3.497 \(\int \tan (c+d x) (a+b \tan (c+d x))^n (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=168 \[ -\frac{(A-i B) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}-\frac{(A+i B) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac{B (a+b \tan (c+d x))^{n+1}}{b d (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.177936, antiderivative size = 168, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {3592, 3539, 3537, 68} \[ -\frac{(A-i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}-\frac{(A+i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac{B (a+b \tan (c+d x))^{n+1}}{b d (n+1)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]*(a + b*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

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

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 3539

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

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 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.208763, size = 125, normalized size = 0.74 \[ \frac{(a+b \tan (c+d x))^{n+1} \left (-\frac{(A-i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )}{a-i b}-\frac{(A+i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )}{a+i b}+\frac{2 B}{b}\right )}{2 d (n+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]*(a + b*Tan[c + d*x])^n*(A + B*Tan[c + d*x]),x]

[Out]

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

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Maple [F]  time = 0.328, size = 0, normalized size = 0. \begin{align*} \int \tan \left ( dx+c \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) ^{n} \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)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int(tan(d*x+c)*(a+b*tan(d*x+c))^n*(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 (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^n*tan(d*x + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \tan \left (d x + c\right )^{2} + A \tan \left (d x + c\right )\right )}{\left (b \tan \left (d x + c\right ) + a\right )}^{n}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

integral((B*tan(d*x + c)^2 + A*tan(d*x + c))*(b*tan(d*x + c) + a)^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \tan{\left (c + d x \right )}\right ) \left (a + b \tan{\left (c + d x \right )}\right )^{n} \tan{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+b*tan(d*x+c))**n*(A+B*tan(d*x+c)),x)

[Out]

Integral((A + B*tan(c + d*x))*(a + b*tan(c + d*x))**n*tan(c + d*x), 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 (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(b*tan(d*x + c) + a)^n*tan(d*x + c), x)