3.226 \(\int (a+a \sec (c+d x))^n \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=106 \[ \frac{2^{n+3} \tan ^3(c+d x) \left (\frac{1}{\sec (c+d x)+1}\right )^{n+3} (a \sec (c+d x)+a)^n F_1\left (\frac{3}{2};n+2,1;\frac{5}{2};-\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{3 d} \]

[Out]

(2^(3 + n)*AppellF1[3/2, 2 + n, 1, 5/2, -((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))^(3 + n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.0553637, 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} \[ \frac{2^{n+3} \tan ^3(c+d x) \left (\frac{1}{\sec (c+d x)+1}\right )^{n+3} (a \sec (c+d x)+a)^n F_1\left (\frac{3}{2};n+2,1;\frac{5}{2};-\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a},\frac{a-a \sec (c+d x)}{\sec (c+d x) a+a}\right )}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^2,x]

[Out]

(2^(3 + n)*AppellF1[3/2, 2 + n, 1, 5/2, -((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))^(3 + n)*(a + a*Sec[c + d*x])^n*Tan[c + d*x]^3)/(3*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 ^2(c+d x) \, dx &=\frac{2^{3+n} F_1\left (\frac{3}{2};2+n,1;\frac{5}{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 )^{3+n} (a+a \sec (c+d x))^n \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [B]  time = 14.9112, size = 2419, normalized size = 22.82 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^n*Tan[c + d*x]^2,x]

[Out]

(2^(3 + n)*Cos[(c + d*x)/2]*Sec[c + d*x]^2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*(a*(1 + Sec[c + d*x]))^n*Sin[(c
+ d*x)/2]^3*(-((Hypergeometric2F1[1/2, 1 + n, 3/2, Tan[(c + d*x)/2]^2] - 2*Hypergeometric2F1[1/2, 2 + n, 3/2,
Tan[(c + d*x)/2]^2])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n) + (3*AppellF1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -
Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2)/(-3*AppellF1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]
+ 2*(AppellF1[3/2, n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)))/(d*(2^n*Sec[(c + d*x)/2]^2*(Cos[(c + d*x)/2]^2*Sec[c
+ d*x])^n*(-((Hypergeometric2F1[1/2, 1 + n, 3/2, Tan[(c + d*x)/2]^2] - 2*Hypergeometric2F1[1/2, 2 + n, 3/2, T
an[(c + d*x)/2]^2])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n) + (3*AppellF1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -Ta
n[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2)/(-3*AppellF1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] +
2*(AppellF1[3/2, n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c +
d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2)) + 2^(1 + n)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^n*Tan[(c
+ d*x)/2]*(-(n*(Hypergeometric2F1[1/2, 1 + n, 3/2, Tan[(c + d*x)/2]^2] - 2*Hypergeometric2F1[1/2, 2 + n, 3/2,
Tan[(c + d*x)/2]^2])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^(-1 + n)*(-(Sec[(c + d*x)/2]^2*Sin[c + d*x]) + Cos[c +
d*x]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])) - (3*AppellF1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2]*Cos[(c + d*x)/2]*Sin[(c + d*x)/2])/(-3*AppellF1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] +
2*(AppellF1[3/2, n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c
+ d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2) + (3*Cos[(c + d*x)/2]^2*(-(AppellF1[3/2, n, 2, 5/2, Tan
[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/3 + (n*AppellF1[3/2, 1 + n, 1, 5/2,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/3))/(-3*AppellF1[1/2, n, 1, 3/2
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(AppellF1[3/2, n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2] - n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2) - (3*Appell
F1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2*(2*(AppellF1[3/2, n, 2, 5/2, Ta
n[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^
2])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2] - 3*(-(AppellF1[3/2, n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^
2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/3 + (n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)
/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/3) + 2*Tan[(c + d*x)/2]^2*((-6*AppellF1[5/2, n, 3, 7/2, Tan[(c + d
*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5 + (3*n*AppellF1[5/2, 1 + n, 2, 7/2, Tan[
(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5 - n*((-3*AppellF1[5/2, 1 + n, 2, 7
/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5 + (3*(1 + n)*AppellF1[5/2,
2 + n, 1, 7/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])/5))))/(-3*Appell
F1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(AppellF1[3/2, n, 2, 5/2, Tan[(c + d*x)/2]^2,
-Tan[(c + d*x)/2]^2] - n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/
2]^2)^2 - (Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n*(-(Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(-Hypergeometric2F1[1/2, 2
+ n, 3/2, Tan[(c + d*x)/2]^2] + (1 - Tan[(c + d*x)/2]^2)^(-2 - n))) + (Csc[(c + d*x)/2]*Sec[(c + d*x)/2]*(-Hyp
ergeometric2F1[1/2, 1 + n, 3/2, Tan[(c + d*x)/2]^2] + (1 - Tan[(c + d*x)/2]^2)^(-1 - n)))/2)) + 2^(1 + n)*n*(C
os[(c + d*x)/2]^2*Sec[c + d*x])^(-1 + n)*Tan[(c + d*x)/2]*(-((Hypergeometric2F1[1/2, 1 + n, 3/2, Tan[(c + d*x)
/2]^2] - 2*Hypergeometric2F1[1/2, 2 + n, 3/2, Tan[(c + d*x)/2]^2])*(Cos[c + d*x]*Sec[(c + d*x)/2]^2)^n) + (3*A
ppellF1[1/2, n, 1, 3/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[(c + d*x)/2]^2)/(-3*AppellF1[1/2, n, 1, 3
/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*(AppellF1[3/2, n, 2, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/
2]^2] - n*AppellF1[3/2, 1 + n, 1, 5/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Tan[(c + d*x)/2]^2))*(-(Cos[(
c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Tan[c + d*x])))

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Maple [F]  time = 0.21, 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 ) ^{2}\, 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)^2,x)

[Out]

int((a+a*sec(d*x+c))^n*tan(d*x+c)^2,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 )^{2}\,{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)^2,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^2, 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 )^{2}, 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)^2,x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)^n*tan(d*x + c)^2, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a \left (\sec{\left (c + d x \right )} + 1\right )\right )^{n} \tan ^{2}{\left (c + d x \right )}\, dx \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)**2,x)

[Out]

Integral((a*(sec(c + d*x) + 1))**n*tan(c + d*x)**2, x)

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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 )^{2}\,{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)^2,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*tan(d*x + c)^2, x)