3.147 \(\int (a+a \sec (c+d x))^n \sin ^3(c+d x) \, dx\)

Optimal. Leaf size=83 \[ \frac{\cos ^3(c+d x) (a \sec (c+d x)+a)^{n+2}}{3 a^2 d}-\frac{(4-n) (a \sec (c+d x)+a)^{n+2} \text{Hypergeometric2F1}(3,n+2,n+3,\sec (c+d x)+1)}{3 a^2 d (n+2)} \]

[Out]

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

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Rubi [A]  time = 0.0726852, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {3873, 78, 65} \[ \frac{\cos ^3(c+d x) (a \sec (c+d x)+a)^{n+2}}{3 a^2 d}-\frac{(4-n) (a \sec (c+d x)+a)^{n+2} \, _2F_1(3,n+2;n+3;\sec (c+d x)+1)}{3 a^2 d (n+2)} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^n*Sin[c + d*x]^3,x]

[Out]

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

Rule 3873

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Dist[(f*b^(p - 1)
)^(-1), Subst[Int[((-a + b*x)^((p - 1)/2)*(a + b*x)^(m + (p - 1)/2))/x^(p + 1), x], x, Csc[e + f*x]], x] /; Fr
eeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0]

Rule 78

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

Rubi steps

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

Mathematica [A]  time = 0.138474, size = 67, normalized size = 0.81 \[ \frac{(\sec (c+d x)+1)^2 (a (\sec (c+d x)+1))^n \left ((n-4) \text{Hypergeometric2F1}(3,n+2,n+3,\sec (c+d x)+1)+(n+2) \cos ^3(c+d x)\right )}{3 d (n+2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^n*Sin[c + d*x]^3,x]

[Out]

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

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Maple [F]  time = 0.581, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n} \left ( \sin \left ( dx+c \right ) \right ) ^{3}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^n*sin(d*x+c)^3,x)

[Out]

int((a+a*sec(d*x+c))^n*sin(d*x+c)^3,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} \sin \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^n*sin(d*x + c)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (\cos \left (d x + c\right )^{2} - 1\right )}{\left (a \sec \left (d x + c\right ) + a\right )}^{n} \sin \left (d x + c\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^3,x, algorithm="fricas")

[Out]

integral(-(cos(d*x + c)^2 - 1)*(a*sec(d*x + c) + a)^n*sin(d*x + c), 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*sin(d*x+c)**3,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} \sin \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^n*sin(d*x+c)^3,x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^n*sin(d*x + c)^3, x)