3.220 \(\int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=97 \[ \frac {(a \sec (c+d x)+a)^{n+4}}{a^4 d (n+4)}-\frac {(a \sec (c+d x)+a)^{n+3} \, _2F_1(1,n+3;n+4;\sec (c+d x)+1)}{a^3 d (n+3)}-\frac {3 (a \sec (c+d x)+a)^{n+3}}{a^3 d (n+3)} \]

[Out]

-3*(a+a*sec(d*x+c))^(3+n)/a^3/d/(3+n)-hypergeom([1, 3+n],[4+n],1+sec(d*x+c))*(a+a*sec(d*x+c))^(3+n)/a^3/d/(3+n
)+(a+a*sec(d*x+c))^(4+n)/a^4/d/(4+n)

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Rubi [A]  time = 0.08, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3880, 88, 65} \[ -\frac {(a \sec (c+d x)+a)^{n+3} \, _2F_1(1,n+3;n+4;\sec (c+d x)+1)}{a^3 d (n+3)}-\frac {3 (a \sec (c+d x)+a)^{n+3}}{a^3 d (n+3)}+\frac {(a \sec (c+d x)+a)^{n+4}}{a^4 d (n+4)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^5,x]

[Out]

(-3*(a + a*Sec[c + d*x])^(3 + n))/(a^3*d*(3 + n)) - (Hypergeometric2F1[1, 3 + n, 4 + n, 1 + Sec[c + d*x]]*(a +
a*Sec[c + d*x])^(3 + n))/(a^3*d*(3 + n)) + (a + a*Sec[c + d*x])^(4 + n)/(a^4*d*(4 + n))

Rule 65

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 3880

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(d*b^(m - 1)
)^(-1), Subst[Int[((-a + b*x)^((m - 1)/2)*(a + b*x)^((m - 1)/2 + n))/x, x], x, Csc[c + d*x]], x] /; FreeQ[{a,
b, c, d, n}, x] && IntegerQ[(m - 1)/2] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x))^n \tan ^5(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {(-a+a x)^2 (a+a x)^{2+n}}{x} \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-3 a^2 (a+a x)^{2+n}+\frac {a^2 (a+a x)^{2+n}}{x}+a (a+a x)^{3+n}\right ) \, dx,x,\sec (c+d x)\right )}{a^4 d}\\ &=-\frac {3 (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}+\frac {(a+a \sec (c+d x))^{4+n}}{a^4 d (4+n)}+\frac {\operatorname {Subst}\left (\int \frac {(a+a x)^{2+n}}{x} \, dx,x,\sec (c+d x)\right )}{a^2 d}\\ &=-\frac {3 (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}-\frac {\, _2F_1(1,3+n;4+n;1+\sec (c+d x)) (a+a \sec (c+d x))^{3+n}}{a^3 d (3+n)}+\frac {(a+a \sec (c+d x))^{4+n}}{a^4 d (4+n)}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 72, normalized size = 0.74 \[ \frac {(\sec (c+d x)+1)^3 (a (\sec (c+d x)+1))^n (-(n+4) \, _2F_1(1,n+3;n+4;\sec (c+d x)+1)+(n+3) \sec (c+d x)-2 n-9)}{d (n+3) (n+4)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^5,x]

[Out]

((1 + Sec[c + d*x])^3*(a*(1 + Sec[c + d*x]))^n*(-9 - 2*n - (4 + n)*Hypergeometric2F1[1, 3 + n, 4 + n, 1 + Sec[
c + d*x]] + (3 + n)*Sec[c + d*x]))/(d*(3 + n)*(4 + n))

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fricas [F]  time = 0.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)

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maple [F]  time = 1.19, size = 0, normalized size = 0.00 \[ \int \left (a +a \sec \left (d x +c \right )\right )^{n} \left (\tan ^{5}\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x)

[Out]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^5,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^5, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^5\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^5*(a + a/cos(c + d*x))^n,x)

[Out]

int(tan(c + d*x)^5*(a + a/cos(c + d*x))^n, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{n} \tan ^{5}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**n*tan(d*x+c)**5,x)

[Out]

Integral((a*(sec(c + d*x) + 1))**n*tan(c + d*x)**5, x)

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