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

Optimal. Leaf size=115 \[ \frac{(a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a+b}\right )}{2 d (m+1) (a+b)}-\frac{(a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a-b}\right )}{2 d (m+1) (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.114643, antiderivative size = 115, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2668, 712, 68} \[ \frac{(a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a+b}\right )}{2 d (m+1) (a+b)}-\frac{(a+b \sin (c+d x))^{m+1} \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a-b}\right )}{2 d (m+1) (a-b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2668

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 712

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m, 1/(a + c*x^2
), x], x] /; FreeQ[{a, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[m]

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

Mathematica [A]  time = 0.11361, size = 99, normalized size = 0.86 \[ -\frac{(a+b \sin (c+d x))^{m+1} \left ((a+b) \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a-b}\right )+(b-a) \, _2F_1\left (1,m+1;m+2;\frac{a+b \sin (c+d x)}{a+b}\right )\right )}{2 d (m+1) (a-b) (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 0.632, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( dx+c \right ) \left ( a+b\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+b*sin(d*x+c))^m,x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \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+b*sin(d*x+c))^m,x, algorithm="maxima")

[Out]

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

[Out]

integral((b*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 + b \sin{\left (c + d x \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+b*sin(d*x+c))**m,x)

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \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+b*sin(d*x+c))^m,x, algorithm="giac")

[Out]

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