3.93 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx\)

Optimal. Leaf size=155 \[ -\frac {64 c^4 \tan (e+f x)}{3 a^2 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^3 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{3 a^2 f}-\frac {4 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2} \]

[Out]

-4*c^2*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f/(a^2+a^2*sec(f*x+e))+2/3*c*(c-c*sec(f*x+e))^(5/2)*tan(f*x+e)/f/(a+a
*sec(f*x+e))^2-64/3*c^4*tan(f*x+e)/a^2/f/(c-c*sec(f*x+e))^(1/2)-16/3*c^3*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/a^2
/f

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Rubi [A]  time = 0.32, antiderivative size = 155, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.088, Rules used = {3954, 3793, 3792} \[ -\frac {64 c^4 \tan (e+f x)}{3 a^2 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^3 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{3 a^2 f}-\frac {4 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{f \left (a^2 \sec (e+f x)+a^2\right )}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{3 f (a \sec (e+f x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(7/2))/(a + a*Sec[e + f*x])^2,x]

[Out]

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

Rule 3792

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

Rule 3793

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

Rule 3954

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

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^2} \, dx &=\frac {2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {(2 c) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{a+a \sec (e+f x)} \, dx}{a}\\ &=-\frac {4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (8 c^2\right ) \int \sec (e+f x) (c-c \sec (e+f x))^{3/2} \, dx}{a^2}\\ &=-\frac {16 c^3 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{3 a^2 f}-\frac {4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}+\frac {\left (32 c^3\right ) \int \sec (e+f x) \sqrt {c-c \sec (e+f x)} \, dx}{3 a^2}\\ &=-\frac {64 c^4 \tan (e+f x)}{3 a^2 f \sqrt {c-c \sec (e+f x)}}-\frac {16 c^3 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{3 a^2 f}-\frac {4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{f \left (a^2+a^2 \sec (e+f x)\right )}+\frac {2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.94, size = 84, normalized size = 0.54 \[ \frac {c^3 (195 \cos (e+f x)+138 \cos (2 (e+f x))+45 \cos (3 (e+f x))+134) \cot \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {c-c \sec (e+f x)}}{6 a^2 f (\cos (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(7/2))/(a + a*Sec[e + f*x])^2,x]

[Out]

(c^3*(134 + 195*Cos[e + f*x] + 138*Cos[2*(e + f*x)] + 45*Cos[3*(e + f*x)])*Cot[(e + f*x)/2]*Sec[e + f*x]*Sqrt[
c - c*Sec[e + f*x]])/(6*a^2*f*(1 + Cos[e + f*x])^2)

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fricas [A]  time = 0.43, size = 103, normalized size = 0.66 \[ \frac {2 \, {\left (45 \, c^{3} \cos \left (f x + e\right )^{3} + 69 \, c^{3} \cos \left (f x + e\right )^{2} + 15 \, c^{3} \cos \left (f x + e\right ) - c^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x, algorithm="fricas")

[Out]

2/3*(45*c^3*cos(f*x + e)^3 + 69*c^3*cos(f*x + e)^2 + 15*c^3*cos(f*x + e) - c^3)*sqrt((c*cos(f*x + e) - c)/cos(
f*x + e))/((a^2*f*cos(f*x + e)^2 + a^2*f*cos(f*x + e))*sin(f*x + e))

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giac [A]  time = 3.38, size = 125, normalized size = 0.81 \[ -\frac {4 \, \sqrt {2} c^{3} {\left (\frac {9 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c + c^{2}}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} a^{2}} - \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} a^{4} c^{2} + 9 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} a^{4} c^{3}}{a^{6} c^{3}}\right )}}{3 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x, algorithm="giac")

[Out]

-4/3*sqrt(2)*c^3*((9*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c + c^2)/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*a^2) - ((c*
tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*a^4*c^2 + 9*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*a^4*c^3)/(a^6*c^3))/f

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maple [A]  time = 1.63, size = 85, normalized size = 0.55 \[ -\frac {2 \left (3 \cos \left (f x +e \right )+1\right ) \left (15 \left (\cos ^{2}\left (f x +e \right )\right )+18 \cos \left (f x +e \right )-1\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}}{3 a^{2} f \sin \left (f x +e \right )^{3} \left (-1+\cos \left (f x +e \right )\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x)

[Out]

-2/3/a^2/f*(3*cos(f*x+e)+1)*(15*cos(f*x+e)^2+18*cos(f*x+e)-1)*cos(f*x+e)^2*(c*(-1+cos(f*x+e))/cos(f*x+e))^(7/2
)/sin(f*x+e)^3/(-1+cos(f*x+e))^2

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maxima [A]  time = 0.50, size = 188, normalized size = 1.21 \[ -\frac {4 \, {\left (16 \, \sqrt {2} c^{\frac {7}{2}} - \frac {56 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {70 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {35 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {4 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} + \frac {\sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}}\right )}}{3 \, a^{2} f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^2,x, algorithm="maxima")

[Out]

-4/3*(16*sqrt(2)*c^(7/2) - 56*sqrt(2)*c^(7/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 70*sqrt(2)*c^(7/2)*sin(f*x
+ e)^4/(cos(f*x + e) + 1)^4 - 35*sqrt(2)*c^(7/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 + 4*sqrt(2)*c^(7/2)*sin(
f*x + e)^8/(cos(f*x + e) + 1)^8 + sqrt(2)*c^(7/2)*sin(f*x + e)^10/(cos(f*x + e) + 1)^10)/(a^2*f*(sin(f*x + e)/
(cos(f*x + e) + 1) + 1)^(7/2)*(sin(f*x + e)/(cos(f*x + e) + 1) - 1)^(7/2))

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mupad [B]  time = 6.02, size = 188, normalized size = 1.21 \[ \frac {2\,c^3\,\sqrt {c-\frac {c}{\frac {{\mathrm {e}}^{-e\,1{}\mathrm {i}-f\,x\,1{}\mathrm {i}}}{2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}}{2}}}\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,138{}\mathrm {i}+{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,195{}\mathrm {i}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,268{}\mathrm {i}+{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,195{}\mathrm {i}+{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}\,138{}\mathrm {i}+{\mathrm {e}}^{e\,6{}\mathrm {i}+f\,x\,6{}\mathrm {i}}\,45{}\mathrm {i}+45{}\mathrm {i}\right )}{3\,a^2\,f\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1\right )}^3\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c - c/cos(e + f*x))^(7/2)/(cos(e + f*x)*(a + a/cos(e + f*x))^2),x)

[Out]

(2*c^3*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*(exp(e*1i + f*x*1i)*138i + exp(e*2i + f*x
*2i)*195i + exp(e*3i + f*x*3i)*268i + exp(e*4i + f*x*4i)*195i + exp(e*5i + f*x*5i)*138i + exp(e*6i + f*x*6i)*4
5i + 45i))/(3*a^2*f*(exp(e*1i + f*x*1i) + 1)^3*(exp(e*1i + f*x*1i) - exp(e*2i + f*x*2i) + exp(e*3i + f*x*3i) -
1))

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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)*(c-c*sec(f*x+e))**(7/2)/(a+a*sec(f*x+e))**2,x)

[Out]

Timed out

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