan (c+d x))}{d (m+1)}-\frac{2 a^4 \left (A \left (2 m^3+19 m^2+60 m+64\right )-i B \left (2 m^3+19 m^2+60 m+67\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+2) (m+3) (m+4)}-\frac{2 \left (A (m+4)^2-i B \left (m^2+8 m+19\right )\right ) \left (a^4+i a^4 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2) (m+3) (m+4)}-\frac{(A (m+4)-i B (m+7)) \left (a^2+i a^2 \tan (c+d x)\right )^2 \tan ^{m+1}(c+d x)}{d (m+3) (m+4)}+\frac{i a B (a+i a \tan (c+d x))^3 \tan ^{m+1}(c+d x)}{d (m+4)} \]

[Out]

(-2*a^4*(A*(64 + 60*m + 19*m^2 + 2*m^3) - I*B*(67 + 60*m + 19*m^2 + 2*m^3))*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(
2 + m)*(3 + m)*(4 + m)) + (8*a^4*(A - I*B)*Hypergeometric2F1[1, 1 + m, 2 + m, I*Tan[c + d*x]]*Tan[c + d*x]^(1
+ m))/(d*(1 + m)) + (I*a*B*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^3)/(d*(4 + m)) - ((A*(4 + m) - I*B*(7 +
m))*Tan[c + d*x]^(1 + m)*(a^2 + I*a^2*Tan[c + d*x])^2)/(d*(3 + m)*(4 + m)) - (2*(A*(4 + m)^2 - I*B*(19 + 8*m
+ m^2))*Tan[c + d*x]^(1 + m)*(a^4 + I*a^4*Tan[c + d*x]))/(d*(2 + m)*(3 + m)*(4 + m))

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Rubi [A]  time = 1.06804, antiderivative size = 290, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {3594, 3592, 3537, 12, 64} \[ \frac{8 a^4 (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}-\frac{2 a^4 \left (A \left (2 m^3+19 m^2+60 m+64\right )-i B \left (2 m^3+19 m^2+60 m+67\right )\right ) \tan ^{m+1}(c+d x)}{d (m+1) (m+2) (m+3) (m+4)}-\frac{2 \left (A (m+4)^2-i B \left (m^2+8 m+19\right )\right ) \left (a^4+i a^4 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2) (m+3) (m+4)}-\frac{(A (m+4)-i B (m+7)) \left (a^2+i a^2 \tan (c+d x)\right )^2 \tan ^{m+1}(c+d x)}{d (m+3) (m+4)}+\frac{i a B (a+i a \tan (c+d x))^3 \tan ^{m+1}(c+d x)}{d (m+4)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

(-2*a^4*(A*(64 + 60*m + 19*m^2 + 2*m^3) - I*B*(67 + 60*m + 19*m^2 + 2*m^3))*Tan[c + d*x]^(1 + m))/(d*(1 + m)*(
2 + m)*(3 + m)*(4 + m)) + (8*a^4*(A - I*B)*Hypergeometric2F1[1, 1 + m, 2 + m, I*Tan[c + d*x]]*Tan[c + d*x]^(1
+ m))/(d*(1 + m)) + (I*a*B*Tan[c + d*x]^(1 + m)*(a + I*a*Tan[c + d*x])^3)/(d*(4 + m)) - ((A*(4 + m) - I*B*(7 +
m))*Tan[c + d*x]^(1 + m)*(a^2 + I*a^2*Tan[c + d*x])^2)/(d*(3 + m)*(4 + m)) - (2*(A*(4 + m)^2 - I*B*(19 + 8*m
+ m^2))*Tan[c + d*x]^(1 + m)*(a^4 + I*a^4*Tan[c + d*x]))/(d*(2 + m)*(3 + m)*(4 + m))

Rule 3594

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

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 3537

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(c*
d)/f, Subst[Int[(a + (b*x)/d)^m/(d^2 + c*x), x], x, d*Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m}, x] &&
NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && EqQ[c^2 + d^2, 0]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 64

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c^n*(b*x)^(m + 1)*Hypergeometric2F1[-n, m +
1, m + 2, -((d*x)/c)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[m] && (IntegerQ[n] || (GtQ[
c, 0] &&  !(EqQ[n, -2^(-1)] && EqQ[c^2 - d^2, 0] && GtQ[-(d/(b*c)), 0])))

