3.4.74 \(\int (b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [374]

Optimal. Leaf size=173 \[ \frac {b C (b \cos (c+d x))^{-1+n} \sin (c+d x)}{d n}-\frac {b (C (1-n)-A n) (b \cos (c+d x))^{-1+n} \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-1+n);\frac {1+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1-n) n \sqrt {\sin ^2(c+d x)}}-\frac {B (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\cos ^2(c+d x)\right ) \sin (c+d x)}{d n \sqrt {\sin ^2(c+d x)}} \]

[Out]

b*C*(b*cos(d*x+c))^(-1+n)*sin(d*x+c)/d/n-b*(C*(1-n)-A*n)*(b*cos(d*x+c))^(-1+n)*hypergeom([1/2, -1/2+1/2*n],[1/
2+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/(1-n)/n/(sin(d*x+c)^2)^(1/2)-B*(b*cos(d*x+c))^n*hypergeom([1/2, 1/2*n],[1+
1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/n/(sin(d*x+c)^2)^(1/2)

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Rubi [A]
time = 0.14, antiderivative size = 173, normalized size of antiderivative = 1.00, 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, 3102, 2827, 2722} \begin {gather*} -\frac {b (C (1-n)-A n) \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) n \sqrt {\sin ^2(c+d x)}}-\frac {B \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {n+2}{2};\cos ^2(c+d x)\right )}{d n \sqrt {\sin ^2(c+d x)}}+\frac {b C \sin (c+d x) (b \cos (c+d x))^{n-1}}{d n} \end {gather*}

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]^2,x]

[Out]

(b*C*(b*Cos[c + d*x])^(-1 + n)*Sin[c + d*x])/(d*n) - (b*(C*(1 - n) - A*n)*(b*Cos[c + d*x])^(-1 + n)*Hypergeome
tric2F1[1/2, (-1 + n)/2, (1 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 - n)*n*Sqrt[Sin[c + d*x]^2]) - (B*(b*C
os[c + d*x])^n*Hypergeometric2F1[1/2, n/2, (2 + n)/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 2827

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 3102

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]

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

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Mathematica [A]
time = 0.32, size = 131, normalized size = 0.76 \begin {gather*} \frac {(b \cos (c+d x))^n \left ((C-A n-C n) \, _2F_1\left (\frac {1}{2},\frac {1}{2} (-1+n);\frac {1+n}{2};\cos ^2(c+d x)\right )-(-1+n) \left (B \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {n}{2};\frac {2+n}{2};\cos ^2(c+d x)\right )-C \sqrt {\sin ^2(c+d x)}\right )\right ) \tan (c+d x)}{d (-1+n) n \sqrt {\sin ^2(c+d x)}} \end {gather*}

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]^2,x]

[Out]

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

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Maple [F]
time = 0.27, size = 0, normalized size = 0.00 \[\int \left (b \cos \left (d x +c \right )\right )^{n} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right ) \left (\sec ^{2}\left (d x +c \right )\right )\, dx\]

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)^2,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)^2,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^2,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)^2, x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^2,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)^2, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)**2,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^2,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)^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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