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

Optimal. Leaf size=322 \[ -\frac{2 a b \sin (e+f x) \cos ^2(e+f x)^{\frac{n p}{2}} \left (c (d \sec (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (n p-2),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{a^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p-1)} \left (c (d \sec (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (n p-3),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{b^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p-1)} \left (c (d \sec (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (n p-1),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]

[Out]

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

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Rubi [A]  time = 0.559336, antiderivative size = 322, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 5, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.185, Rules used = {3948, 3869, 2824, 3189, 429} \[ -\frac{2 a b \sin (e+f x) \cos ^2(e+f x)^{\frac{n p}{2}} \left (c (d \sec (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (n p-2),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{a^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p-1)} \left (c (d \sec (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (n p-3),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2}+\frac{b^2 \sin (e+f x) \cos (e+f x) \cos ^2(e+f x)^{\frac{1}{2} (n p-1)} \left (c (d \sec (e+f x))^p\right )^n F_1\left (\frac{1}{2};\frac{1}{2} (n p-1),2;\frac{3}{2};\sin ^2(e+f x),\frac{a^2 \sin ^2(e+f x)}{a^2-b^2}\right )}{f \left (a^2-b^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3948

Int[((c_.)*((d_.)*sec[(e_.) + (f_.)*(x_)])^(p_))^(n_)*((a_.) + (b_.)*sec[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol]
:> Dist[(c^IntPart[n]*(c*(d*Sec[e + f*x])^p)^FracPart[n])/(d*Sec[e + f*x])^(p*FracPart[n]), Int[(a + b*Sec[e
+ f*x])^m*(d*Sec[e + f*x])^(n*p), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] &&  !IntegerQ[n]

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

Mathematica [B]  time = 32.8527, size = 10678, normalized size = 33.16 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Result too large to show

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Maple [F]  time = 0.154, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( c \left ( d\sec \left ( fx+e \right ) \right ) ^{p} \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((c*(d*sec(f*x+e))^p)^n/(a+b*sec(f*x+e))^2,x)

[Out]

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

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

[Out]

integral(((d*sec(f*x + e))^p*c)^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 (c \left (d \sec{\left (e + f x \right )}\right )^{p}\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((c*(d*sec(f*x+e))**p)**n/(a+b*sec(f*x+e))**2,x)

[Out]

Integral((c*(d*sec(e + f*x))**p)**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 (\left (d \sec \left (f x + e\right )\right )^{p} c\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((c*(d*sec(f*x+e))^p)^n/(a+b*sec(f*x+e))^2,x, algorithm="giac")

[Out]

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