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

Optimal. Leaf size=100 \[ -\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^m}{f \left (4 m^2+8 m+3\right ) \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)} (a \sec (e+f x)+a)^m}{f (2 m+3)} \]

[Out]

-8*c^2*(a+a*sec(f*x+e))^m*tan(f*x+e)/f/(4*m^2+8*m+3)/(c-c*sec(f*x+e))^(1/2)-2*c*(a+a*sec(f*x+e))^m*(c-c*sec(f*
x+e))^(1/2)*tan(f*x+e)/f/(3+2*m)

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Rubi [A]  time = 0.22, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {3955, 3953} \[ -\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^m}{f \left (4 m^2+8 m+3\right ) \sqrt {c-c \sec (e+f x)}}-\frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)} (a \sec (e+f x)+a)^m}{f (2 m+3)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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)]

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])

Rubi steps

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

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Mathematica [F]  time = 38.44, size = 0, normalized size = 0.00 \[ \int \sec (e+f x) (a+a \sec (e+f x))^m (c-c \sec (e+f x))^{3/2} \, dx \]

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])^(3/2),x]

[Out]

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

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fricas [A]  time = 0.44, size = 112, normalized size = 1.12 \[ \frac {2 \, {\left ({\left (2 \, c m + 5 \, c\right )} \cos \left (f x + e\right )^{2} - 2 \, c m + 4 \, c \cos \left (f x + e\right ) - c\right )} \left (\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}\right )^{m} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{{\left (4 \, f m^{2} + 8 \, f m + 3 \, f\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right )} \]

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))^(3/2),x, algorithm="fricas")

[Out]

2*((2*c*m + 5*c)*cos(f*x + e)^2 - 2*c*m + 4*c*cos(f*x + e) - c)*((a*cos(f*x + e) + a)/cos(f*x + e))^m*sqrt((c*
cos(f*x + e) - c)/cos(f*x + e))/((4*f*m^2 + 8*f*m + 3*f)*cos(f*x + e)*sin(f*x + e))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (-c \sec \left (f x + e\right ) + c\right )}^{\frac {3}{2}} {\left (a \sec \left (f x + e\right ) + a\right )}^{m} \sec \left (f x + e\right )\,{d x} \]

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))^(3/2),x, algorithm="giac")

[Out]

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

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maple [F]  time = 1.63, size = 0, normalized size = 0.00 \[ \int \sec \left (f x +e \right ) \left (a +a \sec \left (f x +e \right )\right )^{m} \left (c -c \sec \left (f x +e \right )\right )^{\frac {3}{2}}\, dx \]

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))^(3/2),x)

[Out]

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

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maxima [A]  time = 0.47, size = 171, normalized size = 1.71 \[ -\frac {2 \, {\left (\sqrt {2} 2^{m + 2} \left (-a\right )^{m} c^{\frac {3}{2}} - \frac {\sqrt {2} {\left (2^{m + 2} m + 3 \cdot 2^{m + 1}\right )} \left (-a\right )^{m} c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} e^{\left (-m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right ) - m \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )\right )}}{{\left (4 \, m^{2} + 8 \, m + 3\right )} f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac {3}{2}}} \]

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))^(3/2),x, algorithm="maxima")

[Out]

-2*(sqrt(2)*2^(m + 2)*(-a)^m*c^(3/2) - sqrt(2)*(2^(m + 2)*m + 3*2^(m + 1))*(-a)^m*c^(3/2)*sin(f*x + e)^2/(cos(
f*x + e) + 1)^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))/
((4*m^2 + 8*m + 3)*f*(sin(f*x + e)/(cos(f*x + e) + 1) + 1)^(3/2)*(sin(f*x + e)/(cos(f*x + e) + 1) - 1)^(3/2))

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mupad [B]  time = 3.59, size = 154, normalized size = 1.54 \[ -\frac {2\,c\,{\left (\frac {a\,\left (\cos \left (e+f\,x\right )+1\right )}{\cos \left (e+f\,x\right )}\right )}^m\,\sqrt {\frac {c\,\left (\cos \left (e+f\,x\right )-1\right )}{\cos \left (e+f\,x\right )}}\,\left (5\,\sin \left (e+f\,x\right )-2\,\sin \left (2\,e+2\,f\,x\right )+5\,\sin \left (3\,e+3\,f\,x\right )+2\,m\,\sin \left (e+f\,x\right )-4\,m\,\sin \left (2\,e+2\,f\,x\right )+2\,m\,\sin \left (3\,e+3\,f\,x\right )\right )}{f\,\left (4\,m^2+8\,m+3\right )\,\left (3\,\cos \left (e+f\,x\right )-2\,\cos \left (2\,e+2\,f\,x\right )+\cos \left (3\,e+3\,f\,x\right )-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*c*((a*(cos(e + f*x) + 1))/cos(e + f*x))^m*((c*(cos(e + f*x) - 1))/cos(e + f*x))^(1/2)*(5*sin(e + f*x) - 2*
sin(2*e + 2*f*x) + 5*sin(3*e + 3*f*x) + 2*m*sin(e + f*x) - 4*m*sin(2*e + 2*f*x) + 2*m*sin(3*e + 3*f*x)))/(f*(8
*m + 4*m^2 + 3)*(3*cos(e + f*x) - 2*cos(2*e + 2*f*x) + cos(3*e + 3*f*x) - 2))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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))**(3/2),x)

[Out]

Timed out

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