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

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

[Out]

(2*B*(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
- 2*n)*Sqrt[Cos[c + d*x]]*Sqrt[Sin[c + d*x]^2]) - (2*C*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])

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Rubi [A]  time = 0.133048, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {20, 3010, 2748, 2643} \[ \frac{2 B \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n-1);\frac{1}{4} (2 n+3);\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt{\sin ^2(c+d x)} \sqrt{\cos (c+d x)}}-\frac{2 C \sin (c+d x) \sqrt{\cos (c+d x)} (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+1);\frac{1}{4} (2 n+5);\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt{\sin ^2(c+d x)}} \]

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]^(5/2),x]

[Out]

(2*B*(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
- 2*n)*Sqrt[Cos[c + d*x]]*Sqrt[Sin[c + d*x]^2]) - (2*C*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 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
]

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

Mathematica [A]  time = 0.246398, size = 133, normalized size = 0.82 \[ -\frac{2 \sqrt{\sin ^2(c+d x)} \csc (c+d x) (b \cos (c+d x))^n \left (B (2 n+1) \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n-1);\frac{1}{4} (2 n+3);\cos ^2(c+d x)\right )+C (2 n-1) \cos (c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+1);\frac{1}{4} (2 n+5);\cos ^2(c+d x)\right )\right )}{d \left (4 n^2-1\right ) \sqrt{\cos (c+d x)}} \]

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]^(5/2),x]

[Out]

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

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Maple [F]  time = 0.765, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\cos \left ( dx+c \right ) \right ) ^{n} \left ( B\cos \left ( dx+c \right ) +C \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}}}\, dx \end{align*}

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)^(5/2),x)

[Out]

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

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

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)^(5/2),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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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{3}{2}}}, x\right ) \end{align*}

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)^(5/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c) + B)*(b*cos(d*x + c))^n/cos(d*x + c)^(3/2), 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((b*cos(d*x+c))**n*(B*cos(d*x+c)+C*cos(d*x+c)**2)/cos(d*x+c)**(5/2),x)

[Out]

Timed out

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

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)^(5/2),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)