3.224 \(\int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

Optimal. Leaf size=163 \[ -\frac {2 B \sin (c+d x) \cos ^{\frac {9}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+9);\frac {1}{4} (2 n+13);\cos ^2(c+d x)\right )}{d (2 n+9) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \cos ^{\frac {11}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+11);\frac {1}{4} (2 n+15);\cos ^2(c+d x)\right )}{d (2 n+11) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-2*B*cos(d*x+c)^(9/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 9/4+1/2*n],[13/4+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/(9+2
*n)/(sin(d*x+c)^2)^(1/2)-2*C*cos(d*x+c)^(11/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 11/4+1/2*n],[15/4+1/2*n],cos(d
*x+c)^2)*sin(d*x+c)/d/(11+2*n)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.13, 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} \[ -\frac {2 B \sin (c+d x) \cos ^{\frac {9}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+9);\frac {1}{4} (2 n+13);\cos ^2(c+d x)\right )}{d (2 n+9) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \sin (c+d x) \cos ^{\frac {11}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+11);\frac {1}{4} (2 n+15);\cos ^2(c+d x)\right )}{d (2 n+11) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^n*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(-2*B*Cos[c + d*x]^(9/2)*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (9 + 2*n)/4, (13 + 2*n)/4, Cos[c + d*x]^2]*
Sin[c + d*x])/(d*(9 + 2*n)*Sqrt[Sin[c + d*x]^2]) - (2*C*Cos[c + d*x]^(11/2)*(b*Cos[c + d*x])^n*Hypergeometric2
F1[1/2, (11 + 2*n)/4, (15 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(11 + 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 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 \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \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 {5}{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 {9}{2}+n}(c+d x) \, dx\\ &=-\frac {2 B \cos ^{\frac {9}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (9+2 n);\frac {1}{4} (13+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (9+2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 C \cos ^{\frac {11}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (11+2 n);\frac {1}{4} (15+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (11+2 n) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.28, size = 138, normalized size = 0.85 \[ -\frac {2 \sqrt {\sin ^2(c+d x)} \cos ^{\frac {9}{2}}(c+d x) \csc (c+d x) (b \cos (c+d x))^n \left (B (2 n+11) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+9);\frac {1}{4} (2 n+13);\cos ^2(c+d x)\right )+C (2 n+9) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+11);\frac {1}{4} (2 n+15);\cos ^2(c+d x)\right )\right )}{d (2 n+9) (2 n+11)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^n*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]

[Out]

(-2*Cos[c + d*x]^(9/2)*(b*Cos[c + d*x])^n*Csc[c + d*x]*(B*(11 + 2*n)*Hypergeometric2F1[1/2, (9 + 2*n)/4, (13 +
2*n)/4, Cos[c + d*x]^2] + C*(9 + 2*n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (11 + 2*n)/4, (15 + 2*n)/4, Cos[c +
d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(9 + 2*n)*(11 + 2*n))

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{4} + B \cos \left (d x + c\right )^{3}\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^4 + B*cos(d*x + c)^3)*(b*cos(d*x + c))^n*sqrt(cos(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^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)^(5/2), x)

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maple [F]  time = 0.60, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{n} \left (B \cos \left (d x +c \right )+C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

int(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*(B*cos(d*x+c)+C*cos(d*x+c)^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)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^{5/2}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^n*(B*cos(c + d*x) + C*cos(c + d*x)^2),x)

[Out]

int(cos(c + d*x)^(5/2)*(b*cos(c + d*x))^n*(B*cos(c + d*x) + C*cos(c + d*x)^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(cos(d*x+c)**(5/2)*(b*cos(d*x+c))**n*(B*cos(d*x+c)+C*cos(d*x+c)**2),x)

[Out]

Timed out

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