3.917 \(\int (b \cos (c+d x))^n (A+B \cos (c+d x)) \sec ^3(c+d x) \, dx\)

Optimal. Leaf size=139 \[ \frac {A b^2 \sin (c+d x) (b \cos (c+d x))^{n-2} \, _2F_1\left (\frac {1}{2},\frac {n-2}{2};\frac {n}{2};\cos ^2(c+d x)\right )}{d (2-n) \sqrt {\sin ^2(c+d x)}}+\frac {b B \sin (c+d x) (b \cos (c+d x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {n-1}{2};\frac {n+1}{2};\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {16, 2748, 2643} \[ \frac {A b^2 \sin (c+d x) (b \cos (c+d x))^{n-2} \, _2F_1\left (\frac {1}{2},\frac {n-2}{2};\frac {n}{2};\cos ^2(c+d x)\right )}{d (2-n) \sqrt {\sin ^2(c+d x)}}+\frac {b B \sin (c+d x) (b \cos (c+d x))^{n-1} \, _2F_1\left (\frac {1}{2},\frac {n-1}{2};\frac {n+1}{2};\cos ^2(c+d x)\right )}{d (1-n) \sqrt {\sin ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(A*b^2*(b*Cos[c + d*x])^(-2 + n)*Hypergeometric2F1[1/2, (-2 + n)/2, n/2, Cos[c + d*x]^2]*Sin[c + d*x])/(d*(2 -
n)*Sqrt[Sin[c + d*x]^2]) + (b*B*(b*Cos[c + d*x])^(-1 + n)*Hypergeometric2F1[1/2, (-1 + n)/2, (1 + n)/2, Cos[c
+ d*x]^2]*Sin[c + d*x])/(d*(1 - n)*Sqrt[Sin[c + d*x]^2])

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

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

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Mathematica [A]  time = 0.18, size = 118, normalized size = 0.85 \[ -\frac {\sqrt {\sin ^2(c+d x)} \csc (c+d x) \sec ^2(c+d x) (b \cos (c+d x))^n \left (A (n-1) \, _2F_1\left (\frac {1}{2},\frac {n-2}{2};\frac {n}{2};\cos ^2(c+d x)\right )+B (n-2) \cos (c+d x) \, _2F_1\left (\frac {1}{2},\frac {n-1}{2};\frac {n+1}{2};\cos ^2(c+d x)\right )\right )}{d (n-2) (n-1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.75, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B \cos \left (d x + c\right ) + A\right )} \left (b \cos \left (d x + c\right )\right )^{n} \sec \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 1.37, size = 0, normalized size = 0.00 \[ \int \left (b \cos \left (d x +c \right )\right )^{n} \left (A +B \cos \left (d x +c \right )\right ) \left (\sec ^{3}\left (d x +c \right )\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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