3.142 \(\int \frac{(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=205 \[ -\frac{(5-2 n) \tan (e+f x) (c-c \sec (e+f x))^n \text{Hypergeometric2F1}\left (1,n+\frac{1}{2},n+\frac{3}{2},\frac{1}{2} (1-\sec (e+f x))\right )}{4 a f (2 n+1) \sqrt{a \sec (e+f x)+a}}+\frac{2 \tan (e+f x) (c-c \sec (e+f x))^n \text{Hypergeometric2F1}\left (1,n+\frac{1}{2},n+\frac{3}{2},1-\sec (e+f x)\right )}{a f (2 n+1) \sqrt{a \sec (e+f x)+a}}-\frac{\tan (e+f x) (c-c \sec (e+f x))^n}{2 a f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}} \]

[Out]

-((5 - 2*n)*Hypergeometric2F1[1, 1/2 + n, 3/2 + n, (1 - Sec[e + f*x])/2]*(c - c*Sec[e + f*x])^n*Tan[e + f*x])/
(4*a*f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]]) + (2*Hypergeometric2F1[1, 1/2 + n, 3/2 + n, 1 - Sec[e + f*x]]*(c -
c*Sec[e + f*x])^n*Tan[e + f*x])/(a*f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]]) - ((c - c*Sec[e + f*x])^n*Tan[e + f*x
])/(2*a*f*(1 + Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]])

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Rubi [A]  time = 0.167874, antiderivative size = 205, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {3912, 103, 156, 65, 68} \[ -\frac{(5-2 n) \tan (e+f x) (c-c \sec (e+f x))^n \, _2F_1\left (1,n+\frac{1}{2};n+\frac{3}{2};\frac{1}{2} (1-\sec (e+f x))\right )}{4 a f (2 n+1) \sqrt{a \sec (e+f x)+a}}+\frac{2 \tan (e+f x) (c-c \sec (e+f x))^n \, _2F_1\left (1,n+\frac{1}{2};n+\frac{3}{2};1-\sec (e+f x)\right )}{a f (2 n+1) \sqrt{a \sec (e+f x)+a}}-\frac{\tan (e+f x) (c-c \sec (e+f x))^n}{2 a f (\sec (e+f x)+1) \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(c - c*Sec[e + f*x])^n/(a + a*Sec[e + f*x])^(3/2),x]

[Out]

-((5 - 2*n)*Hypergeometric2F1[1, 1/2 + n, 3/2 + n, (1 - Sec[e + f*x])/2]*(c - c*Sec[e + f*x])^n*Tan[e + f*x])/
(4*a*f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]]) + (2*Hypergeometric2F1[1, 1/2 + n, 3/2 + n, 1 - Sec[e + f*x]]*(c -
c*Sec[e + f*x])^n*Tan[e + f*x])/(a*f*(1 + 2*n)*Sqrt[a + a*Sec[e + f*x]]) - ((c - c*Sec[e + f*x])^n*Tan[e + f*x
])/(2*a*f*(1 + Sec[e + f*x])*Sqrt[a + a*Sec[e + f*x]])

Rule 3912

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Di
st[(a*c*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[c + d*Csc[e + f*x]]), Subst[Int[((a + b*x)^(m - 1/2)*(c
+ d*x)^(n - 1/2))/x, x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && E
qQ[a^2 - b^2, 0]

Rule 103

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[1/((m + 1)*(b*
c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*(m + 1) - b*(d*e*(m + n + 2) +
c*f*(m + p + 2)) - b*d*f*(m + n + p + 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && LtQ[m, -1] &&
IntegerQ[m] && (IntegerQ[n] || IntegersQ[2*n, 2*p])

Rule 156

Int[(((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :>
Dist[(b*g - a*h)/(b*c - a*d), Int[(e + f*x)^p/(a + b*x), x], x] - Dist[(d*g - c*h)/(b*c - a*d), Int[(e + f*x)
^p/(c + d*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, h}, x]

