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

Optimal. Leaf size=88 \[ \frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{5 f (a \sec (e+f x)+a)^3}-\frac {4 c^2 \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}} \]

[Out]

-4/15*c^2*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(1/2)+2/5*c*(c-c*sec(f*x+e))^(1/2)*tan(f*x+e)/f/(
a+a*sec(f*x+e))^3

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Rubi [A]  time = 0.22, antiderivative size = 88, 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 = {3954, 3953} \[ \frac {2 c \tan (e+f x) \sqrt {c-c \sec (e+f x)}}{5 f (a \sec (e+f x)+a)^3}-\frac {4 c^2 \tan (e+f x)}{15 a f (a \sec (e+f x)+a)^2 \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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))^{3/2}}{(a+a \sec (e+f x))^3} \, dx &=\frac {2 c \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}-\frac {(2 c) \int \frac {\sec (e+f x) \sqrt {c-c \sec (e+f x)}}{(a+a \sec (e+f x))^2} \, dx}{5 a}\\ &=-\frac {4 c^2 \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2 \sqrt {c-c \sec (e+f x)}}+\frac {2 c \sqrt {c-c \sec (e+f x)} \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}\\ \end {align*}

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Mathematica [A]  time = 0.32, size = 60, normalized size = 0.68 \[ -\frac {2 c (\cos (e+f x)-5) \cot \left (\frac {1}{2} (e+f x)\right ) \sec (e+f x) \sqrt {c-c \sec (e+f x)}}{15 a^3 f (\sec (e+f x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*(-5 + Cos[e + f*x])*Cot[(e + f*x)/2]*Sec[e + f*x]*Sqrt[c - c*Sec[e + f*x]])/(15*a^3*f*(1 + Sec[e + f*x])
^3)

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fricas [A]  time = 0.45, size = 88, normalized size = 1.00 \[ -\frac {2 \, {\left (c \cos \left (f x + e\right )^{3} - 5 \, c \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{2} + 2 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )} \sin \left (f x + e\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^3,x, algorithm="fricas")

[Out]

-2/15*(c*cos(f*x + e)^3 - 5*c*cos(f*x + e)^2)*sqrt((c*cos(f*x + e) - c)/cos(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 = 3.49, size = 60, normalized size = 0.68 \[ -\frac {\sqrt {2} {\left (3 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {5}{2}} + 5 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} c\right )}}{30 \, a^{3} c f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^3,x, algorithm="giac")

[Out]

-1/30*sqrt(2)*(3*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2) + 5*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c)/(a^3*c*f)

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maple [A]  time = 1.82, size = 63, normalized size = 0.72 \[ -\frac {2 \left (5+\cos ^{2}\left (f x +e \right )-6 \cos \left (f x +e \right )\right ) \left (\cos ^{3}\left (f x +e \right )\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{15 a^{3} f \sin \left (f x +e \right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15/a^3/f*(5+cos(f*x+e)^2-6*cos(f*x+e))*cos(f*x+e)^3*(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x+e)^5

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maxima [B]  time = 0.44, size = 163, normalized size = 1.85 \[ -\frac {2 \, \sqrt {2} c^{\frac {3}{2}} - \frac {3 \, \sqrt {2} c^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {3 \, \sqrt {2} c^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {7 \, \sqrt {2} c^{\frac {3}{2}} \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - \frac {3 \, \sqrt {2} c^{\frac {3}{2}} \sin \left (f x + e\right )^{8}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{8}}}{30 \, a^{3} f {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}^{\frac {3}{2}} {\left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)*(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^3,x, algorithm="maxima")

[Out]

-1/30*(2*sqrt(2)*c^(3/2) - 3*sqrt(2)*c^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 3*sqrt(2)*c^(3/2)*sin(f*x +
e)^4/(cos(f*x + e) + 1)^4 + 7*sqrt(2)*c^(3/2)*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 3*sqrt(2)*c^(3/2)*sin(f*x
+ e)^8/(cos(f*x + e) + 1)^8)/(a^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 = 7.72, size = 446, normalized size = 5.07 \[ -\frac {c\,\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}}}\,2{}\mathrm {i}}{15\,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 )}+\frac {c\,\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}}}\,28{}\mathrm {i}}{15\,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\,\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}}}\,76{}\mathrm {i}}{15\,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\,\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}}}\,32{}\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\,\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 )}^5} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(c*(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)*28i)/(15*a^3*f*(exp(
e*1i + f*x*1i) - 1)*(exp(e*1i + f*x*1i) + 1)^2) - (c*(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)*2i)/(15*a^3*f*(exp(e*1i + f*x*1i) - 1)*(exp(e*1i + f*x*1i) + 1)) - (c*(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)*76i)/(15*a^3*f*(exp(e*1i + f*x*1i
) - 1)*(exp(e*1i + f*x*1i) + 1)^3) + (c*(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)*32i)/(5*a^3*f*(exp(e*1i + f*x*1i) - 1)*(exp(e*1i + f*x*1i) + 1)^4) - (c*(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)^5)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\right )\, dx}{a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1),
x) + Integral(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**2/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f
*x) + 1), x))/a**3

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