3.879 \(\int \frac{(d \cos (e+f x))^n}{(a+b \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=309 \[ \frac{a^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (-n-1)} (d \cos (e+f x))^n F_1\left (\frac{1}{2};\frac{1}{2} (-n-3),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{b^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (-n-1)} (d \cos (e+f x))^n F_1\left (\frac{1}{2};\frac{1}{2} (-n-1),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac{2 a b \sin (e+f x) \cos ^2(e+f x)^{-n/2} (d \cos (e+f x))^n F_1\left (\frac{1}{2};\frac{1}{2} (-n-2),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]

[Out]

(a^2*AppellF1[1/2, (-3 - n)/2, 2, 3/2, Sin[e + f*x]^2, (a^2*Sin[e + f*x]^2)/(a^2 - b^2)]*Cos[e + f*x]*(d*Cos[e
+ f*x])^n*(Cos[e + f*x]^2)^((-1 - n)/2)*Sin[e + f*x])/((a^2 - b^2)^2*f) + (b^2*AppellF1[1/2, (-1 - n)/2, 2, 3
/2, Sin[e + f*x]^2, (a^2*Sin[e + f*x]^2)/(a^2 - b^2)]*Cos[e + f*x]*(d*Cos[e + f*x])^n*(Cos[e + f*x]^2)^((-1 -
n)/2)*Sin[e + f*x])/((a^2 - b^2)^2*f) - (2*a*b*AppellF1[1/2, (-2 - n)/2, 2, 3/2, Sin[e + f*x]^2, (a^2*Sin[e +
f*x]^2)/(a^2 - b^2)]*(d*Cos[e + f*x])^n*Sin[e + f*x])/((a^2 - b^2)^2*f*(Cos[e + f*x]^2)^(n/2))

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Rubi [A]  time = 0.513906, antiderivative size = 309, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {4264, 3869, 2824, 3189, 429} \[ \frac{a^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (-n-1)} (d \cos (e+f x))^n F_1\left (\frac{1}{2};\frac{1}{2} (-n-3),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{b^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (-n-1)} (d \cos (e+f x))^n F_1\left (\frac{1}{2};\frac{1}{2} (-n-1),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}-\frac{2 a b \sin (e+f x) \cos ^2(e+f x)^{-n/2} (d \cos (e+f x))^n F_1\left (\frac{1}{2};\frac{1}{2} (-n-2),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[(d*Cos[e + f*x])^n/(a + b*Sec[e + f*x])^2,x]

[Out]

(a^2*AppellF1[1/2, (-3 - n)/2, 2, 3/2, Sin[e + f*x]^2, (a^2*Sin[e + f*x]^2)/(a^2 - b^2)]*Cos[e + f*x]*(d*Cos[e
+ f*x])^n*(Cos[e + f*x]^2)^((-1 - n)/2)*Sin[e + f*x])/((a^2 - b^2)^2*f) + (b^2*AppellF1[1/2, (-1 - n)/2, 2, 3
/2, Sin[e + f*x]^2, (a^2*Sin[e + f*x]^2)/(a^2 - b^2)]*Cos[e + f*x]*(d*Cos[e + f*x])^n*(Cos[e + f*x]^2)^((-1 -
n)/2)*Sin[e + f*x])/((a^2 - b^2)^2*f) - (2*a*b*AppellF1[1/2, (-2 - n)/2, 2, 3/2, Sin[e + f*x]^2, (a^2*Sin[e +
f*x]^2)/(a^2 - b^2)]*(d*Cos[e + f*x])^n*Sin[e + f*x])/((a^2 - b^2)^2*f*(Cos[e + f*x]^2)^(n/2))

Rule 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 3869

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Dist[Sin[
e + f*x]^n*(d*Csc[e + f*x])^n, Int[(b + a*Sin[e + f*x])^m/Sin[e + f*x]^(m + n), x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[m]

Rule 2824

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Int[Expan
dTrig[(d*sin[e + f*x])^n/((a - b*sin[e + f*x])^m/(a^2 - b^2*sin[e + f*x]^2)^m), x], x] /; FreeQ[{a, b, d, e, f
, n}, x] && NeQ[a^2 - b^2, 0] && ILtQ[m, -1]

Rule 3189

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff
= FreeFactors[Cos[e + f*x], x]}, -Dist[(ff*d^(2*IntPart[(m - 1)/2] + 1)*(d*Sin[e + f*x])^(2*FracPart[(m - 1)/
2]))/(f*(Sin[e + f*x]^2)^FracPart[(m - 1)/2]), Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - b*ff^2*x^2)^p, x]
, x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e, f, m, p}, x] &&  !IntegerQ[m]

