3.206 \(\int \tan ^m(c+d x) (a+i a \tan (c+d x))^2 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=132 \[ \frac {2 a^2 (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}+\frac {i a^2 (B+(m+2) (B+i A)) \tan ^{m+1}(c+d x)}{d (m+1) (m+2)}+\frac {i B \left (a^2+i a^2 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2)} \]

[Out]

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

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Rubi [A]  time = 0.36, antiderivative size = 132, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {2 a^2 (A-i B) \tan ^{m+1}(c+d x) \, _2F_1(1,m+1;m+2;i \tan (c+d x))}{d (m+1)}+\frac {i a^2 (B+(m+2) (B+i A)) \tan ^{m+1}(c+d x)}{d (m+1) (m+2)}+\frac {i B \left (a^2+i a^2 \tan (c+d x)\right ) \tan ^{m+1}(c+d x)}{d (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])))

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 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 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]

Rubi steps

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

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Mathematica [B]  time = 10.80, size = 1587, normalized size = 12.02 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(I*A*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*Cos[c + d*x]^3*(2^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*Hypergeometri
c2F1[m, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/2])*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(2^m*d*m*(Cos[d
*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (B*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(
c + d*x))))^m*Cos[c + d*x]^3*(2^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*Hypergeometric2F1[m, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/2])*(a + I*
a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(2^m*d*m*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))
- ((2*I)*(A - I*B)*(-1 + E^((2*I)*(c + d*x)))^m*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^
m*Cos[c + d*x]^3*(-(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*m)) - ((1 + E^((2*I)*c))*(-1 + E^((2*I)*(c + d*x)))*(1 + E^((2*I)*(c + d*x)))^(-1 - m)*
Hypergeometric2F1[1, 1 + m, 2 + m, (1 - E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))])/(1 + m) + Hypergeomet
ric2F1[m, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/2]/(2^m*m))*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(d*E^
((2*I)*c)*(1 + E^((2*I)*c))*((-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x))))^m*(Cos[d*x] + I*Sin[d*x])^2
*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (A*(((-I)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*Cos[c
+ d*x]^3*(2^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*Hypergeometric2F1[m, m, 1 + m, (1 - E^((2*I)*(c + d*x)))/2])*Tan[c]*(a + I*a*Tan[c + d*
x])^2*(A + B*Tan[c + d*x]))/(2^m*d*m*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x])) - (I*B*(((-I
)*(-1 + E^((2*I)*(c + d*x))))/(1 + E^((2*I)*(c + d*x))))^m*Cos[c + d*x]^3*(2^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*Hypergeometric2F1[m, m,
1 + m, (1 - E^((2*I)*(c + d*x)))/2])*Tan[c]*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(2^m*d*m*(Cos[d*x]
+ I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^3*(((B*Cos[c - d*x] - B*Cos[c + d*x])*Sec[
c]^2*Sec[c + d*x]*(Cos[2*c]/2 - (I/2)*Sin[2*c]))/(1 + m) + ((-3 - 2*m + Cos[2*c])*Sec[c]^2*(-1/2*(B*Cos[2*c])
+ (I/2)*B*Sin[2*c]))/((1 + m)*(2 + m)) + (Sec[c + d*x]^2*(-(B*Cos[2*c]) + I*B*Sin[2*c]))/(2 + m))*Tan[c + d*x]
^m*(a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x
])) + (Cos[c + d*x]^3*((Sec[c]^2*Sec[c + d*x]*(Cos[2*c]/2 - (I/2)*Sin[2*c])*(-(B*Cos[c - d*x]) + B*Cos[c + d*x
] + A*Sin[c - d*x] - (2*I)*B*Sin[c - d*x] - A*Sin[c + d*x] + (2*I)*B*Sin[c + d*x]))/(1 + m) + (Sec[c]*(A*Cos[c
] - (2*I)*B*Cos[c] + B*Sin[c])*(-Cos[2*c] + I*Sin[2*c])*Tan[c])/(1 + m))*Tan[c + d*x]^m*(a + I*a*Tan[c + d*x])
^2*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [F]  time = 0.54, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {4 \, {\left ({\left (A - i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (A + i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )}\right )} \left (\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{m}}{e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.23, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{m}\left (d x +c \right )\right ) \left (a +i a \tan \left (d x +c \right )\right )^{2} \left (A +B \tan \left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^m\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2 \,d x \]

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 \[ - a^{2} \left (\int \left (- A \tan ^{m}{\left (c + d x \right )}\right )\, dx + \int A \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- B \tan {\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx + \int B \tan ^{3}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\, dx + \int \left (- 2 i A \tan {\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx + \int \left (- 2 i B \tan ^{2}{\left (c + d x \right )} \tan ^{m}{\left (c + d x \right )}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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