3.777 \(\int (a+b \cos (c+d x)) \sec ^m(c+d x) \, dx\)

Optimal. Leaf size=143 \[ -\frac {a \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(c+d x)\right )}{d (1-m) \sqrt {\sin ^2(c+d x)}}-\frac {b \sin (c+d x) \sec ^{m-2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {2-m}{2};\frac {4-m}{2};\cos ^2(c+d x)\right )}{d (2-m) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-b*hypergeom([1/2, 1-1/2*m],[2-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-2+m)*sin(d*x+c)/d/(2-m)/(sin(d*x+c)^2)^(1/2)-
a*hypergeom([1/2, 1/2-1/2*m],[3/2-1/2*m],cos(d*x+c)^2)*sec(d*x+c)^(-1+m)*sin(d*x+c)/d/(1-m)/(sin(d*x+c)^2)^(1/
2)

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Rubi [A]  time = 0.11, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3238, 3787, 3772, 2643} \[ -\frac {a \sin (c+d x) \sec ^{m-1}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1-m}{2};\frac {3-m}{2};\cos ^2(c+d x)\right )}{d (1-m) \sqrt {\sin ^2(c+d x)}}-\frac {b \sin (c+d x) \sec ^{m-2}(c+d x) \, _2F_1\left (\frac {1}{2},\frac {2-m}{2};\frac {4-m}{2};\cos ^2(c+d x)\right )}{d (2-m) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2643

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Sin[c + d*x])^(n + 1)*Hypergeomet
ric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2])/(b*d*(n + 1)*Sqrt[Cos[c + d*x]^2]), x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3238

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Csc[e + f*x])^(m - n*p)*(b + a*Csc[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rule 3772

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x])^(n - 1)*((Sin[c + d*x]/b)^(n - 1)
*Int[1/(Sin[c + d*x]/b)^n, x]), x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 3787

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 107, normalized size = 0.75 \[ \frac {\sqrt {-\tan ^2(c+d x)} \csc (c+d x) \sec ^{m-1}(c+d x) \left (a (m-1) \, _2F_1\left (\frac {1}{2},\frac {m}{2};\frac {m+2}{2};\sec ^2(c+d x)\right )+b m \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {m-1}{2};\frac {m+1}{2};\sec ^2(c+d x)\right )\right )}{d (m-1) m} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[c + d*x]*(b*m*Cos[c + d*x]*Hypergeometric2F1[1/2, (-1 + m)/2, (1 + m)/2, Sec[c + d*x]^2] + a*(-1 + m)*Hyp
ergeometric2F1[1/2, m/2, (2 + m)/2, Sec[c + d*x]^2])*Sec[c + d*x]^(-1 + m)*Sqrt[-Tan[c + d*x]^2])/(d*(-1 + m)*
m)

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.10, size = 0, normalized size = 0.00 \[ \int \left (a +b \cos \left (d x +c \right )\right ) \left (\sec ^{m}\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{m}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^m\,\left (a+b\,\cos \left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((1/cos(c + d*x))^m*(a + b*cos(c + d*x)),x)

[Out]

int((1/cos(c + d*x))^m*(a + b*cos(c + d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \cos {\left (c + d x \right )}\right ) \sec ^{m}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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