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

Optimal. Leaf size=163 \[ -\frac{2 A \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+7);\frac{1}{4} (2 n+11);\cos ^2(c+d x)\right )}{d (2 n+7) \sqrt{\sin ^2(c+d x)}}-\frac{2 B \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+9);\frac{1}{4} (2 n+13);\cos ^2(c+d x)\right )}{d (2 n+9) \sqrt{\sin ^2(c+d x)}} \]

[Out]

(-2*A*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*(7 + 2*n)*Sqrt[Sin[c + d*x]^2]) - (2*B*Cos[c + d*x]^(9/2)*(b*Cos[c + d*x])^n*Hypergeometric2F
1[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])

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Rubi [A]  time = 0.0946512, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.097, Rules used = {20, 2748, 2643} \[ -\frac{2 A \sin (c+d x) \cos ^{\frac{7}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+7);\frac{1}{4} (2 n+11);\cos ^2(c+d x)\right )}{d (2 n+7) \sqrt{\sin ^2(c+d x)}}-\frac{2 B \sin (c+d x) \cos ^{\frac{9}{2}}(c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+9);\frac{1}{4} (2 n+13);\cos ^2(c+d x)\right )}{d (2 n+9) \sqrt{\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*A*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*(7 + 2*n)*Sqrt[Sin[c + d*x]^2]) - (2*B*Cos[c + d*x]^(9/2)*(b*Cos[c + d*x])^n*Hypergeometric2F
1[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])

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.40253, size = 138, normalized size = 0.85 \[ -\frac{2 \sqrt{\sin ^2(c+d x)} \cos ^{\frac{7}{2}}(c+d x) \csc (c+d x) (b \cos (c+d x))^n \left (A (2 n+9) \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+7);\frac{1}{4} (2 n+11);\cos ^2(c+d x)\right )+B (2 n+7) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+9);\frac{1}{4} (2 n+13);\cos ^2(c+d x)\right )\right )}{d (2 n+7) (2 n+9)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Cos[c + d*x]^(7/2)*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(9 + 2*n)*Hypergeometric2F1[1/2, (7 + 2*n)/4, (11 +
2*n)/4, Cos[c + d*x]^2] + B*(7 + 2*n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (9 + 2*n)/4, (13 + 2*n)/4, Cos[c + d
*x]^2])*Sqrt[Sin[c + d*x]^2])/(d*(7 + 2*n)*(9 + 2*n))

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Maple [F]  time = 0.356, size = 0, normalized size = 0. \begin{align*} \int \left ( \cos \left ( dx+c \right ) \right ) ^{{\frac{5}{2}}} \left ( b\cos \left ( dx+c \right ) \right ) ^{n} \left ( A+B\cos \left ( dx+c \right ) \right ) \, dx \end{align*}

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+B*cos(d*x+c)),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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{5}{2}}\,{d x} \end{align*}

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+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \cos \left (d x + c\right )^{3} + 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 ) \end{align*}

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+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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

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+B*cos(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (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{5}{2}}\,{d x} \end{align*}

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+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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