3.2.97 \(\int \frac {(b \cos (c+d x))^n (A+C \cos ^2(c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [197]

Optimal. Leaf size=142 \[ -\frac {2 C (b \cos (c+d x))^n \sin (c+d x)}{d (5-2 n) \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (A (5-2 n)+C (7-2 n)) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-7+2 n);\frac {1}{4} (-3+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5-2 n) (7-2 n) \cos ^{\frac {7}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-2*C*(b*cos(d*x+c))^n*sin(d*x+c)/d/(5-2*n)/cos(d*x+c)^(7/2)+2*(A*(5-2*n)+C*(7-2*n))*(b*cos(d*x+c))^n*hypergeom
([1/2, -7/4+1/2*n],[-3/4+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/(4*n^2-24*n+35)/cos(d*x+c)^(7/2)/(sin(d*x+c)^2)^(1/
2)

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Rubi [A]
time = 0.08, 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, 3093, 2722} \begin {gather*} \frac {2 \left (\frac {A}{7-2 n}+\frac {C}{5-2 n}\right ) \sin (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-3);\cos ^2(c+d x)\right )}{d \sqrt {\sin ^2(c+d x)} \cos ^{\frac {7}{2}}(c+d x)}-\frac {2 C \sin (c+d x) (b \cos (c+d x))^n}{d (5-2 n) \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^n*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(-2*C*(b*Cos[c + d*x])^n*Sin[c + d*x])/(d*(5 - 2*n)*Cos[c + d*x]^(7/2)) + (2*(C/(5 - 2*n) + A/(7 - 2*n))*(b*Co
s[c + d*x])^n*Hypergeometric2F1[1/2, (-7 + 2*n)/4, (-3 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*Cos[c + d*x]
^(7/2)*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 2722

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
)*Sqrt[Cos[c + d*x]^2]))*Hypergeometric2F1[1/2, (n + 1)/2, (n + 3)/2, Sin[c + d*x]^2], x] /; FreeQ[{b, c, d, n
}, x] &&  !IntegerQ[2*n]

Rule 3093

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 \frac {(b \cos (c+d x))^n \left (A+C \cos ^2(c+d x)\right )}{\cos ^{\frac {9}{2}}(c+d x)} \, dx &=\left (\cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {9}{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 (5-2 n) \cos ^{\frac {7}{2}}(c+d x)}+\frac {\left (\left (C \left (-\frac {7}{2}+n\right )+A \left (-\frac {5}{2}+n\right )\right ) \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {9}{2}+n}(c+d x) \, dx}{-\frac {5}{2}+n}\\ &=-\frac {2 C (b \cos (c+d x))^n \sin (c+d x)}{d (5-2 n) \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (A (5-2 n)+C (7-2 n)) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-7+2 n);\frac {1}{4} (-3+2 n);\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5-2 n) (7-2 n) \cos ^{\frac {7}{2}}(c+d x) \sqrt {\sin ^2(c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.20, size = 140, normalized size = 0.99 \begin {gather*} -\frac {2 (b \cos (c+d x))^n \csc (c+d x) \left (A (-3+2 n) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-7+2 n);\frac {1}{4} (-3+2 n);\cos ^2(c+d x)\right )+C (-7+2 n) \cos ^2(c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (-3+2 n);\frac {1}{4} (1+2 n);\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (-7+2 n) (-3+2 n) \cos ^{\frac {7}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^n*(A + C*Cos[c + d*x]^2))/Cos[c + d*x]^(9/2),x]

[Out]

(-2*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(-3 + 2*n)*Hypergeometric2F1[1/2, (-7 + 2*n)/4, (-3 + 2*n)/4, Cos[c + d
*x]^2] + C*(-7 + 2*n)*Cos[c + d*x]^2*Hypergeometric2F1[1/2, (-3 + 2*n)/4, (1 + 2*n)/4, Cos[c + d*x]^2])*Sqrt[S
in[c + d*x]^2])/(d*(-7 + 2*n)*(-3 + 2*n)*Cos[c + d*x]^(7/2))

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Maple [F]
time = 0.22, size = 0, normalized size = 0.00 \[\int \frac {\left (b \cos \left (d x +c \right )\right )^{n} \left (A +C \left (\cos ^{2}\left (d x +c \right )\right )\right )}{\cos \left (d x +c \right )^{\frac {9}{2}}}\, dx\]

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)^(9/2),x)

[Out]

int((b*cos(d*x+c))^n*(A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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)^(9/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(9/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(9/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(9/2), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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)**(9/2),x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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)^(9/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(9/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n}{{\cos \left (c+d\,x\right )}^{9/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^n)/cos(c + d*x)^(9/2),x)

[Out]

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^n)/cos(c + d*x)^(9/2), x)

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