3.10.19 \(\int \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx\) [919]

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

[Out]

-2*A*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/(7+2
*n)/(sin(d*x+c)^2)^(1/2)-2*B*cos(d*x+c)^(9/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 9/4+1/2*n],[13/4+1/2*n],cos(d*x
+c)^2)*sin(d*x+c)/d/(9+2*n)/(sin(d*x+c)^2)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.07, antiderivative size = 163, normalized size of antiderivative = 1.00, 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, 2827, 2722} \begin {gather*} -\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)}} \end {gather*}

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 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 2827

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]

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 = 0.45, size = 138, normalized size = 0.85 \begin {gather*} -\frac {2 \cos ^{\frac {7}{2}}(c+d x) (b \cos (c+d x))^n \csc (c+d x) \left (A (9+2 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 )+B (7+2 n) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (9+2 n);\frac {1}{4} (13+2 n);\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (7+2 n) (9+2 n)} \end {gather*}

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))

________________________________________________________________________________________

Maple [F]
time = 0.16, size = 0, normalized size = 0.00 \[\int \left (\cos ^{\frac {5}{2}}\left (d x +c \right )\right ) \left (b \cos \left (d x +c \right )\right )^{n} \left (A +B \cos \left (d x +c \right )\right )\, dx\]

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+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)

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^{5/2}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (A+B\,\cos \left (c+d\,x\right )\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________