3.231 \(\int \frac {(b \cos (c+d x))^n (B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=163 \[ \frac {2 B \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-5);\frac {1}{4} (2 n-1);\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)} \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-3);\frac {1}{4} (2 n+1);\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)} \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {20, 3010, 2748, 2643} \[ \frac {2 B \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-5);\frac {1}{4} (2 n-1);\cos ^2(c+d x)\right )}{d (5-2 n) \sqrt {\sin ^2(c+d x)} \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 C \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-3);\frac {1}{4} (2 n+1);\cos ^2(c+d x)\right )}{d (3-2 n) \sqrt {\sin ^2(c+d x)} \cos ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^n*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(2*B*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (-5 + 2*n)/4, (-1 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(5
- 2*n)*Cos[c + d*x]^(5/2)*Sqrt[Sin[c + d*x]^2]) + (2*C*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (-3 + 2*n)/4
, (1 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(3 - 2*n)*Cos[c + d*x]^(3/2)*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 3010

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x
_Symbol] :> Dist[1/b, Int[(b*Sin[e + f*x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x
]

Rubi steps

\begin {align*} \int \frac {(b \cos (c+d x))^n \left (B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\left (\cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {9}{2}+n}(c+d x) \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx\\ &=\left (\cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {7}{2}+n}(c+d x) (B+C \cos (c+d x)) \, dx\\ &=\left (B \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {7}{2}+n}(c+d x) \, dx+\left (C \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {5}{2}+n}(c+d x) \, dx\\ &=\frac {2 B (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-5+2 n);\frac {1}{4} (-1+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5-2 n) \cos ^{\frac {5}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}+\frac {2 C (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-3+2 n);\frac {1}{4} (1+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3-2 n) \cos ^{\frac {3}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.23, size = 138, normalized size = 0.85 \[ -\frac {2 \sqrt {\sin ^2(c+d x)} \csc (c+d x) (b \cos (c+d x))^n \left (B (2 n-3) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-5);\frac {1}{4} (2 n-1);\cos ^2(c+d x)\right )+C (2 n-5) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-3);\frac {1}{4} (2 n+1);\cos ^2(c+d x)\right )\right )}{d (2 n-5) (2 n-3) \cos ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^n*(B*Cos[c + d*x] + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

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

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right ) + B\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {7}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c) + B)*(b*cos(d*x + c))^n/cos(d*x + c)^(7/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^n/cos(d*x + c)^(9/2), x)

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maple [F]  time = 0.63, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \cos \left (d x +c \right )\right )^{n} \left (B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\cos \left (d x +c \right )^{\frac {9}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x)

[Out]

int((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*(b*cos(d*x + c))^n/cos(d*x + c)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*cos(c + d*x))^n*(B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(9/2),x)

[Out]

int(((b*cos(c + d*x))^n*(B*cos(c + d*x) + C*cos(c + d*x)^2))/cos(c + d*x)^(9/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**n*(B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2),x)

[Out]

Timed out

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