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

Optimal. Leaf size=188 \[ \frac {a \tan (e+f x) \log (1-\cos (e+f x))}{c^3 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x)}{c^2 f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}-\frac {a \tan (e+f x)}{2 c f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{3 f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{7/2}} \]

[Out]

-1/3*a*tan(f*x+e)/f/(c-c*sec(f*x+e))^(7/2)/(a+a*sec(f*x+e))^(1/2)-1/2*a*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^(5/2)/
(a+a*sec(f*x+e))^(1/2)-a*tan(f*x+e)/c^2/f/(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(1/2)+a*ln(1-cos(f*x+e))*tan
(f*x+e)/c^3/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3907, 3911, 31} \[ -\frac {a \tan (e+f x)}{c^2 f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}}+\frac {a \tan (e+f x) \log (1-\cos (e+f x))}{c^3 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {a \tan (e+f x)}{2 c f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{3 f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a*Tan[e + f*x])/(3*f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(7/2)) - (a*Tan[e + f*x])/(2*c*f*Sqrt[a +
 a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(5/2)) - (a*Tan[e + f*x])/(c^2*f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e +
 f*x])^(3/2)) + (a*Log[1 - Cos[e + f*x]]*Tan[e + f*x])/(c^3*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]
])

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 3907

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

Rule 3911

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

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{7/2}} \, dx &=-\frac {a \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}+\frac {\int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{5/2}} \, dx}{c}\\ &=-\frac {a \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}-\frac {a \tan (e+f x)}{2 c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {\int \frac {\sqrt {a+a \sec (e+f x)}}{(c-c \sec (e+f x))^{3/2}} \, dx}{c^2}\\ &=-\frac {a \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}-\frac {a \tan (e+f x)}{2 c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {\int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx}{c^3}\\ &=-\frac {a \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}-\frac {a \tan (e+f x)}{2 c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {(a \tan (e+f x)) \operatorname {Subst}\left (\int \frac {1}{-c+c x} \, dx,x,\cos (e+f x)\right )}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {a \tan (e+f x)}{3 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{7/2}}-\frac {a \tan (e+f x)}{2 c f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}-\frac {a \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a \log (1-\cos (e+f x)) \tan (e+f x)}{c^3 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 1.94, size = 198, normalized size = 1.05 \[ \frac {\tan \left (\frac {1}{2} (e+f x)\right ) \sqrt {a (\sec (e+f x)+1)} \left (-60 \log \left (1-e^{i (e+f x)}\right )-3 i f x \cos (3 (e+f x))+18 i \left (2 i \log \left (1-e^{i (e+f x)}\right )+f x+i\right ) \cos (2 (e+f x))+6 \log \left (1-e^{i (e+f x)}\right ) \cos (3 (e+f x))+9 \left (10 \log \left (1-e^{i (e+f x)}\right )-5 i f x+6\right ) \cos (e+f x)+30 i f x-40\right )}{12 c^3 f (\cos (e+f x)-1)^3 \sqrt {c-c \sec (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-40 + (30*I)*f*x - (3*I)*f*x*Cos[3*(e + f*x)] + (18*I)*Cos[2*(e + f*x)]*(I + f*x + (2*I)*Log[1 - E^(I*(e + f
*x))]) - 60*Log[1 - E^(I*(e + f*x))] + 6*Cos[3*(e + f*x)]*Log[1 - E^(I*(e + f*x))] + 9*Cos[e + f*x]*(6 - (5*I)
*f*x + 10*Log[1 - E^(I*(e + f*x))]))*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(12*c^3*f*(-1 + Cos[e + f*x]
)^3*Sqrt[c - c*Sec[e + f*x]])

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {a \sec \left (f x + e\right ) + a} \sqrt {-c \sec \left (f x + e\right ) + c}}{c^{4} \sec \left (f x + e\right )^{4} - 4 \, c^{4} \sec \left (f x + e\right )^{3} + 6 \, c^{4} \sec \left (f x + e\right )^{2} - 4 \, c^{4} \sec \left (f x + e\right ) + c^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(a*sec(f*x + e) + a)*sqrt(-c*sec(f*x + e) + c)/(c^4*sec(f*x + e)^4 - 4*c^4*sec(f*x + e)^3 + 6*c^4
*sec(f*x + e)^2 - 4*c^4*sec(f*x + e) + c^4), x)

