3.100 \(\int \frac {\sec (e+f x) (c-c \sec (e+f x))^{7/2}}{(a+a \sec (e+f x))^3} \, dx\)

Optimal. Leaf size=169 \[ \frac {32 c^4 \tan (e+f x)}{5 a^3 f \sqrt {c-c \sec (e+f x)}}+\frac {16 c^3 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{5 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac {4 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{5 a f (a \sec (e+f x)+a)^2}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{5 f (a \sec (e+f x)+a)^3} \]

[Out]

-4/5*c^2*(c-c*sec(f*x+e))^(3/2)*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^2+2/5*c*(c-c*sec(f*x+e))^(5/2)*tan(f*x+e)/f/(a
+a*sec(f*x+e))^3+32/5*c^4*tan(f*x+e)/a^3/f/(c-c*sec(f*x+e))^(1/2)+16/5*c^3*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f
/(a^3+a^3*sec(f*x+e))

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Rubi [A]  time = 0.40, antiderivative size = 169, 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 = {3954, 3792} \[ \frac {32 c^4 \tan (e+f x)}{5 a^3 f \sqrt {c-c \sec (e+f x)}}+\frac {16 c^3 \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{5 f \left (a^3 \sec (e+f x)+a^3\right )}-\frac {4 c^2 \tan (e+f x) (c-c \sec (e+f x))^{3/2}}{5 a f (a \sec (e+f x)+a)^2}+\frac {2 c \tan (e+f x) (c-c \sec (e+f x))^{5/2}}{5 f (a \sec (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(7/2))/(a + a*Sec[e + f*x])^3,x]

[Out]

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

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 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))^3} \, dx &=\frac {2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {(6 c) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^2} \, dx}{5 a}\\ &=-\frac {4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac {2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {\left (8 c^2\right ) \int \frac {\sec (e+f x) (c-c \sec (e+f x))^{3/2}}{a+a \sec (e+f x)} \, dx}{5 a^2}\\ &=\frac {16 c^3 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{5 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac {2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {\left (16 c^3\right ) \int \sec (e+f x) \sqrt {c-c \sec (e+f x)} \, dx}{5 a^3}\\ &=\frac {32 c^4 \tan (e+f x)}{5 a^3 f \sqrt {c-c \sec (e+f x)}}+\frac {16 c^3 \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{5 f \left (a^3+a^3 \sec (e+f x)\right )}-\frac {4 c^2 (c-c \sec (e+f x))^{3/2} \tan (e+f x)}{5 a f (a+a \sec (e+f x))^2}+\frac {2 c (c-c \sec (e+f x))^{5/2} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}\\ \end {align*}

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Mathematica [A]  time = 0.63, size = 78, normalized size = 0.46 \[ -\frac {c^3 (249 \cos (e+f x)+110 \cos (2 (e+f x))+23 \cos (3 (e+f x))+130) \cot \left (\frac {1}{2} (e+f x)\right ) \sqrt {c-c \sec (e+f x)}}{10 a^3 f (\cos (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[e + f*x]*(c - c*Sec[e + f*x])^(7/2))/(a + a*Sec[e + f*x])^3,x]

[Out]

-1/10*(c^3*(130 + 249*Cos[e + f*x] + 110*Cos[2*(e + f*x)] + 23*Cos[3*(e + f*x)])*Cot[(e + f*x)/2]*Sqrt[c - c*S
ec[e + f*x]])/(a^3*f*(1 + Cos[e + f*x])^3)

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fricas [A]  time = 0.43, size = 109, normalized size = 0.64 \[ -\frac {2 \, {\left (23 \, c^{3} \cos \left (f x + e\right )^{3} + 55 \, c^{3} \cos \left (f x + e\right )^{2} + 45 \, c^{3} \cos \left (f x + e\right ) + 5 \, c^{3}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{5 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} + 2 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )} \sin \left (f x + e\right )} \]

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))^3,x, algorithm="fricas")

[Out]

-2/5*(23*c^3*cos(f*x + e)^3 + 55*c^3*cos(f*x + e)^2 + 45*c^3*cos(f*x + e) + 5*c^3)*sqrt((c*cos(f*x + e) - c)/c
os(f*x + e))/((a^3*f*cos(f*x + e)^2 + 2*a^3*f*cos(f*x + e) + a^3*f)*sin(f*x + e))

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giac [A]  time = 4.47, size = 130, normalized size = 0.77 \[ \frac {2 \, \sqrt {2} c^{3} {\left (\frac {5 \, c}{\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} a^{3}} - \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {5}{2}} a^{12} c^{8} + 5 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} a^{12} c^{9} + 15 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} a^{12} c^{10}}{a^{15} c^{10}}\right )}}{5 \, f} \]

