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

Optimal. Leaf size=171 \[ -\frac {256 c^4 \tan (e+f x) (a \sec (e+f x)+a)^2}{1155 f \sqrt {c-c \sec (e+f x)}}-\frac {64 c^3 \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}{231 f}-\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}{33 f}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{5/2}}{11 f} \]

[Out]

-8/33*c^2*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/f-2/11*c*(a+a*sec(f*x+e))^2*(c-c*sec(f*x+e))^(5
/2)*tan(f*x+e)/f-256/1155*c^4*(a+a*sec(f*x+e))^2*tan(f*x+e)/f/(c-c*sec(f*x+e))^(1/2)-64/231*c^3*(a+a*sec(f*x+e
))^2*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f

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Rubi [A]  time = 0.45, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.059, Rules used = {3955, 3953} \[ -\frac {256 c^4 \tan (e+f x) (a \sec (e+f x)+a)^2}{1155 f \sqrt {c-c \sec (e+f x)}}-\frac {64 c^3 \tan (e+f x) (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}}{231 f}-\frac {8 c^2 \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}{33 f}-\frac {2 c \tan (e+f x) (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{5/2}}{11 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-256*c^4*(a + a*Sec[e + f*x])^2*Tan[e + f*x])/(1155*f*Sqrt[c - c*Sec[e + f*x]]) - (64*c^3*(a + a*Sec[e + f*x]
)^2*Sqrt[c - c*Sec[e + f*x]]*Tan[e + f*x])/(231*f) - (8*c^2*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)*
Tan[e + f*x])/(33*f) - (2*c*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(5/2)*Tan[e + f*x])/(11*f)

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))^2 (c-c \sec (e+f x))^{7/2} \, dx &=-\frac {2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{11 f}+\frac {1}{11} (12 c) \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \, dx\\ &=-\frac {8 c^2 (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{33 f}-\frac {2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{11 f}+\frac {1}{33} \left (32 c^2\right ) \int \sec (e+f x) (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \, dx\\ &=-\frac {64 c^3 (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{231 f}-\frac {8 c^2 (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{33 f}-\frac {2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{11 f}+\frac {1}{231} \left (128 c^3\right ) \int \sec (e+f x) (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)} \, dx\\ &=-\frac {256 c^4 (a+a \sec (e+f x))^2 \tan (e+f x)}{1155 f \sqrt {c-c \sec (e+f x)}}-\frac {64 c^3 (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{231 f}-\frac {8 c^2 (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{33 f}-\frac {2 c (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{11 f}\\ \end {align*}

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Mathematica [A]  time = 1.64, size = 88, normalized size = 0.51 \[ \frac {2 a^2 c^3 \cos ^4\left (\frac {1}{2} (e+f x)\right ) (3419 \cos (e+f x)-1510 \cos (2 (e+f x))+533 \cos (3 (e+f x))-1930) \cot \left (\frac {1}{2} (e+f x)\right ) \sec ^5(e+f x) \sqrt {c-c \sec (e+f x)}}{1155 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*c^3*Cos[(e + f*x)/2]^4*(-1930 + 3419*Cos[e + f*x] - 1510*Cos[2*(e + f*x)] + 533*Cos[3*(e + f*x)])*Cot[(
e + f*x)/2]*Sec[e + f*x]^5*Sqrt[c - c*Sec[e + f*x]])/(1155*f)

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fricas [A]  time = 0.45, size = 147, normalized size = 0.86 \[ \frac {2 \, {\left (533 \, a^{2} c^{3} \cos \left (f x + e\right )^{6} + 844 \, a^{2} c^{3} \cos \left (f x + e\right )^{5} - 211 \, a^{2} c^{3} \cos \left (f x + e\right )^{4} - 472 \, a^{2} c^{3} \cos \left (f x + e\right )^{3} + 295 \, a^{2} c^{3} \cos \left (f x + e\right )^{2} + 140 \, a^{2} c^{3} \cos \left (f x + e\right ) - 105 \, a^{2} c^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{1155 \, f \cos \left (f x + e\right )^{5} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/1155*(533*a^2*c^3*cos(f*x + e)^6 + 844*a^2*c^3*cos(f*x + e)^5 - 211*a^2*c^3*cos(f*x + e)^4 - 472*a^2*c^3*cos
(f*x + e)^3 + 295*a^2*c^3*cos(f*x + e)^2 + 140*a^2*c^3*cos(f*x + e) - 105*a^2*c^3)*sqrt((c*cos(f*x + e) - c)/c
os(f*x + e))/(f*cos(f*x + e)^5*sin(f*x + e))