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giac [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((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 2.18, size = 288, normalized size = 1.53 \[ -\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (12 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{3}\left (f x +e \right )\right )-6 \left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-36 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-7 \left (\cos ^{3}\left (f x +e \right )\right )+18 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )+36 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \cos \left (f x +e \right )+3 \left (\cos ^{2}\left (f x +e \right )\right )-18 \cos \left (f x +e \right ) \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-12 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+6 \cos \left (f x +e \right )+6 \ln \left (\frac {2}{1+\cos \left (f x +e \right )}\right )-4\right ) \sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}}{6 f \sin \left (f x +e \right ) \cos \left (f x +e \right )^{3} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(7/2),x)

[Out]

-1/6/f*(-1+cos(f*x+e))*(12*ln(-(-1+cos(f*x+e))/sin(f*x+e))*cos(f*x+e)^3-6*cos(f*x+e)^3*ln(2/(1+cos(f*x+e)))-36
*cos(f*x+e)^2*ln(-(-1+cos(f*x+e))/sin(f*x+e))-7*cos(f*x+e)^3+18*ln(2/(1+cos(f*x+e)))*cos(f*x+e)^2+36*ln(-(-1+c
os(f*x+e))/sin(f*x+e))*cos(f*x+e)+3*cos(f*x+e)^2-18*cos(f*x+e)*ln(2/(1+cos(f*x+e)))-12*ln(-(-1+cos(f*x+e))/sin
(f*x+e))+6*cos(f*x+e)+6*ln(2/(1+cos(f*x+e)))-4)*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)/sin(f*x+e)/cos(f*x+e)^3/(c
*(-1+cos(f*x+e))/cos(f*x+e))^(7/2)

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maxima [B]  time = 4.63, size = 2444, normalized size = 13.00 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

