3.708 \(\int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx\)

Optimal. Leaf size=192 \[ -\frac {a (a+b \tan (c+d x))^{n+1}}{b^2 d (n+1) (n+2)}+\frac {(a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}+\frac {(a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac {\tan (c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.28, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3566, 3630, 12, 3539, 3537, 68} \[ -\frac {a (a+b \tan (c+d x))^{n+1}}{b^2 d (n+1) (n+2)}+\frac {(a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )}{2 d (n+1) (a-i b)}+\frac {(a+b \tan (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )}{2 d (n+1) (a+i b)}+\frac {\tan (c+d x) (a+b \tan (c+d x))^{n+1}}{b d (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^3*(a + b*Tan[c + d*x])^n,x]

[Out]

-((a*(a + b*Tan[c + d*x])^(1 + n))/(b^2*d*(1 + n)*(2 + n))) + (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)) + (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)) + (Tan[c + d*x]*(a + b
*Tan[c + d*x])^(1 + n))/(b*d*(2 + n))

Rule 12

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

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]

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 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 3566

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(m + n - 1)), x] + Dist[1/(d*(m + n -
1)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^n*Simp[a^3*d*(m + n - 1) - b^2*(b*c*(m - 2) + a*d*(
1 + n)) + b*d*(m + n - 1)*(3*a^2 - b^2)*Tan[e + f*x] - b^2*(b*c*(m - 2) - a*d*(3*m + 2*n - 4))*Tan[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& IntegerQ[2*m] && GtQ[m, 2] && (GeQ[n, -1] || IntegerQ[m]) &&  !(IGtQ[n, 2] && ( !IntegerQ[m] || (EqQ[c, 0]
&& NeQ[a, 0])))

Rule 3630

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(C*(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*Tan[e + f*x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && NeQ[A*b^2 - a*b*B + a^2*C, 0]
&&  !LeQ[m, -1]

Rubi steps

\begin {align*} \int \tan ^3(c+d x) (a+b \tan (c+d x))^n \, dx &=\frac {\tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}+\frac {\int (a+b \tan (c+d x))^n \left (-a-b (2+n) \tan (c+d x)-a \tan ^2(c+d x)\right ) \, dx}{b (2+n)}\\ &=-\frac {a (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {\tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}-\frac {\int b (2+n) \tan (c+d x) (a+b \tan (c+d x))^n \, dx}{b (2+n)}\\ &=-\frac {a (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {\tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}-\int \tan (c+d x) (a+b \tan (c+d x))^n \, dx\\ &=-\frac {a (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {\tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}-\frac {1}{2} i \int (1-i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx+\frac {1}{2} i \int (1+i \tan (c+d x)) (a+b \tan (c+d x))^n \, dx\\ &=-\frac {a (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {\tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}-\frac {\operatorname {Subst}\left (\int \frac {(a-i b x)^n}{-1+x} \, dx,x,i \tan (c+d x)\right )}{2 d}-\frac {\operatorname {Subst}\left (\int \frac {(a+i b x)^n}{-1+x} \, dx,x,-i \tan (c+d x)\right )}{2 d}\\ &=-\frac {a (a+b \tan (c+d x))^{1+n}}{b^2 d (1+n) (2+n)}+\frac {\, _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 {\, _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 {\tan (c+d x) (a+b \tan (c+d x))^{1+n}}{b d (2+n)}\\ \end {align*}

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Mathematica [A]  time = 1.05, size = 135, normalized size = 0.70 \[ \frac {(a+b \tan (c+d x))^{n+1} \left (\frac {2 \left (b \tan (c+d x)-\frac {a}{n+1}\right )}{b^2 (n+2)}+\frac {\, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a-i b}\right )}{(n+1) (a-i b)}+\frac {\, _2F_1\left (1,n+1;n+2;\frac {a+b \tan (c+d x)}{a+i b}\right )}{(n+1) (a+i b)}\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^3*(a + b*Tan[c + d*x])^n,x]

[Out]

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

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fricas [F]  time = 1.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="fricas")

[Out]

integral((b*tan(d*x + c) + a)^n*tan(d*x + c)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)^n*tan(d*x + c)^3, x)

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maple [F]  time = 0.83, size = 0, normalized size = 0.00 \[ \int \left (\tan ^{3}\left (d x +c \right )\right ) \left (a +b \tan \left (d x +c \right )\right )^{n}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x)

[Out]

int(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{3}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3*(a+b*tan(d*x+c))^n,x, algorithm="maxima")

[Out]

integrate((b*tan(d*x + c) + a)^n*tan(d*x + c)^3, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^3\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^3*(a + b*tan(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^3*(a + b*tan(c + d*x))^n, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \tan {\left (c + d x \right )}\right )^{n} \tan ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3*(a+b*tan(d*x+c))**n,x)

[Out]

Integral((a + b*tan(c + d*x))**n*tan(c + d*x)**3, x)

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