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

Optimal. Leaf size=177 \[ -\frac{(a+b \sec (c+d x))^{n+1} \text{Hypergeometric2F1}\left (1,n+1,n+2,\frac{b \sec (c+d x)}{a}+1\right )}{a d (n+1)}-\frac{a \left (a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+1}}{b^4 d (n+1)}+\frac{\left (3 a^2-2 b^2\right ) (a+b \sec (c+d x))^{n+2}}{b^4 d (n+2)}-\frac{3 a (a+b \sec (c+d x))^{n+3}}{b^4 d (n+3)}+\frac{(a+b \sec (c+d x))^{n+4}}{b^4 d (n+4)} \]

[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))

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Rubi [A]  time = 0.201419, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, 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 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]

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 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])

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*}

Mathematica [A]  time = 2.92646, size = 298, normalized size = 1.68 \[ -\frac{\sec ^8\left (\frac{1}{2} (c+d x)\right ) (a+b \sec (c+d x))^n \left (n (a \cos (c+d x)+b) \left (3 a \left (3 a^2+b^2 \left (n^2-n-8\right )\right ) \cos (c+d x)+2 b (n+1) \left (b^2 \left (n^2+7 n+12\right )-3 a^2\right ) \cos (2 (c+d x))-6 a^2 b n-6 a^2 b+3 a^3 \cos (3 (c+d x))-a b^2 n^2 \cos (3 (c+d x))-7 a b^2 n \cos (3 (c+d x))-12 a b^2 \cos (3 (c+d x))+4 b^3 n^2+16 b^3 n+12 b^3\right )-2 b^4 \left (n^4+10 n^3+35 n^2+50 n+24\right ) \cos ^4(c+d x) \text{Hypergeometric2F1}\left (1,-n,1-n,\frac{a \cos (c+d x)}{a \cos (c+d x)+b}\right )\right )}{2 b^4 d n (n+1) (n+2) (n+3) (n+4) \left (\tan ^2\left (\frac{1}{2} (c+d x)\right )-1\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((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])])*Sec[(c
+ d*x)/2]^8*(a + b*Sec[c + d*x])^n)/(2*b^4*d*n*(1 + n)*(2 + n)*(3 + n)*(4 + n)*(-1 + Tan[(c + d*x)/2]^2)^4)

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Maple [F]  time = 0.402, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b\sec \left ( dx+c \right ) \right ) ^{n} \left ( \tan \left ( dx+c \right ) \right ) ^{5}\, dx \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}, x\right ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{5}\,{d x} \end{align*}

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)