3.156 \(\int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=160 \[ -\frac{16 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)} (a \sec (e+f x)+a)^m}{f \left (4 m^2+16 m+15\right )}-\frac{64 c^3 \tan (e+f x) (a \sec (e+f x)+a)^m}{f (2 m+5) \left (4 m^2+8 m+3\right ) \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2} (a \sec (e+f x)+a)^m}{f (2 m+5)} \]

[Out]

(-64*c^3*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(5 + 2*m)*(3 + 8*m + 4*m^2)*Sqrt[c - c*Sec[e + f*x]]) - (16*c
^2*(a + a*Sec[e + f*x])^m*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(f*(15 + 16*m + 4*m^2)) - (2*c*(a + a*Sec[e +
f*x])^m*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(5 + 2*m))

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Rubi [A]  time = 0.380242, antiderivative size = 160, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.059, Rules used = {3955, 3953} \[ -\frac{16 c^2 \tan (e+f x) \sqrt{c-c \sec (e+f x)} (a \sec (e+f x)+a)^m}{f \left (4 m^2+16 m+15\right )}-\frac{64 c^3 \tan (e+f x) (a \sec (e+f x)+a)^m}{f (2 m+5) \left (4 m^2+8 m+3\right ) \sqrt{c-c \sec (e+f x)}}-\frac{2 c \tan (e+f x) (c-c \sec (e+f x))^{3/2} (a \sec (e+f x)+a)^m}{f (2 m+5)} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]*(a + a*Sec[e + f*x])^m*(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(-64*c^3*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(f*(5 + 2*m)*(3 + 8*m + 4*m^2)*Sqrt[c - c*Sec[e + f*x]]) - (16*c
^2*(a + a*Sec[e + f*x])^m*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(f*(15 + 16*m + 4*m^2)) - (2*c*(a + a*Sec[e +
f*x])^m*(c - c*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(5 + 2*m))

Rule 3955

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

Rule 3953

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

Rubi steps

\begin{align*} \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx &=-\frac{2 c (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (5+2 m)}+\frac{(8 c) \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{3/2} \, dx}{5+2 m}\\ &=-\frac{16 c^2 (a+a \sec (e+f x))^m \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{f \left (15+16 m+4 m^2\right )}-\frac{2 c (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (5+2 m)}+\frac{\left (32 c^2\right ) \int \sec (e+f x) (a+a \sec (e+f x))^m \sqrt{c-c \sec (e+f x)} \, dx}{15+16 m+4 m^2}\\ &=-\frac{64 c^3 (a+a \sec (e+f x))^m \tan (e+f x)}{f \left (15+46 m+36 m^2+8 m^3\right ) \sqrt{c-c \sec (e+f x)}}-\frac{16 c^2 (a+a \sec (e+f x))^m \sqrt{c-c \sec (e+f x)} \tan (e+f x)}{f \left (15+16 m+4 m^2\right )}-\frac{2 c (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f (5+2 m)}\\ \end{align*}

Mathematica [F]  time = 15.7782, size = 0, normalized size = 0. \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{5/2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^m*(c - c*Sec[e + f*x])^(5/2),x]

[Out]

Integrate[Sec[e + f*x]*(a + a*Sec[e + f*x])^m*(c - c*Sec[e + f*x])^(5/2), x]

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Maple [F]  time = 0.265, size = 0, normalized size = 0. \begin{align*} \int \sec \left ( fx+e \right ) \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m} \left ( c-c\sec \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(5/2),x)

[Out]

int(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(5/2),x)

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Maxima [A]  time = 1.56329, size = 308, normalized size = 1.92 \begin{align*} -\frac{2 \,{\left (\frac{\sqrt{2}{\left (2^{m + 5} m + 5 \cdot 2^{m + 4}\right )} \left (-a\right )^{m} c^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac{\sqrt{2}{\left (2^{m + 4} m^{2} + 2^{m + 6} m + 15 \cdot 2^{m + 2}\right )} \left (-a\right )^{m} c^{\frac{5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - 2^{m + \frac{11}{2}} \left (-a\right )^{m} c^{\frac{5}{2}}\right )} e^{\left (-m \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )\right )}}{{\left (8 \, m^{3} + 36 \, m^{2} + 46 \, m + 15\right )} f{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac{5}{2}}{\left (\frac{\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-2*(sqrt(2)*(2^(m + 5)*m + 5*2^(m + 4))*(-a)^m*c^(5/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - sqrt(2)*(2^(m + 4
)*m^2 + 2^(m + 6)*m + 15*2^(m + 2))*(-a)^m*c^(5/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - 2^(m + 11/2)*(-a)^m*c
^(5/2))*e^(-m*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1) - m*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1))/((8*m^3 +
36*m^2 + 46*m + 15)*f*(sin(f*x + e)/(cos(f*x + e) + 1) + 1)^(5/2)*(sin(f*x + e)/(cos(f*x + e) + 1) - 1)^(5/2)
)

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Fricas [A]  time = 0.52022, size = 435, normalized size = 2.72 \begin{align*} \frac{2 \,{\left (4 \, c^{2} m^{2} +{\left (4 \, c^{2} m^{2} + 24 \, c^{2} m + 43 \, c^{2}\right )} \cos \left (f x + e\right )^{3} + 8 \, c^{2} m -{\left (4 \, c^{2} m^{2} + 8 \, c^{2} m - 29 \, c^{2}\right )} \cos \left (f x + e\right )^{2} + 3 \, c^{2} -{\left (4 \, c^{2} m^{2} + 24 \, c^{2} m + 11 \, c^{2}\right )} \cos \left (f x + e\right )\right )} \left (\frac{a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}\right )^{m} \sqrt{\frac{c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{{\left (8 \, f m^{3} + 36 \, f m^{2} + 46 \, f m + 15 \, f\right )} \cos \left (f x + e\right )^{2} \sin \left (f x + e\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

2*(4*c^2*m^2 + (4*c^2*m^2 + 24*c^2*m + 43*c^2)*cos(f*x + e)^3 + 8*c^2*m - (4*c^2*m^2 + 8*c^2*m - 29*c^2)*cos(f
*x + e)^2 + 3*c^2 - (4*c^2*m^2 + 24*c^2*m + 11*c^2)*cos(f*x + e))*((a*cos(f*x + e) + a)/cos(f*x + e))^m*sqrt((
c*cos(f*x + e) - c)/cos(f*x + e))/((8*f*m^3 + 36*f*m^2 + 46*f*m + 15*f)*cos(f*x + e)^2*sin(f*x + e))

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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(sec(f*x+e)*(a+a*sec(f*x+e))**m*(c-c*sec(f*x+e))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac{5}{2}}{\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(a+a*sec(f*x+e))^m*(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((-c*sec(f*x + e) + c)^(5/2)*(a*sec(f*x + e) + a)^m*sec(f*x + e), x)