3.2.42 \(\int \frac {(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx\) [142]

Optimal. Leaf size=205 \[ -\frac {(5-2 n) \, _2F_1\left (1,\frac {1}{2}+n;\frac {3}{2}+n;\frac {1}{2} (1-\sec (e+f x))\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{4 a f (1+2 n) \sqrt {a+a \sec (e+f x)}}+\frac {2 \, _2F_1\left (1,\frac {1}{2}+n;\frac {3}{2}+n;1-\sec (e+f x)\right ) (c-c \sec (e+f x))^n \tan (e+f x)}{a f (1+2 n) \sqrt {a+a \sec (e+f x)}}-\frac {(c-c \sec (e+f x))^n \tan (e+f x)}{2 a f (1+\sec (e+f x)) \sqrt {a+a \sec (e+f x)}} \]

[Out]

-1/4*(5-2*n)*hypergeom([1, 1/2+n],[3/2+n],1/2-1/2*sec(f*x+e))*(c-c*sec(f*x+e))^n*tan(f*x+e)/a/f/(1+2*n)/(a+a*s
ec(f*x+e))^(1/2)+2*hypergeom([1, 1/2+n],[3/2+n],1-sec(f*x+e))*(c-c*sec(f*x+e))^n*tan(f*x+e)/a/f/(1+2*n)/(a+a*s
ec(f*x+e))^(1/2)-1/2*(c-c*sec(f*x+e))^n*tan(f*x+e)/a/f/(1+sec(f*x+e))/(a+a*sec(f*x+e))^(1/2)

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Rubi [A]
time = 0.13, antiderivative size = 205, normalized size of antiderivative = 1.00, 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 = {3997, 105, 162, 67, 70} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*((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])/(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 67

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

Rule 70

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

Rule 105

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] && ILtQ[m, -1] &
& (IntegerQ[n] || IntegersQ[2*n, 2*p] || ILtQ[m + n + p + 3, 0])

Rule 162

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 3997

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]

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)) \text {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) \text {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)) \text {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)) \text {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*}

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Mathematica [F]
time = 1.49, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {(c-c \sec (e+f x))^n}{(a+a \sec (e+f x))^{3/2}} \, dx \end {gather*}

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 = 0.17, size = 0, normalized size = 0.00 \[\int \frac {\left (c -c \sec \left (f x +e \right )\right )^{n}}{\left (a +a \sec \left (f x +e \right )\right )^{\frac {3}{2}}}\, dx\]

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (- c \left (\sec {\left (e + f x \right )} - 1\right )\right )^{n}}{\left (a \left (\sec {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

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]

Integral((-c*(sec(e + f*x) - 1))**n/(a*(sec(e + f*x) + 1))**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^n}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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