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

Optimal. Leaf size=140 \[ \frac{2 (-2 A n+A+C (3-2 n)) \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n-3);\frac{1}{4} (2 n+1);\cos ^2(c+d x)\right )}{d (1-2 n) (3-2 n) \sqrt{\sin ^2(c+d x)} \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 C \sin (c+d x) (b \cos (c+d x))^n}{d (1-2 n) \cos ^{\frac{3}{2}}(c+d x)} \]

[Out]

(-2*C*(b*Cos[c + d*x])^n*Sin[c + d*x])/(d*(1 - 2*n)*Cos[c + d*x]^(3/2)) + (2*(A + C*(3 - 2*n) - 2*A*n)*(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*(1 - 2*n)*(3 -
2*n)*Cos[c + d*x]^(3/2)*Sqrt[Sin[c + d*x]^2])

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Rubi [A]  time = 0.112909, antiderivative size = 132, normalized size of antiderivative = 0.94, 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} \[ \frac{2 \left (\frac{A}{3-2 n}+\frac{C}{1-2 n}\right ) \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n-3);\frac{1}{4} (2 n+1);\cos ^2(c+d x)\right )}{d \sqrt{\sin ^2(c+d x)} \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 C \sin (c+d x) (b \cos (c+d x))^n}{d (1-2 n) \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*C*(b*Cos[c + d*x])^n*Sin[c + d*x])/(d*(1 - 2*n)*Cos[c + d*x]^(3/2)) + (2*(C/(1 - 2*n) + A/(3 - 2*n))*(b*Co
s[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*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 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]

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

Mathematica [A]  time = 0.166223, size = 140, normalized size = 1. \[ -\frac{2 \sqrt{\sin ^2(c+d x)} \csc (c+d x) (b \cos (c+d x))^n \left (A (2 n+1) \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n-3);\frac{1}{4} (2 n+1);\cos ^2(c+d x)\right )+C (2 n-3) \cos ^2(c+d x) \, _2F_1\left (\frac{1}{2},\frac{1}{4} (2 n+1);\frac{1}{4} (2 n+5);\cos ^2(c+d x)\right )\right )}{d (2 n-3) (2 n+1) \cos ^{\frac{3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

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Maple [F]  time = 0.612, size = 0, normalized size = 0. \begin{align*} \int{ \left ( b\cos \left ( dx+c \right ) \right ) ^{n} \left ( A+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*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),x)

[Out]

int((b*cos(d*x+c))^n*(A+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} + 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((b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(5/2),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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\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}}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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