3.380 \(\int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=217 \[ \frac {2 (2 A n+A-C (1-2 n)) \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-1);\frac {1}{4} (2 n+3);\cos ^2(c+d x)\right )}{d \left (1-4 n^2\right ) \sqrt {\sin ^2(c+d x)} \sqrt {\cos (c+d x)}}-\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+1);\frac {1}{4} (2 n+5);\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^n}{d (2 n+1) \sqrt {\cos (c+d x)}} \]

[Out]

2*C*(b*cos(d*x+c))^n*sin(d*x+c)/d/(1+2*n)/cos(d*x+c)^(1/2)+2*(A-C*(1-2*n)+2*A*n)*(b*cos(d*x+c))^n*hypergeom([1
/2, -1/4+1/2*n],[3/4+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/(-4*n^2+1)/cos(d*x+c)^(1/2)/(sin(d*x+c)^2)^(1/2)-2*B*(b
*cos(d*x+c))^n*hypergeom([1/2, 1/4+1/2*n],[5/4+1/2*n],cos(d*x+c)^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/d/(1+2*n)/(sin
(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.20, antiderivative size = 217, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.098, Rules used = {20, 3023, 2748, 2643} \[ \frac {2 (2 A n+A-C (1-2 n)) \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-1);\frac {1}{4} (2 n+3);\cos ^2(c+d x)\right )}{d \left (1-4 n^2\right ) \sqrt {\sin ^2(c+d x)} \sqrt {\cos (c+d x)}}-\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+1);\frac {1}{4} (2 n+5);\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^n}{d (2 n+1) \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*(b*Cos[c + d*x])^n*Sin[c + d*x])/(d*(1 + 2*n)*Sqrt[Cos[c + d*x]]) + (2*(A - C*(1 - 2*n) + 2*A*n)*(b*Cos[c
+ d*x])^n*Hypergeometric2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 - 4*n^2)*Sqrt
[Cos[c + d*x]]*Sqrt[Sin[c + d*x]^2]) - (2*B*Sqrt[Cos[c + d*x]]*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (1 +
2*n)/4, (5 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(1 + 2*n)*Sqrt[Sin[c + d*x]^2])

Rule 20

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

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 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 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]

Rubi steps

\begin {align*} \int \frac {(b \cos (c+d x))^n \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {3}{2}+n}(c+d x) \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 C (b \cos (c+d x))^n \sin (c+d x)}{d (1+2 n) \sqrt {\cos (c+d x)}}+\frac {\left (2 \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {3}{2}+n}(c+d x) \left (\frac {1}{2} \left (-2 C \left (\frac {1}{2}-n\right )+2 A \left (\frac {1}{2}+n\right )\right )+\frac {1}{2} B (1+2 n) \cos (c+d x)\right ) \, dx}{1+2 n}\\ &=\frac {2 C (b \cos (c+d x))^n \sin (c+d x)}{d (1+2 n) \sqrt {\cos (c+d x)}}+\left (B \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {1}{2}+n}(c+d x) \, dx+\frac {\left ((A-C (1-2 n)+2 A n) \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {3}{2}+n}(c+d x) \, dx}{1+2 n}\\ &=\frac {2 C (b \cos (c+d x))^n \sin (c+d x)}{d (1+2 n) \sqrt {\cos (c+d x)}}+\frac {2 (A-C (1-2 n)+2 A n) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-1+2 n);\frac {1}{4} (3+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d \left (1-4 n^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (1+2 n);\frac {1}{4} (5+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (1+2 n) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 157, normalized size = 0.72 \[ -\frac {2 \sin (c+d x) (b \cos (c+d x))^n \left ((2 A n+A+C (2 n-1)) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-1);\frac {1}{4} (2 n+3);\cos ^2(c+d x)\right )+(2 n-1) \left (B \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+1);\frac {1}{4} (2 n+5);\cos ^2(c+d x)\right )-C \sqrt {\sin ^2(c+d x)}\right )\right )}{d \left (4 n^2-1\right ) \sqrt {\sin ^2(c+d x)} \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(b*Cos[c + d*x])^n*Sin[c + d*x]*((A + 2*A*n + C*(-1 + 2*n))*Hypergeometric2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)
/4, Cos[c + d*x]^2] + (-1 + 2*n)*(B*Cos[c + d*x]*Hypergeometric2F1[1/2, (1 + 2*n)/4, (5 + 2*n)/4, Cos[c + d*x]
^2] - C*Sqrt[Sin[c + d*x]^2])))/(d*(-1 + 4*n^2)*Sqrt[Cos[c + d*x]]*Sqrt[Sin[c + d*x]^2])

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fricas [F]  time = 0.86, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

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)/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

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)/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 0.62, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \cos \left (d x +c \right )\right )^{n} \left (A +B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\cos \left (d x +c \right )^{\frac {3}{2}}}\, dx \]

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)/cos(d*x+c)^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

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)/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A\right )}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]

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)^(3/2),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)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \cos {\left (c + d x \right )}\right )^{n} \left (A + B \cos {\left (c + d x \right )} + C \cos ^{2}{\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

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)/cos(d*x+c)**(3/2),x)

[Out]

Integral((b*cos(c + d*x))**n*(A + B*cos(c + d*x) + C*cos(c + d*x)**2)/cos(c + d*x)**(3/2), x)

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