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giac [A]  time = 4.14, size = 113, normalized size = 0.66 \[ -\frac {64 \, \sqrt {2} {\left (231 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{3} c^{3} + 495 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} c^{4} + 385 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{5} + 105 \, c^{6}\right )} a^{2} c^{3}}{1155 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {11}{2}} f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-64/1155*sqrt(2)*(231*(c*tan(1/2*f*x + 1/2*e)^2 - c)^3*c^3 + 495*(c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^4 + 385*(c
*tan(1/2*f*x + 1/2*e)^2 - c)*c^5 + 105*c^6)*a^2*c^3/((c*tan(1/2*f*x + 1/2*e)^2 - c)^(11/2)*f)

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maple [A]  time = 1.65, size = 85, normalized size = 0.50 \[ -\frac {2 a^{2} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}} \left (\sin ^{5}\left (f x +e \right )\right ) \left (533 \left (\cos ^{3}\left (f x +e \right )\right )-755 \left (\cos ^{2}\left (f x +e \right )\right )+455 \cos \left (f x +e \right )-105\right )}{1155 f \left (-1+\cos \left (f x +e \right )\right )^{6} \cos \left (f x +e \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/1155*a^2/f*(c*(-1+cos(f*x+e))/cos(f*x+e))^(7/2)*sin(f*x+e)^5*(533*cos(f*x+e)^3-755*cos(f*x+e)^2+455*cos(f*x
+e)-105)/(-1+cos(f*x+e))^6/cos(f*x+e)^2

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

[Out]

Timed out

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mupad [B]  time = 14.40, size = 606, normalized size = 3.54 \[ \frac {\left (\frac {a^2\,c^3\,2{}\mathrm {i}}{f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,1066{}\mathrm {i}}{1155\,f}\right )\,\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}}}}{{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1}+\frac {\left (\frac {a^2\,c^3\,64{}\mathrm {i}}{11\,f}-\frac {a^2\,c^3\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,64{}\mathrm {i}}{11\,f}\right )\,\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}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^5}-\frac {\left (\frac {a^2\,c^3\,32{}\mathrm {i}}{3\,f}-\frac {a^2\,c^3\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,608{}\mathrm {i}}{33\,f}\right )\,\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}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^4}-\frac {\left (\frac {a^2\,c^3\,4{}\mathrm {i}}{f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,2932{}\mathrm {i}}{1155\,f}\right )\,\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}}-1\right )\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}+\frac {\left (\frac {a^2\,c^3\,16{}\mathrm {i}}{5\,f}+\frac {a^2\,c^3\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,4272{}\mathrm {i}}{385\,f}\right )\,\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}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\left (\frac {a^2\,c^3\,32{}\mathrm {i}}{7\,f}-\frac {a^2\,c^3\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,4640{}\mathrm {i}}{231\,f}\right )\,\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}}-1\right )\,{\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(((a^2*c^3*2i)/f + (a^2*c^3*exp(e*1i + f*x*1i)*1066i)/(1155*f))*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*
x*1i)/2))^(1/2))/(exp(e*1i + f*x*1i) - 1) + (((a^2*c^3*64i)/(11*f) - (a^2*c^3*exp(e*1i + f*x*1i)*64i)/(11*f))*
(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) +
1)^5) - (((a^2*c^3*32i)/(3*f) - (a^2*c^3*exp(e*1i + f*x*1i)*608i)/(33*f))*(c - c/(exp(- e*1i - f*x*1i)/2 + ex
p(e*1i + f*x*1i)/2))^(1/2))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^4) - (((a^2*c^3*4i)/f + (a^2*c^
3*exp(e*1i + f*x*1i)*2932i)/(1155*f))*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2))/((exp(e*1
i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)) + (((a^2*c^3*16i)/(5*f) + (a^2*c^3*exp(e*1i + f*x*1i)*4272i)/(385*f
))*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i
) + 1)^2) + (((a^2*c^3*32i)/(7*f) - (a^2*c^3*exp(e*1i + f*x*1i)*4640i)/(231*f))*(c - c/(exp(- e*1i - f*x*1i)/2
+ exp(e*1i + f*x*1i)/2))^(1/2))/((exp(e*1i + f*x*1i) - 1)*(exp(e*2i + f*x*2i) + 1)^3)

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

[Out]

Timed out

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