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

Optimal. Leaf size=177 \[ -\frac {a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+1}}{b^4 d (n+1)}+\frac {\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+2}}{b^4 d (n+2)}-\frac {3 a (a+b \sec (c+d x))^{n+3}}{b^4 d (n+3)}+\frac {(a+b \sec (c+d x))^{n+4}}{b^4 d (n+4)}-\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)} \]

[Out]

-a*(a^2-2*b^2)*(a+b*sec(d*x+c))^(1+n)/b^4/d/(1+n)-hypergeom([1, 1+n],[2+n],1+b*sec(d*x+c)/a)*(a+b*sec(d*x+c))^
(1+n)/a/d/(1+n)+(3*a^2-2*b^2)*(a+b*sec(d*x+c))^(2+n)/b^4/d/(2+n)-3*a*(a+b*sec(d*x+c))^(3+n)/b^4/d/(3+n)+(a+b*s
ec(d*x+c))^(4+n)/b^4/d/(4+n)

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Rubi [A]  time = 0.20, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3885, 952, 1620, 65} \[ -\frac {a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+1}}{b^4 d (n+1)}+\frac {\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+2}}{b^4 d (n+2)}-\frac {3 a (a+b \sec (c+d x))^{n+3}}{b^4 d (n+3)}+\frac {(a+b \sec (c+d x))^{n+4}}{b^4 d (n+4)}-\frac {(a+b \sec (c+d x))^{n+1} \, _2F_1\left (1,n+1;n+2;\frac {b \sec (c+d x)}{a}+1\right )}{a d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*(a^2 - 2*b^2)*(a + b*Sec[c + d*x])^(1 + n))/(b^4*d*(1 + n))) - (Hypergeometric2F1[1, 1 + n, 2 + n, 1 + (b
*Sec[c + d*x])/a]*(a + b*Sec[c + d*x])^(1 + n))/(a*d*(1 + n)) + ((3*a^2 - 2*b^2)*(a + b*Sec[c + d*x])^(2 + n))
/(b^4*d*(2 + n)) - (3*a*(a + b*Sec[c + d*x])^(3 + n))/(b^4*d*(3 + n)) + (a + b*Sec[c + d*x])^(4 + n)/(b^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 952

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(c^p*(d
+ e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m + n + 2*p + 1)),
Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + c*x^2)^p - c^p*(d + e*x)^(2*p)) - c
^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0]
&& NeQ[c*d^2 + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

Rule 1620

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rule 3885

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

Rubi steps

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

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Mathematica [A]  time = 3.31, size = 298, normalized size = 1.68 \[ -\frac {\sec ^8\left (\frac {1}{2} (c+d x)\right ) (a+b \sec (c+d x))^n \left (n (a \cos (c+d x)+b) \left (3 a^3 \cos (3 (c+d x))+3 a \left (3 a^2+b^2 \left (n^2-n-8\right )\right ) \cos (c+d x)+2 b (n+1) \left (b^2 \left (n^2+7 n+12\right )-3 a^2\right ) \cos (2 (c+d x))-6 a^2 b n-6 a^2 b-a b^2 n^2 \cos (3 (c+d x))-7 a b^2 n \cos (3 (c+d x))-12 a b^2 \cos (3 (c+d x))+4 b^3 n^2+16 b^3 n+12 b^3\right )-2 b^4 \left (n^4+10 n^3+35 n^2+50 n+24\right ) \cos ^4(c+d x) \, _2F_1\left (1,-n;1-n;\frac {a \cos (c+d x)}{b+a \cos (c+d x)}\right )\right )}{2 b^4 d n (n+1) (n+2) (n+3) (n+4) \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*((n*(b + a*Cos[c + d*x])*(-6*a^2*b + 12*b^3 - 6*a^2*b*n + 16*b^3*n + 4*b^3*n^2 + 3*a*(3*a^2 + b^2*(-8 - n
+ n^2))*Cos[c + d*x] + 2*b*(1 + n)*(-3*a^2 + b^2*(12 + 7*n + n^2))*Cos[2*(c + d*x)] + 3*a^3*Cos[3*(c + d*x)]
- 12*a*b^2*Cos[3*(c + d*x)] - 7*a*b^2*n*Cos[3*(c + d*x)] - a*b^2*n^2*Cos[3*(c + d*x)]) - 2*b^4*(24 + 50*n + 35
*n^2 + 10*n^3 + n^4)*Cos[c + d*x]^4*Hypergeometric2F1[1, -n, 1 - n, (a*Cos[c + d*x])/(b + a*Cos[c + d*x])])*Se
c[(c + d*x)/2]^8*(a + b*Sec[c + d*x])^n)/(b^4*d*n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(-1 + Tan[(c + d*x)/2]^2)^4)

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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*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 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.24, size = 0, normalized size = 0.00 \[ \int \left (a +b \sec \left (d x +c \right )\right )^{n} \left (\tan ^{5}\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

int((a+b*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 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*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 \[ \int {\mathrm {tan}\left (c+d\,x\right )}^5\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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