3.918 \(\int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\)

Optimal. Leaf size=141 \[ \frac{A b^3 \sin (c+d x) (b \cos (c+d x))^{n-3} \, _2F_1\left (\frac{1}{2},\frac{n-3}{2};\frac{n-1}{2};\cos ^2(c+d x)\right )}{d (3-n) \sqrt{\sin ^2(c+d x)}}+\frac{b^2 B \sin (c+d x) (b \cos (c+d x))^{n-2} \, _2F_1\left (\frac{1}{2},\frac{n-2}{2};\frac{n}{2};\cos ^2(c+d x)\right )}{d (2-n) \sqrt{\sin ^2(c+d x)}} \]

[Out]

(A*b^3*(b*Cos[c + d*x])^(-3 + n)*Hypergeometric2F1[1/2, (-3 + n)/2, (-1 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/
(d*(3 - n)*Sqrt[Sin[c + d*x]^2]) + (b^2*B*(b*Cos[c + d*x])^(-2 + n)*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Co
s[c + d*x]^2]*Sin[c + d*x])/(d*(2 - n)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.119445, antiderivative size = 141, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {16, 2748, 2643} \[ \frac{A b^3 \sin (c+d x) (b \cos (c+d x))^{n-3} \, _2F_1\left (\frac{1}{2},\frac{n-3}{2};\frac{n-1}{2};\cos ^2(c+d x)\right )}{d (3-n) \sqrt{\sin ^2(c+d x)}}+\frac{b^2 B \sin (c+d x) (b \cos (c+d x))^{n-2} \, _2F_1\left (\frac{1}{2},\frac{n-2}{2};\frac{n}{2};\cos ^2(c+d x)\right )}{d (2-n) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]

[Out]

(A*b^3*(b*Cos[c + d*x])^(-3 + n)*Hypergeometric2F1[1/2, (-3 + n)/2, (-1 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/
(d*(3 - n)*Sqrt[Sin[c + d*x]^2]) + (b^2*B*(b*Cos[c + d*x])^(-2 + n)*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Co
s[c + d*x]^2]*Sin[c + d*x])/(d*(2 - n)*Sqrt[Sin[c + d*x]^2])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.160283, size = 118, normalized size = 0.84 \[ -\frac{\sqrt{\sin ^2(c+d x)} \csc (c+d x) \sec ^3(c+d x) (b \cos (c+d x))^n \left (A (n-2) \, _2F_1\left (\frac{1}{2},\frac{n-3}{2};\frac{n-1}{2};\cos ^2(c+d x)\right )+B (n-3) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{n-2}{2};\frac{n}{2};\cos ^2(c+d x)\right )\right )}{d (n-3) (n-2)} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]

[Out]

-(((b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(-2 + n)*Hypergeometric2F1[1/2, (-3 + n)/2, (-1 + n)/2, Cos[c + d*x]^2]
+ B*(-3 + n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Cos[c + d*x]^2])*Sec[c + d*x]^3*Sqrt[Sin[c +
d*x]^2])/(d*(-3 + n)*(-2 + n)))

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Maple [F]  time = 1.182, size = 0, normalized size = 0. \begin{align*} \int \left ( b\cos \left ( dx+c \right ) \right ) ^{n} \left ( A+B\cos \left ( dx+c \right ) \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{4}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x)

[Out]

int((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^4, 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((b*cos(d*x+c))**n*(A+B*cos(d*x+c))*sec(d*x+c)**4,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^4, x)