3.347 \(\int \sec (c+d x) (a+a \sin (c+d x))^m \, dx\)

Optimal. Leaf size=40 \[ \frac{(a \sin (c+d x)+a)^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (\sin (c+d x)+1)\right )}{2 d m} \]

[Out]

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

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Rubi [A]  time = 0.0487234, antiderivative size = 40, 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 = {2667, 68} \[ \frac{(a \sin (c+d x)+a)^m \, _2F_1\left (1,m;m+1;\frac{1}{2} (\sin (c+d x)+1)\right )}{2 d m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

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

Mathematica [A]  time = 0.0941581, size = 63, normalized size = 1.58 \[ \frac{(a (\sin (c+d x)+1))^m \left (m (\sin (c+d x)+1) \, _2F_1\left (1,m+1;m+2;\frac{1}{2} (\sin (c+d x)+1)\right )+2 (m+1)\right )}{4 d m (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a*(1 + Sin[c + d*x]))^m*(2*(1 + m) + m*Hypergeometric2F1[1, 1 + m, 2 + m, (1 + Sin[c + d*x])/2]*(1 + Sin[c +
d*x])))/(4*d*m*(1 + m))

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Maple [F]  time = 0.54, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( dx+c \right ) \left ( a+a\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((a*sin(d*x + c) + a)^m*sec(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a*(sin(c + d*x) + 1))**m*sec(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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