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

Optimal. Leaf size=119 \[ \frac{2 a^2 \tan (e+f x) (c-c \sec (e+f x))^n \text{Hypergeometric2F1}\left (1,n+\frac{1}{2},n+\frac{3}{2},1-\sec (e+f x)\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \tan (e+f x) (c-c \sec (e+f x))^n}{f (2 n+1) \sqrt{a \sec (e+f x)+a}} \]

[Out]

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

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Rubi [A]  time = 0.162068, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.107, Rules used = {3909, 3912, 65} \[ \frac{2 a^2 \tan (e+f x) (c-c \sec (e+f x))^n \, _2F_1\left (1,n+\frac{1}{2};n+\frac{3}{2};1-\sec (e+f x)\right )}{f (2 n+1) \sqrt{a \sec (e+f x)+a}}+\frac{2 a^2 \tan (e+f x) (c-c \sec (e+f x))^n}{f (2 n+1) \sqrt{a \sec (e+f x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3909

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Si
mp[(-2*a^2*Cot[e + f*x]*(c + d*Csc[e + f*x])^n)/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]]), x] + Dist[a, Int[Sqrt[
a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && EqQ[b*c + a*d, 0] &&
EqQ[a^2 - b^2, 0] &&  !LeQ[n, -2^(-1)]

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 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])

Rubi steps

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

Mathematica [F]  time = 11.8065, size = 0, normalized size = 0. \[ \int (a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^n \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [F]  time = 0.302, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( c-c\sec \left ( fx+e \right ) \right ) ^{n}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out