Rubi steps

\begin{align*} \int \tan ^m(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx &=\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}+\frac{\int \tan ^m(c+d x) (a+i a \tan (c+d x))^3 (-a (i B (1+m)-A (4+m))+a (i A (4+m)+B (7+m)) \tan (c+d x)) \, dx}{4+m}\\ &=\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac{(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}+\frac{\int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 \left (-2 a^2 \left (i B \left (5+6 m+m^2\right )-A \left (8+6 m+m^2\right )\right )+2 a^2 \left (i A (4+m)^2+B \left (19+8 m+m^2\right )\right ) \tan (c+d x)\right ) \, dx}{12+7 m+m^2}\\ &=\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac{(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac{2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}+\frac{\int \tan ^m(c+d x) (a+i a \tan (c+d x)) \left (-2 a^3 \left (i B \left (29+44 m+17 m^2+2 m^3\right )-A \left (32+44 m+17 m^2+2 m^3\right )\right )+2 a^3 \left (i A \left (64+60 m+19 m^2+2 m^3\right )+B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=-\frac{2 a^4 \left (A \left (64+60 m+19 m^2+2 m^3\right )-i B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac{(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac{2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}+\frac{\int \tan ^m(c+d x) \left (8 a^4 (A-i B) (2+m) (3+m) (4+m)+8 a^4 (i A+B) (2+m) (3+m) (4+m) \tan (c+d x)\right ) \, dx}{24+26 m+9 m^2+m^3}\\ &=-\frac{2 a^4 \left (A \left (64+60 m+19 m^2+2 m^3\right )-i B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac{(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac{2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}+\frac{\left (64 i a^8 (A-i B)^2 (2+m) (3+m) (4+m)\right ) \operatorname{Subst}\left (\int \frac{8^{-m} \left (\frac{x}{a^4 (i A+B) (2+m) (3+m) (4+m)}\right )^m}{64 a^8 (i A+B)^2 (2+m)^2 (3+m)^2 (4+m)^2+8 a^4 (A-i B) (2+m) (3+m) (4+m) x} \, dx,x,8 a^4 (i A+B) (2+m) (3+m) (4+m) \tan (c+d x)\right )}{d}\\ &=-\frac{2 a^4 \left (A \left (64+60 m+19 m^2+2 m^3\right )-i B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac{(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac{2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}+\frac{\left (i 8^{2-m} a^8 (A-i B)^2 (2+m) (3+m) (4+m)\right ) \operatorname{Subst}\left (\int \frac{\left (\frac{x}{a^4 (i A+B) (2+m) (3+m) (4+m)}\right )^m}{64 a^8 (i A+B)^2 (2+m)^2 (3+m)^2 (4+m)^2+8 a^4 (A-i B) (2+m) (3+m) (4+m) x} \, dx,x,8 a^4 (i A+B) (2+m) (3+m) (4+m) \tan (c+d x)\right )}{d}\\ &=-\frac{2 a^4 \left (A \left (64+60 m+19 m^2+2 m^3\right )-i B \left (67+60 m+19 m^2+2 m^3\right )\right ) \tan ^{1+m}(c+d x)}{d (1+m) \left (24+26 m+9 m^2+m^3\right )}+\frac{8 a^4 (A-i B) \, _2F_1(1,1+m;2+m;i \tan (c+d x)) \tan ^{1+m}(c+d x)}{d (1+m)}+\frac{i a B \tan ^{1+m}(c+d x) (a+i a \tan (c+d x))^3}{d (4+m)}-\frac{(A (4+m)-i B (7+m)) \tan ^{1+m}(c+d x) \left (a^2+i a^2 \tan (c+d x)\right )^2}{d (3+m) (4+m)}-\frac{2 \left (A (4+m)^2-i B \left (19+8 m+m^2\right )\right ) \tan ^{1+m}(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d (2+m) \left (12+7 m+m^2\right )}\\ \end{align*}

Mathematica [F]  time = 19.6468, size = 0, normalized size = 0. \[ \int \tan ^m(c+d x) (a+i a \tan (c+d x))^4 (A+B \tan (c+d x)) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]),x]

[Out]

Integrate[Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])^4*(A + B*Tan[c + d*x]), x]

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Maple [F]  time = 0.532, size = 0, normalized size = 0. \begin{align*} \int \left ( \tan \left ( dx+c \right ) \right ) ^{m} \left ( a+ia\tan \left ( dx+c \right ) \right ) ^{4} \left ( A+B\tan \left ( dx+c \right ) \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^4*tan(d*x + c)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{16 \,{\left ({\left (A - i \, B\right )} a^{4} e^{\left (10 i \, d x + 10 i \, c\right )} +{\left (A + i \, B\right )} a^{4} e^{\left (8 i \, d x + 8 i \, c\right )}\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 (10 i \, d x + 10 i \, c\right )} + 5 \, e^{\left (8 i \, d x + 8 i \, c\right )} + 10 \, e^{\left (6 i \, d x + 6 i \, c\right )} + 10 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 5 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral(16*((A - I*B)*a^4*e^(10*I*d*x + 10*I*c) + (A + I*B)*a^4*e^(8*I*d*x + 8*I*c))*((-I*e^(2*I*d*x + 2*I*c)
+ I)/(e^(2*I*d*x + 2*I*c) + 1))^m/(e^(10*I*d*x + 10*I*c) + 5*e^(8*I*d*x + 8*I*c) + 10*e^(6*I*d*x + 6*I*c) + 1
0*e^(4*I*d*x + 4*I*c) + 5*e^(2*I*d*x + 2*I*c) + 1), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{4} \left (\int A \tan ^{m}{\left (c + d x \right )}\, dx + \int - 6 A \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int A \tan ^{4}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int B \tan{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int - 6 B \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int B \tan ^{5}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int 4 i A \tan{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int - 4 i A \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int 4 i B \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int - 4 i B \tan ^{4}{\left (c + d x \right )} \tan ^{m}{\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)**m*(a+I*a*tan(d*x+c))**4*(A+B*tan(d*x+c)),x)

[Out]

a**4*(Integral(A*tan(c + d*x)**m, x) + Integral(-6*A*tan(c + d*x)**2*tan(c + d*x)**m, x) + Integral(A*tan(c +
d*x)**4*tan(c + d*x)**m, x) + Integral(B*tan(c + d*x)*tan(c + d*x)**m, x) + Integral(-6*B*tan(c + d*x)**3*tan(
c + d*x)**m, x) + Integral(B*tan(c + d*x)**5*tan(c + d*x)**m, x) + Integral(4*I*A*tan(c + d*x)*tan(c + d*x)**m
, x) + Integral(-4*I*A*tan(c + d*x)**3*tan(c + d*x)**m, x) + Integral(4*I*B*tan(c + d*x)**2*tan(c + d*x)**m, x
) + Integral(-4*I*B*tan(c + d*x)**4*tan(c + d*x)**m, x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} \tan \left (d x + c\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^m*(a+I*a*tan(d*x+c))^4*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^4*tan(d*x + c)^m, x)