-1/3*(3*(f*x + e)*cos(6*f*x + 6*e)^2 + 108*(f*x + e)*cos(5*f*x + 5*e)^2 + 675*(f*x + e)*cos(4*f*x + 4*e)^2 + 1
200*(f*x + e)*cos(3*f*x + 3*e)^2 + 675*(f*x + e)*cos(2*f*x + 2*e)^2 + 108*(f*x + e)*cos(f*x + e)^2 + 3*(f*x +
e)*sin(6*f*x + 6*e)^2 + 108*(f*x + e)*sin(5*f*x + 5*e)^2 + 675*(f*x + e)*sin(4*f*x + 4*e)^2 + 1200*(f*x + e)*s
in(3*f*x + 3*e)^2 + 675*(f*x + e)*sin(2*f*x + 2*e)^2 + 108*(f*x + e)*sin(f*x + e)^2 + 3*f*x + 6*(2*(6*cos(5*f*
x + 5*e) - 15*cos(4*f*x + 4*e) + 20*cos(3*f*x + 3*e) - 15*cos(2*f*x + 2*e) + 6*cos(f*x + e) - 1)*cos(6*f*x + 6
*e) - cos(6*f*x + 6*e)^2 + 12*(15*cos(4*f*x + 4*e) - 20*cos(3*f*x + 3*e) + 15*cos(2*f*x + 2*e) - 6*cos(f*x + e
) + 1)*cos(5*f*x + 5*e) - 36*cos(5*f*x + 5*e)^2 + 30*(20*cos(3*f*x + 3*e) - 15*cos(2*f*x + 2*e) + 6*cos(f*x +
e) - 1)*cos(4*f*x + 4*e) - 225*cos(4*f*x + 4*e)^2 + 40*(15*cos(2*f*x + 2*e) - 6*cos(f*x + e) + 1)*cos(3*f*x +
3*e) - 400*cos(3*f*x + 3*e)^2 + 30*(6*cos(f*x + e) - 1)*cos(2*f*x + 2*e) - 225*cos(2*f*x + 2*e)^2 - 36*cos(f*x
 + e)^2 + 2*(6*sin(5*f*x + 5*e) - 15*sin(4*f*x + 4*e) + 20*sin(3*f*x + 3*e) - 15*sin(2*f*x + 2*e) + 6*sin(f*x
+ e))*sin(6*f*x + 6*e) - sin(6*f*x + 6*e)^2 + 12*(15*sin(4*f*x + 4*e) - 20*sin(3*f*x + 3*e) + 15*sin(2*f*x + 2
*e) - 6*sin(f*x + e))*sin(5*f*x + 5*e) - 36*sin(5*f*x + 5*e)^2 + 30*(20*sin(3*f*x + 3*e) - 15*sin(2*f*x + 2*e)
 + 6*sin(f*x + e))*sin(4*f*x + 4*e) - 225*sin(4*f*x + 4*e)^2 + 120*(5*sin(2*f*x + 2*e) - 2*sin(f*x + e))*sin(3
*f*x + 3*e) - 400*sin(3*f*x + 3*e)^2 - 225*sin(2*f*x + 2*e)^2 + 180*sin(2*f*x + 2*e)*sin(f*x + e) - 36*sin(f*x
 + e)^2 + 12*cos(f*x + e) - 1)*arctan2(sin(f*x + e), cos(f*x + e) - 1) + 2*(3*f*x - 18*(f*x + e)*cos(5*f*x + 5
*e) + 45*(f*x + e)*cos(4*f*x + 4*e) - 60*(f*x + e)*cos(3*f*x + 3*e) + 45*(f*x + e)*cos(2*f*x + 2*e) - 18*(f*x
+ e)*cos(f*x + e) + 3*e + 9*sin(5*f*x + 5*e) - 27*sin(4*f*x + 4*e) + 40*sin(3*f*x + 3*e) - 27*sin(2*f*x + 2*e)
 + 9*sin(f*x + e))*cos(6*f*x + 6*e) - 6*(6*f*x + 90*(f*x + e)*cos(4*f*x + 4*e) - 120*(f*x + e)*cos(3*f*x + 3*e
) + 90*(f*x + e)*cos(2*f*x + 2*e) - 36*(f*x + e)*cos(f*x + e) + 6*e - 9*sin(4*f*x + 4*e) + 20*sin(3*f*x + 3*e)
 - 9*sin(2*f*x + 2*e))*cos(5*f*x + 5*e) + 6*(15*f*x - 300*(f*x + e)*cos(3*f*x + 3*e) + 225*(f*x + e)*cos(2*f*x
 + 2*e) - 90*(f*x + e)*cos(f*x + e) + 15*e + 20*sin(3*f*x + 3*e) - 9*sin(f*x + e))*cos(4*f*x + 4*e) - 120*(f*x
 + 15*(f*x + e)*cos(2*f*x + 2*e) - 6*(f*x + e)*cos(f*x + e) + e + sin(2*f*x + 2*e) - sin(f*x + e))*cos(3*f*x +
 3*e) + 18*(5*f*x - 30*(f*x + e)*cos(f*x + e) + 5*e - 3*sin(f*x + e))*cos(2*f*x + 2*e) - 36*(f*x + e)*cos(f*x
+ e) - 2*(18*(f*x + e)*sin(5*f*x + 5*e) - 45*(f*x + e)*sin(4*f*x + 4*e) + 60*(f*x + e)*sin(3*f*x + 3*e) - 45*(
