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

Optimal. Leaf size=143 \[ \frac{(A-i B) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}+\frac{(-B+i A) (a+b \tan (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)} \]

[Out]

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

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Rubi [A]  time = 0.130205, antiderivative size = 143, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3539, 3537, 68} \[ \frac{(A-i B) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (b+i a)}+\frac{(-B+i A) (a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac{a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.14948, size = 120, normalized size = 0.84 \[ \frac{i (a+b \tan (c+d x))^{n+1} \left (\frac{(A+i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a+i b}\right )}{a+i b}-\frac{(A-i B) \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{a+b \tan (c+d x)}{a-i b}\right )}{a-i b}\right )}{2 d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((I/2)*(-(((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)
)/(d*(1 + n))

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Maple [F]  time = 0.553, size = 0, normalized size = 0. \begin{align*} \int \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((a+b*tan(d*x+c))^n*(A+B*tan(d*x+c)),x)

[Out]

int((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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((B*tan(d*x + c) + A)*(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}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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, 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}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((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, x)