3.923 \(\int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.09, 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, 2748, 2643} \[ \frac {2 A \sin (c+d x) (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-1);\frac {1}{4} (2 n+3);\cos ^2(c+d x)\right )}{d (1-2 n) \sqrt {\sin ^2(c+d x)} \sqrt {\cos (c+d x)}}-\frac {2 B \sin (c+d x) \sqrt {\cos (c+d x)} (b \cos (c+d x))^n \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+1);\frac {1}{4} (2 n+5);\cos ^2(c+d x)\right )}{d (2 n+1) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[((b*Cos[c + d*x])^n*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

(2*A*(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*B*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 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]

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]

Rubi steps

\begin {align*} \int \frac {(b \cos (c+d x))^n (A+B \cos (c+d x))}{\cos ^{\frac {3}{2}}(c+d x)} \, dx &=\left (\cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {3}{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 {3}{2}+n}(c+d x) \, dx+\left (B \cos ^{-n}(c+d x) (b \cos (c+d x))^n\right ) \int \cos ^{-\frac {1}{2}+n}(c+d x) \, dx\\ &=\frac {2 A (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 B \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*}

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Mathematica [A]  time = 0.27, size = 133, normalized size = 0.82 \[ -\frac {2 \sqrt {\sin ^2(c+d x)} \csc (c+d x) (b \cos (c+d x))^n \left (A (2 n+1) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n-1);\frac {1}{4} (2 n+3);\cos ^2(c+d x)\right )+B (2 n-1) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {1}{4} (2 n+1);\frac {1}{4} (2 n+5);\cos ^2(c+d x)\right )\right )}{d \left (4 n^2-1\right ) \sqrt {\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((b*Cos[c + d*x])^n*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(3/2),x]

[Out]

(-2*(b*Cos[c + d*x])^n*Csc[c + d*x]*(A*(1 + 2*n)*Hypergeometric2F1[1/2, (-1 + 2*n)/4, (3 + 2*n)/4, Cos[c + d*x
]^2] + B*(-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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fricas [F]  time = 0.98, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(3/2), x)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \cos \left (d x +c \right )\right )^{n} \left (A +B \cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*cos(d*x+c))^n*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x)

[Out]

int((b*cos(d*x+c))^n*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n}}{\cos \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))^n*(A+B*cos(d*x+c))/cos(d*x+c)^(3/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c))^n/cos(d*x + c)^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^n\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\cos \left (c+d\,x\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x)))/cos(c + d*x)^(3/2),x)

[Out]

int(((b*cos(c + d*x))^n*(A + B*cos(c + d*x)))/cos(c + d*x)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \cos {\left (c + d x \right )}\right )^{n} \left (A + B \cos {\left (c + d x \right )}\right )}{\cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*cos(d*x+c))**n*(A+B*cos(d*x+c))/cos(d*x+c)**(3/2),x)

[Out]

Integral((b*cos(c + d*x))**n*(A + B*cos(c + d*x))/cos(c + d*x)**(3/2), x)

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