\(\int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx\) [921]

Optimal result

Integrand size = 31, antiderivative size = 163 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=-\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3+2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+2 n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

-2*A*cos(d*x+c)^(3/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 3/4+1/2*n],[7/4+1/2*n],cos(d*x+c)^2)*sin(d*x+c)/d/(3+2*
n)/(sin(d*x+c)^2)^(1/2)-2*B*cos(d*x+c)^(5/2)*(b*cos(d*x+c))^n*hypergeom([1/2, 5/4+1/2*n],[9/4+1/2*n],cos(d*x+c
)^2)*sin(d*x+c)/d/(5+2*n)/(sin(d*x+c)^2)^(1/2)

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {20, 2827, 2722} \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=-\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+3),\frac {1}{4} (2 n+7),\cos ^2(c+d x)\right )}{d (2 n+3) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (2 n+5),\frac {1}{4} (2 n+9),\cos ^2(c+d x)\right )}{d (2 n+5) \sqrt {\sin ^2(c+d x)}} \]

[In]

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

[Out]

(-2*A*Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^n*Hypergeometric2F1[1/2, (3 + 2*n)/4, (7 + 2*n)/4, Cos[c + d*x]^2]*S
in[c + d*x])/(d*(3 + 2*n)*Sqrt[Sin[c + d*x]^2]) - (2*B*Cos[c + d*x]^(5/2)*(b*Cos[c + d*x])^n*Hypergeometric2F1
[1/2, (5 + 2*n)/4, (9 + 2*n)/4, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(5 + 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*} \text {integral}& = \left (\cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{\frac {1}{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 {1}{2}+n}(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 \\ & = -\frac {2 A \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (3+2 n) \sqrt {\sin ^2(c+d x)}}-\frac {2 B \cos ^{\frac {5}{2}}(c+d x) (b \cos (c+d x))^n \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\cos ^2(c+d x)\right ) \sin (c+d x)}{d (5+2 n) \sqrt {\sin ^2(c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.85 \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=-\frac {2 \cos ^{\frac {3}{2}}(c+d x) (b \cos (c+d x))^n \csc (c+d x) \left (A (5+2 n) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (3+2 n),\frac {1}{4} (7+2 n),\cos ^2(c+d x)\right )+B (3+2 n) \cos (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {1}{4} (5+2 n),\frac {1}{4} (9+2 n),\cos ^2(c+d x)\right )\right ) \sqrt {\sin ^2(c+d x)}}{d (3+2 n) (5+2 n)} \]

[In]

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

[Out]

(-2*Cos[c + d*x]^(3/2)*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(5 + 2*n)*Hypergeometric2F1[1/2, (3 + 2*n)/4, (7 + 2
*n)/4, Cos[c + d*x]^2] + B*(3 + 2*n)*Cos[c + d*x]*Hypergeometric2F1[1/2, (5 + 2*n)/4, (9 + 2*n)/4, Cos[c + d*x
]^2])*Sqrt[Sin[c + d*x]^2])/(d*(3 + 2*n)*(5 + 2*n))

Maple [F]

\[\int \left (\cos \left (d x +c \right ) b \right )^{n} \left (A +B \cos \left (d x +c \right )\right ) \left (\sqrt {\cos }\left (d x +c \right )\right )d x\]

[In]

int((cos(d*x+c)*b)^n*(A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x)

[Out]

int((cos(d*x+c)*b)^n*(A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x)

Fricas [F]

\[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sqrt {\cos \left (d x + c\right )} \,d x } \]

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=\int \left (b \cos {\left (c + d x \right )}\right )^{n} \left (A + B \cos {\left (c + d x \right )}\right ) \sqrt {\cos {\left (c + d x \right )}}\, dx \]

[In]

integrate((b*cos(d*x+c))**n*(A+B*cos(d*x+c))*cos(d*x+c)**(1/2),x)

[Out]

Integral((b*cos(c + d*x))**n*(A + B*cos(c + d*x))*sqrt(cos(c + d*x)), x)

Maxima [F]

\[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sqrt {\cos \left (d x + c\right )} \,d x } \]

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sqrt(cos(d*x + c)), x)

Giac [F]

\[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=\int { {\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sqrt {\cos \left (d x + c\right )} \,d x } \]

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))*cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n*sqrt(cos(d*x + c)), x)

Mupad [F(-1)]

Timed out. \[ \int \sqrt {\cos (c+d x)} (b \cos (c+d x))^n (A+B \cos (c+d x)) \, dx=\int \sqrt {\cos \left (c+d\,x\right )}\,{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (A+B\,\cos \left (c+d\,x\right )\right ) \,d x \]

[In]

int(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^n*(A + B*cos(c + d*x)),x)

[Out]

int(cos(c + d*x)^(1/2)*(b*cos(c + d*x))^n*(A + B*cos(c + d*x)), x)