Rule 65

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((c + d*x)^(n + 1)*Hypergeometric2F1[-m, n +
1, n + 2, 1 + (d*x)/c])/(d*(n + 1)*(-(d/(b*c)))^m), x] /; FreeQ[{b, c, d, m, n}, x] &&  !IntegerQ[n] && (Inte
gerQ[m] || GtQ[-(d/(b*c)), 0])

Rule 68

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((b*c - a*d)^n*(a + b*x)^(m + 1)*Hype
rgeometric2F1[-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b^(n + 1)*(m + 1)), x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rubi steps

\begin{align*} \int \frac{(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx &=-\frac{(a c \tan (e+f x)) \operatorname{Subst}\left (\int \frac{(c-c x)^{-\frac{1}{2}+n}}{x (a+a x)^2} \, dx,x,\sec (e+f x)\right )}{f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{(c-c \sec (e+f x))^n \tan (e+f x)}{2 a f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}-\frac{\tan (e+f x) \operatorname{Subst}\left (\int \frac{(c-c x)^{-\frac{1}{2}+n} \left (2 a c-\frac{1}{2} a c (1-2 n) x\right )}{x (a+a x)} \, dx,x,\sec (e+f x)\right )}{2 a f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{(c-c \sec (e+f x))^n \tan (e+f x)}{2 a f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}-\frac{(c \tan (e+f x)) \operatorname{Subst}\left (\int \frac{(c-c x)^{-\frac{1}{2}+n}}{x} \, dx,x,\sec (e+f x)\right )}{a f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}+\frac{(c (5-2 n) \tan (e+f x)) \operatorname{Subst}\left (\int \frac{(c-c x)^{-\frac{1}{2}+n}}{a+a x} \, dx,x,\sec (e+f x)\right )}{4 f \sqrt{a+a \sec (e+f x)} \sqrt{c-c \sec (e+f x)}}\\ &=-\frac{(5-2 n) \, _2F_1\left (1,\frac{1}{2}+n;\frac{3}{2}+n;\frac{1}{2} (1-\sec (e+f x))\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{4 a f (1+2 n) \sqrt{a+a \sec (e+f x)}}+\frac{2 \, _2F_1\left (1,\frac{1}{2}+n;\frac{3}{2}+n;1-\sec (e+f x)\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{a f (1+2 n) \sqrt{a+a \sec (e+f x)}}-\frac{(c-c \sec (e+f x))^n \tan (e+f x)}{2 a f (1+\sec (e+f x)) \sqrt{a+a \sec (e+f x)}}\\ \end{align*}

Mathematica [F]  time = 1.6534, size = 0, normalized size = 0. \[ \int \frac{(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(c - c*Sec[e + f*x])^n/(a + a*Sec[e + f*x])^(3/2),x]

[Out]

Integrate[(c - c*Sec[e + f*x])^n/(a + a*Sec[e + f*x])^(3/2), x]

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Maple [F]  time = 0.264, size = 0, normalized size = 0. \begin{align*} \int{ \left ( c-c\sec \left ( fx+e \right ) \right ) ^{n} \left ( a+a\sec \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(3/2),x)

[Out]

int((c-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (-c \sec \left (f x + e\right ) + c\right )}^{n}}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((-c*sec(f*x + e) + c)^n/(a*sec(f*x + e) + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{a \sec \left (f x + e\right ) + a}{\left (-c \sec \left (f x + e\right ) + c\right )}^{n}}{a^{2} \sec \left (f x + e\right )^{2} + 2 \, a^{2} \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-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*sec(f*x + e) + a)*(-c*sec(f*x + e) + c)^n/(a^2*sec(f*x + e)^2 + 2*a^2*sec(f*x + e) + a^2), x)

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Sympy [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-c*sec(f*x+e))**n/(a+a*sec(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [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-c*sec(f*x+e))^n/(a+a*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

Timed out