Rule 429

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*x*AppellF1[1/n, -p,
-q, 1 + 1/n, -((b*x^n)/a), -((d*x^n)/c)], x] /; FreeQ[{a, b, c, d, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[n
, -1] && (IntegerQ[p] || GtQ[a, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rubi steps

\begin{align*} \int \frac{(d \cos (e+f x))^n}{(a+b \sec (e+f x))^2} \, dx &=\left ((d \cos (e+f x))^n (d \sec (e+f x))^n\right ) \int \frac{(d \sec (e+f x))^{-n}}{(a+b \sec (e+f x))^2} \, dx\\ &=\left (\cos ^{-n}(e+f x) (d \cos (e+f x))^n\right ) \int \frac{\cos ^{2+n}(e+f x)}{(b+a \cos (e+f x))^2} \, dx\\ &=\left (\cos ^{-n}(e+f x) (d \cos (e+f x))^n\right ) \int \left (\frac{b^2 \cos ^{2+n}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2}-\frac{2 a b \cos ^{3+n}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2}+\frac{a^2 \cos ^{4+n}(e+f x)}{\left (-b^2+a^2 \cos ^2(e+f x)\right )^2}\right ) \, dx\\ &=\left (a^2 \cos ^{-n}(e+f x) (d \cos (e+f x))^n\right ) \int \frac{\cos ^{4+n}(e+f x)}{\left (-b^2+a^2 \cos ^2(e+f x)\right )^2} \, dx-\left (2 a b \cos ^{-n}(e+f x) (d \cos (e+f x))^n\right ) \int \frac{\cos ^{3+n}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2} \, dx+\left (b^2 \cos ^{-n}(e+f x) (d \cos (e+f x))^n\right ) \int \frac{\cos ^{2+n}(e+f x)}{\left (b^2-a^2 \cos ^2(e+f x)\right )^2} \, dx\\ &=\frac{\left (a^2 \cos ^{2 \left (\frac{1}{2}+\frac{n}{2}\right )-n}(e+f x) (d \cos (e+f x))^n \cos ^2(e+f x)^{-\frac{1}{2}-\frac{n}{2}}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{\frac{3+n}{2}}}{\left (a^2-b^2-a^2 x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}+\frac{\left (b^2 \cos ^{2 \left (\frac{1}{2}+\frac{n}{2}\right )-n}(e+f x) (d \cos (e+f x))^n \cos ^2(e+f x)^{-\frac{1}{2}-\frac{n}{2}}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{\frac{1+n}{2}}}{\left (-a^2+b^2+a^2 x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}-\frac{\left (2 a b (d \cos (e+f x))^n \cos ^2(e+f x)^{-n/2}\right ) \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^{\frac{2+n}{2}}}{\left (-a^2+b^2+a^2 x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{a^2 F_1\left (\frac{1}{2};\frac{1}{2} (-3-n),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \cos (e+f x))^n \cos ^2(e+f x)^{\frac{1}{2} (-1-n)} \sin (e+f x)}{\left (a^2-b^2\right )^2 f}+\frac{b^2 F_1\left (\frac{1}{2};\frac{1}{2} (-1-n),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) \cos (e+f x) (d \cos (e+f x))^n \cos ^2(e+f x)^{\frac{1}{2} (-1-n)} \sin (e+f x)}{\left (a^2-b^2\right )^2 f}-\frac{2 a b F_1\left (\frac{1}{2};\frac{1}{2} (-2-n),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right ) (d \cos (e+f x))^n \cos ^2(e+f x)^{-n/2} \sin (e+f x)}{\left (a^2-b^2\right )^2 f}\\ \end{align*}

Mathematica [B]  time = 32.6211, size = 10296, normalized size = 33.32 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Cos[e + f*x])^n/(a + b*Sec[e + f*x])^2,x]

[Out]

Result too large to show

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Maple [F]  time = 0.276, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( d\cos \left ( fx+e \right ) \right ) ^{n}}{ \left ( a+b\sec \left ( fx+e \right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*cos(f*x+e))^n/(a+b*sec(f*x+e))^2,x)

[Out]

int((d*cos(f*x+e))^n/(a+b*sec(f*x+e))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \cos \left (f x + e\right )\right )^{n}}{{\left (b \sec \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^n/(a+b*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((d*cos(f*x + e))^n/(b*sec(f*x + e) + a)^2, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\left (d \cos \left (f x + e\right )\right )^{n}}{b^{2} \sec \left (f x + e\right )^{2} + 2 \, a b \sec \left (f x + e\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^n/(a+b*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

integral((d*cos(f*x + e))^n/(b^2*sec(f*x + e)^2 + 2*a*b*sec(f*x + e) + a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \cos{\left (e + f x \right )}\right )^{n}}{\left (a + b \sec{\left (e + f x \right )}\right )^{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))**n/(a+b*sec(f*x+e))**2,x)

[Out]

Integral((d*cos(e + f*x))**n/(a + b*sec(e + f*x))**2, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \cos \left (f x + e\right )\right )^{n}}{{\left (b \sec \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*cos(f*x+e))^n/(a+b*sec(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*cos(f*x + e))^n/(b*sec(f*x + e) + a)^2, x)