3.375 \(\int (b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=194 \[ \frac{b^2 (A (1-n)+C (2-n)) \sin (c+d x) (b \cos (c+d x))^{n-2} \, _2F_1\left (\frac{1}{2},\frac{n-2}{2};\frac{n}{2};\cos ^2(c+d x)\right )}{d (1-n) (2-n) \sqrt{\sin ^2(c+d x)}}-\frac{b^2 C \sin (c+d x) (b \cos (c+d x))^{n-2}}{d (1-n)}+\frac{b B \sin (c+d x) (b \cos (c+d x))^{n-1} \, _2F_1\left (\frac{1}{2},\frac{n-1}{2};\frac{n+1}{2};\cos ^2(c+d x)\right )}{d (1-n) \sqrt{\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.255924, antiderivative size = 194, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {16, 3023, 2748, 2643} \[ \frac{b^2 (A (1-n)+C (2-n)) \sin (c+d x) (b \cos (c+d x))^{n-2} \, _2F_1\left (\frac{1}{2},\frac{n-2}{2};\frac{n}{2};\cos ^2(c+d x)\right )}{d (1-n) (2-n) \sqrt{\sin ^2(c+d x)}}-\frac{b^2 C \sin (c+d x) (b \cos (c+d x))^{n-2}}{d (1-n)}+\frac{b B \sin (c+d x) (b \cos (c+d x))^{n-1} \, _2F_1\left (\frac{1}{2},\frac{n-1}{2};\frac{n+1}{2};\cos ^2(c+d x)\right )}{d (1-n) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

-((b^2*C*(b*Cos[c + d*x])^(-2 + n)*Sin[c + d*x])/(d*(1 - n))) + (b^2*(A*(1 - n) + C*(2 - n))*(b*Cos[c + d*x])^
(-2 + n)*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 - n)*(2 - n)*Sqrt[Sin[c +
d*x]^2]) + (b*B*(b*Cos[c + d*x])^(-1 + n)*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Cos[c + d*x]^2]*Sin[c
+ d*x])/(d*(1 - 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 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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

Mathematica [A]  time = 0.457482, size = 137, normalized size = 0.71 \[ -\frac{b \tan (c+d x) (b \cos (c+d x))^{n-1} \left ((A (n-1)+C (n-2)) \, _2F_1\left (\frac{1}{2},\frac{n-2}{2};\frac{n}{2};\cos ^2(c+d x)\right )+(n-2) \left (B \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{n-1}{2};\frac{n+1}{2};\cos ^2(c+d x)\right )-C \sqrt{\sin ^2(c+d x)}\right )\right )}{d (n-2) (n-1) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*Cos[c + d*x])^n*(A + B*Cos[c + d*x] + C*Cos[c + d*x]^2)*Sec[c + d*x]^3,x]

[Out]

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

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Maple [F]  time = 1.602, 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 ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}\, 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)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)

[Out]

int((b*cos(d*x+c))^n*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + 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 )^{3}\,{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)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^3, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C \cos \left (d x + c\right )^{2} + 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 )^{3}, 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)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^3, 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)+C*cos(d*x+c)**2)*sec(d*x+c)**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + 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 )^{3}\,{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)+C*cos(d*x+c)^2)*sec(d*x+c)^3,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sec(d*x + c)^3, x)