f*x + e)*sin(2*f*x + 2*e) + 18*(f*x + e)*sin(f*x + e) + 9*cos(5*f*x + 5*e) - 27*cos(4*f*x + 4*e) + 40*cos(3*f*
x + 3*e) - 27*cos(2*f*x + 2*e) + 9*cos(f*x + e))*sin(6*f*x + 6*e) - 6*(90*(f*x + e)*sin(4*f*x + 4*e) - 120*(f*
x + e)*sin(3*f*x + 3*e) + 90*(f*x + e)*sin(2*f*x + 2*e) - 36*(f*x + e)*sin(f*x + e) + 9*cos(4*f*x + 4*e) - 20*
cos(3*f*x + 3*e) + 9*cos(2*f*x + 2*e) - 3)*sin(5*f*x + 5*e) - 6*(300*(f*x + e)*sin(3*f*x + 3*e) - 225*(f*x + e
)*sin(2*f*x + 2*e) + 90*(f*x + e)*sin(f*x + e) + 20*cos(3*f*x + 3*e) - 9*cos(f*x + e) + 9)*sin(4*f*x + 4*e) -
40*(45*(f*x + e)*sin(2*f*x + 2*e) - 18*(f*x + e)*sin(f*x + e) - 3*cos(2*f*x + 2*e) + 3*cos(f*x + e) - 2)*sin(3
*f*x + 3*e) - 54*(10*(f*x + e)*sin(f*x + e) - cos(f*x + e) + 1)*sin(2*f*x + 2*e) + 3*e + 18*sin(f*x + e))*sqrt
(a)*sqrt(c)/((c^4*cos(6*f*x + 6*e)^2 + 36*c^4*cos(5*f*x + 5*e)^2 + 225*c^4*cos(4*f*x + 4*e)^2 + 400*c^4*cos(3*
f*x + 3*e)^2 + 225*c^4*cos(2*f*x + 2*e)^2 + 36*c^4*cos(f*x + e)^2 + c^4*sin(6*f*x + 6*e)^2 + 36*c^4*sin(5*f*x
+ 5*e)^2 + 225*c^4*sin(4*f*x + 4*e)^2 + 400*c^4*sin(3*f*x + 3*e)^2 + 225*c^4*sin(2*f*x + 2*e)^2 - 180*c^4*sin(
2*f*x + 2*e)*sin(f*x + e) + 36*c^4*sin(f*x + e)^2 - 12*c^4*cos(f*x + e) + c^4 - 2*(6*c^4*cos(5*f*x + 5*e) - 15
*c^4*cos(4*f*x + 4*e) + 20*c^4*cos(3*f*x + 3*e) - 15*c^4*cos(2*f*x + 2*e) + 6*c^4*cos(f*x + e) - c^4)*cos(6*f*
x + 6*e) - 12*(15*c^4*cos(4*f*x + 4*e) - 20*c^4*cos(3*f*x + 3*e) + 15*c^4*cos(2*f*x + 2*e) - 6*c^4*cos(f*x + e
) + c^4)*cos(5*f*x + 5*e) - 30*(20*c^4*cos(3*f*x + 3*e) - 15*c^4*cos(2*f*x + 2*e) + 6*c^4*cos(f*x + e) - c^4)*
cos(4*f*x + 4*e) - 40*(15*c^4*cos(2*f*x + 2*e) - 6*c^4*cos(f*x + e) + c^4)*cos(3*f*x + 3*e) - 30*(6*c^4*cos(f*
x + e) - c^4)*cos(2*f*x + 2*e) - 2*(6*c^4*sin(5*f*x + 5*e) - 15*c^4*sin(4*f*x + 4*e) + 20*c^4*sin(3*f*x + 3*e)
 - 15*c^4*sin(2*f*x + 2*e) + 6*c^4*sin(f*x + e))*sin(6*f*x + 6*e) - 12*(15*c^4*sin(4*f*x + 4*e) - 20*c^4*sin(3
*f*x + 3*e) + 15*c^4*sin(2*f*x + 2*e) - 6*c^4*sin(f*x + e))*sin(5*f*x + 5*e) - 30*(20*c^4*sin(3*f*x + 3*e) - 1
5*c^4*sin(2*f*x + 2*e) + 6*c^4*sin(f*x + e))*sin(4*f*x + 4*e) - 120*(5*c^4*sin(2*f*x + 2*e) - 2*c^4*sin(f*x +
e))*sin(3*f*x + 3*e))*f)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{7/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(1/2)/(c - c/cos(e + f*x))^(7/2),x)

[Out]

int((a + a/cos(e + f*x))^(1/2)/(c - c/cos(e + f*x))^(7/2), x)

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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((a+a*sec(f*x+e))**(1/2)/(c-c*sec(f*x+e))**(7/2),x)

[Out]

Timed out

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