3.225 \(\int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=106 \[ \frac{2^{n+5} \tan ^5(c+d x) \left (\frac{1}{\sec (c+d x)+1}\right )^{n+5} (a \sec (c+d x)+a)^n F_1\left (\frac{5}{2};n+4,1;\frac{7}{2};-\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{5 d} \]

[Out]

(2^(5 + n)*AppellF1[5/2, 4 + n, 1, 7/2, -((a - a*Sec[c + d*x])/(a + a*Sec[c + d*x])), (a - a*Sec[c + d*x])/(a
+ a*Sec[c + d*x])]*((1 + Sec[c + d*x])^(-1))^(5 + n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x]^5)/(5*d)

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Rubi [A]  time = 0.0553183, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.048, Rules used = {3889} \[ \frac{2^{n+5} \tan ^5(c+d x) \left (\frac{1}{\sec (c+d x)+1}\right )^{n+5} (a \sec (c+d x)+a)^n F_1\left (\frac{5}{2};n+4,1;\frac{7}{2};-\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^4,x]

[Out]

(2^(5 + n)*AppellF1[5/2, 4 + n, 1, 7/2, -((a - a*Sec[c + d*x])/(a + a*Sec[c + d*x])), (a - a*Sec[c + d*x])/(a
+ a*Sec[c + d*x])]*((1 + Sec[c + d*x])^(-1))^(5 + n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x]^5)/(5*d)

Rule 3889

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(2^(m
+ n + 1)*(e*Cot[c + d*x])^(m + 1)*(a + b*Csc[c + d*x])^n*(a/(a + b*Csc[c + d*x]))^(m + n + 1)*AppellF1[(m + 1
)/2, m + n, 1, (m + 3)/2, -((a - b*Csc[c + d*x])/(a + b*Csc[c + d*x])), (a - b*Csc[c + d*x])/(a + b*Csc[c + d*
x])])/(d*e*(m + 1)), x] /; FreeQ[{a, b, c, d, e, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !IntegerQ[n]

Rubi steps

\begin{align*} \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx &=\frac{2^{5+n} F_1\left (\frac{5}{2};4+n,1;\frac{7}{2};-\frac{a-a \sec (c+d x)}{a+a \sec (c+d x)},\frac{a-a \sec (c+d x)}{a+a \sec (c+d x)}\right ) \left (\frac{1}{1+\sec (c+d x)}\right )^{5+n} (a+a \sec (c+d x))^n \tan ^5(c+d x)}{5 d}\\ \end{align*}

Mathematica [F]  time = 1.15102, size = 0, normalized size = 0. \[ \int (a+a \sec (c+d x))^n \tan ^4(c+d x) \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^4,x]

[Out]

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^4, x]

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Maple [F]  time = 0.272, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( \tan \left ( dx+c \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x)

[Out]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*tan(d*x + c)^4, 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+a*sec(d*x+c))**n*tan(d*x+c)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \tan \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*tan(d*x+c)^4,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^4, x)