3.2.6 \(\int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx\) [106]

Optimal. Leaf size=100 \[ -\frac {2 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {a^3 \log (1-\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 100, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3995, 3996, 31} \begin {gather*} \frac {a^3 \tan (e+f x) \log (1-\cos (e+f x))}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {2 a^3 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(-2*a^3*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(5/2)) + (a^3*Log[1 - Cos[e + f*x]]*Tan
[e + f*x])/(c^2*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 3995

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(5/2)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Si
mp[-8*a^3*Cot[e + f*x]*((c + d*Csc[e + f*x])^n/(f*(2*n + 1)*Sqrt[a + b*Csc[e + f*x]])), x] + Dist[a^2/c^2, Int
[Sqrt[a + b*Csc[e + f*x]]*(c + d*Csc[e + f*x])^(n + 2), 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 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Dist
[(-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 {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx &=-\frac {2 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {a^2 \int \frac {\sqrt {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx}{c^2}\\ &=-\frac {2 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {\left (a^3 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{-c+c x} \, dx,x,\cos (e+f x)\right )}{c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {2 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{5/2}}+\frac {a^3 \log (1-\cos (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 1.44, size = 155, normalized size = 1.55 \begin {gather*} \frac {a^2 \left (4-3 i f x+\cos (e+f x) \left (-8+4 i f x-8 \log \left (1-e^{i (e+f x)}\right )\right )+6 \log \left (1-e^{i (e+f x)}\right )+\cos (2 (e+f x)) \left (-i f x+2 \log \left (1-e^{i (e+f x)}\right )\right )\right ) \sqrt {a (1+\sec (e+f x))} \tan \left (\frac {1}{2} (e+f x)\right )}{2 c^2 f (-1+\cos (e+f x))^2 \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

(a^2*(4 - (3*I)*f*x + Cos[e + f*x]*(-8 + (4*I)*f*x - 8*Log[1 - E^(I*(e + f*x))]) + 6*Log[1 - E^(I*(e + f*x))]
+ Cos[2*(e + f*x)]*((-I)*f*x + 2*Log[1 - E^(I*(e + f*x))]))*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(2*c^
2*f*(-1 + Cos[e + f*x])^2*Sqrt[c - c*Sec[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(228\) vs. \(2(92)=184\).
time = 0.25, size = 229, normalized size = 2.29

method	result	size
default	\(\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-4 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-4 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+3 \left (\cos ^{2}\left (f x +e \right )\right )+8 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+2 \cos \left (f x +e \right )-4 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-1\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, a^{2}}{2 f \sin \left (f x +e \right ) \cos \left (f x +e \right )^{2} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}}}\)	\(229\)
risch	\(\frac {a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) x}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}-\frac {2 a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left (f x +e \right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}+\frac {8 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{3 i \left (f x +e \right )}-{\mathrm e}^{2 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}-\frac {2 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(430\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/2/f*(-1+cos(f*x+e))*(2*cos(f*x+e)^2*ln(2/(cos(f*x+e)+1))-4*cos(f*x+e)^2*ln(-(-1+cos(f*x+e))/sin(f*x+e))-4*co
s(f*x+e)*ln(2/(cos(f*x+e)+1))+3*cos(f*x+e)^2+8*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))+2*ln(2/(cos(f*x+e)+1
))+2*cos(f*x+e)-4*ln(-(-1+cos(f*x+e))/sin(f*x+e))-1)*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)/sin(f*x+e)/cos(f*x+e)
^2/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)*a^2

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Maxima [A]
time = 0.52, size = 147, normalized size = 1.47 \begin {gather*} \frac {\frac {4 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{c^{\frac {5}{2}}} - \frac {2 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{c^{\frac {5}{2}}} - \frac {{\left (\sqrt {-a} a^{2} \sqrt {c} - \frac {2 \, \sqrt {-a} a^{2} \sqrt {c} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{4}}{c^{3} \sin \left (f x + e\right )^{4}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

1/2*(4*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f*x + e) + 1))/c^(5/2) - 2*sqrt(-a)*a^2*log(sin(f*x + e)^2/(cos(f*x
+ e) + 1)^2 + 1)/c^(5/2) - (sqrt(-a)*a^2*sqrt(c) - 2*sqrt(-a)*a^2*sqrt(c)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)
*(cos(f*x + e) + 1)^4/(c^3*sin(f*x + e)^4))/f

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**(5/2)/(c-c*sec(f*x+e))**(5/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 deep

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Giac [A]
time = 2.04, size = 170, normalized size = 1.70 \begin {gather*} -\frac {\frac {2 \, \sqrt {-a c} a^{3} \log \left ({\left | a \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right )}{c^{3} {\left | a \right |}} - \frac {2 \, \sqrt {-a c} a^{3} \log \left ({\left | -a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a \right |}\right )}{c^{3} {\left | a \right |}} - \frac {3 \, {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a\right )}^{2} \sqrt {-a c} a^{3} + 4 \, {\left (a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - a\right )} \sqrt {-a c} a^{4} + 2 \, \sqrt {-a c} a^{5}}{a^{2} c^{3} {\left | a \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

-1/2*(2*sqrt(-a*c)*a^3*log(abs(a)*tan(1/2*f*x + 1/2*e)^2)/(c^3*abs(a)) - 2*sqrt(-a*c)*a^3*log(abs(-a*tan(1/2*f
*x + 1/2*e)^2 - a))/(c^3*abs(a)) - (3*(a*tan(1/2*f*x + 1/2*e)^2 - a)^2*sqrt(-a*c)*a^3 + 4*(a*tan(1/2*f*x + 1/2
*e)^2 - a)*sqrt(-a*c)*a^4 + 2*sqrt(-a*c)*a^5)/(a^2*c^3*abs(a)*tan(1/2*f*x + 1/2*e)^4))/f

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(5/2)/(c - c/cos(e + f*x))^(5/2),x)

[Out]

int((a + a/cos(e + f*x))^(5/2)/(c - c/cos(e + f*x))^(5/2), x)

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