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

Optimal. Leaf size=42 \[ \frac{(a \sec (c+d x)+a)^{n+1} \text{Hypergeometric2F1}(2,n+1,n+2,\sec (c+d x)+1)}{a d (n+1)} \]

[Out]

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

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Rubi [A]  time = 0.0373294, antiderivative size = 42, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3873, 65} \[ \frac{(a \sec (c+d x)+a)^{n+1} \, _2F_1(2,n+1;n+2;\sec (c+d x)+1)}{a d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[2, 1 + n, 2 + n, 1 + Sec[c + d*x]]*(a + a*Sec[c + d*x])^(1 + n))/(a*d*(1 + 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 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 (c+d x) \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{(a-a x)^n}{x^2} \, dx,x,-\sec (c+d x)\right )}{d}\\ &=\frac{\, _2F_1(2,1+n;2+n;1+\sec (c+d x)) (a+a \sec (c+d x))^{1+n}}{a d (1+n)}\\ \end{align*}

Mathematica [A]  time = 0.0370683, size = 42, normalized size = 1. \[ \frac{(a (\sec (c+d x)+1))^{n+1} \text{Hypergeometric2F1}(2,n+1,n+2,\sec (c+d x)+1)}{a d (n+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.337, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{n}\sin \left ( dx+c \right ) \, 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),x)

[Out]

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

[Out]

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

[Out]

integral((a*sec(d*x + c) + a)^n*sin(d*x + c), 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} \sin{\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*sin(d*x+c),x)

[Out]

Integral((a*(sec(c + d*x) + 1))**n*sin(c + d*x), 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} \sin \left (d x + c\right )\,{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),x, algorithm="giac")

[Out]

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