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

Optimal. Leaf size=142 \[ \frac {2 C \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n}{d (2 n+9)}-\frac {2 (A (2 n+9)+C (2 n+7)) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+7);\frac {1}{4} (2 n+11);\cos ^2(c+d x)\right )}{d (2 n+7) (2 n+9) \sqrt {\sin ^2(c+d x)}} \]

[Out]

2*C*cos(d*x+c)^(7/2)*(b*cos(d*x+c))^n*sin(d*x+c)/d/(9+2*n)-2*(C*(7+2*n)+A*(9+2*n))*cos(d*x+c)^(7/2)*(b*cos(d*x
+c))^n*hypergeom([1/2, 7/4+1/2*n],[11/4+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/(4*n^2+32*n+63)/(sin(d*x+c)^2)^(1/2)

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Rubi [A]  time = 0.11, antiderivative size = 132, normalized size of antiderivative = 0.93, number of steps used = 3, number of rules used = 3, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {20, 3014, 2643} \[ \frac {2 C \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n}{d (2 n+9)}-\frac {2 \left (\frac {A}{2 n+7}+\frac {C}{2 n+9}\right ) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+7);\frac {1}{4} (2 n+11);\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*C*Cos[c + d*x]^(7/2)*(b*Cos[c + d*x])^n*Sin[c + d*x])/(d*(9 + 2*n)) - (2*(A/(7 + 2*n) + C/(9 + 2*n))*Cos[c
+ d*x]^(7/2)*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (7 + 2*n)/4, (11 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x]
)/(d*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 3014

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

Rubi steps

\begin {align*} \int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \left (A+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 (A+C \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 C \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \sin (c+d x)}{d (9+2 n)}+\frac {\left (\left (C \left (\frac {7}{2}+n\right )+A \left (\frac {9}{2}+n\right )\right ) \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{\frac {5}{2}+n}(c+d x) \, dx}{\frac {9}{2}+n}\\ &=\frac {2 C \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \sin (c+d x)}{d (9+2 n)}-\frac {2 (C (7+2 n)+A (9+2 n)) \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (7+2 n);\frac {1}{4} (11+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (7+2 n) (9+2 n) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.25, size = 140, normalized size = 0.99 \[ -\frac {2 \sqrt {\sin ^2(c+d x)} \cos ^{\frac {7}{2}}(c+d x) \csc (c+d x) (b \cos (c+d x))^n \left (A (2 n+11) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+7);\frac {1}{4} (2 n+11);\cos ^2(c+d x)\right )+C (2 n+7) \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+11);\frac {1}{4} (2 n+15);\cos ^2(c+d x)\right )\right )}{d (2 n+7) (2 n+11)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]^(7/2)*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(11 + 2*n)*Hypergeometric2F1[1/2, (7 + 2*n)/4, (11 +
2*n)/4, Cos[c + d*x]^2] + C*(7 + 2*n)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (11 + 2*n)/4, (15 + 2*n)/4, Cos[c
+ d*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(7 + 2*n)*(11 + 2*n))

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fricas [F]  time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C \cos \left (d x + c\right )^{4} + A \cos \left (d x + c\right )^{2}\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^4 + A*cos(d*x + c)^2)*(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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.49, size = 0, normalized size = 0.00 \[ \int \left (\cos ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{n} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*cos(d*x + c)^2 + A)*(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 \[ \int {\cos \left (c+d\,x\right )}^{5/2}\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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