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))^3,x, algorithm="giac")

[Out]

2/5*sqrt(2)*c^3*(5*c/(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*a^3) - ((c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2)*a^12*c^8
+ 5*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*a^12*c^9 + 15*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*a^12*c^10)/(a^15*c^
10))/f

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maple [A]  time = 1.81, size = 85, normalized size = 0.50 \[ -\frac {2 \left (23 \left (\cos ^{3}\left (f x +e \right )\right )+55 \left (\cos ^{2}\left (f x +e \right )\right )+45 \cos \left (f x +e \right )+5\right ) \left (\cos ^{3}\left (f x +e \right )\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}}{5 a^{3} f \sin \left (f x +e \right )^{5} \left (-1+\cos \left (f x +e \right )\right )} \]

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))^3,x)

[Out]

-2/5/a^3/f*(23*cos(f*x+e)^3+55*cos(f*x+e)^2+45*cos(f*x+e)+5)*cos(f*x+e)^3*(c*(-1+cos(f*x+e))/cos(f*x+e))^(7/2)
/sin(f*x+e)^5/(-1+cos(f*x+e))

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maxima [A]  time = 0.44, size = 214, normalized size = 1.27 \[ \frac {2 \, {\left (16 \, \sqrt {2} c^{\frac {7}{2}} - \frac {56 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {70 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {35 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} + \frac {5 \, \sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}} - \frac {\sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{10}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{10}} + \frac {\sqrt {2} c^{\frac {7}{2}} \sin \left (f x + e\right )^{12}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{12}}\right )}}{5 \, a^{3} f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac {7}{2}}} \]

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))^3,x, algorithm="maxima")

[Out]

2/5*(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 + 5*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 + sqrt(2)*c^(7/2)*sin(f
*x + e)^12/(cos(f*x + e) + 1)^12)/(a^3*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 = 10.24, size = 492, normalized size = 2.91 \[ -\frac {\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 (\frac {c^3\,46{}\mathrm {i}}{5\,a^3\,f}+\frac {c^3\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,4{}\mathrm {i}}{a^3\,f}+\frac {c^3\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,46{}\mathrm {i}}{5\,a^3\,f}\right )}{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1\right )}-\frac {c^3\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\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}}}\,16{}\mathrm {i}}{5\,a^3\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1\right )}^2}-\frac {c^3\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\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}}}\,48{}\mathrm {i}}{5\,a^3\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1\right )}^3}+\frac {c^3\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\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}}}\,128{}\mathrm {i}}{5\,a^3\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1\right )}^4}-\frac {c^3\,\left ({\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}+1\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}}}\,64{}\mathrm {i}}{5\,a^3\,f\,\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}-1\right )\,{\left ({\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}+1\right )}^5} \]

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))^3),x)

[Out]

(c^3*(exp(e*2i + f*x*2i) + 1)*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*128i)/(5*a^3*f*(ex
p(e*1i + f*x*1i) - 1)*(exp(e*1i + f*x*1i) + 1)^4) - (c^3*(exp(e*2i + f*x*2i) + 1)*(c - c/(exp(- e*1i - f*x*1i)
/2 + exp(e*1i + f*x*1i)/2))^(1/2)*16i)/(5*a^3*f*(exp(e*1i + f*x*1i) - 1)*(exp(e*1i + f*x*1i) + 1)^2) - (c^3*(e
xp(e*2i + f*x*2i) + 1)*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*48i)/(5*a^3*f*(exp(e*1i +
f*x*1i) - 1)*(exp(e*1i + f*x*1i) + 1)^3) - ((c - c/(exp(- e*1i - f*x*1i)/2 + exp(e*1i + f*x*1i)/2))^(1/2)*((c
^3*46i)/(5*a^3*f) + (c^3*exp(e*1i + f*x*1i)*4i)/(a^3*f) + (c^3*exp(e*2i + f*x*2i)*46i)/(5*a^3*f)))/((exp(e*1i
+ f*x*1i) - 1)*(exp(e*1i + f*x*1i) + 1)) - (c^3*(exp(e*2i + f*x*2i) + 1)*(c - c/(exp(- e*1i - f*x*1i)/2 + exp(
e*1i + f*x*1i)/2))^(1/2)*64i)/(5*a^3*f*(exp(e*1i + f*x*1i) - 1)*(exp(e*1i + f*x*1i) + 1)^5)

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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))**3,x)

[Out]

Timed out

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