3.2.25 \(\int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx\) [125]

Optimal. Leaf size=98 \[ -\frac {2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^3 \log (1+\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.12, antiderivative size = 98, 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 {c^3 \tan (e+f x) \log (\cos (e+f x)+1)}{a^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {2 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{5/2} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^3*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])^(5/2)*Sqrt[c - c*Sec[e + f*x]]) + (c^3*Log[1 + Cos[e + f*x]]*Tan
[e + f*x])/(a^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 {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac {2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^2 \int \frac {\sqrt {c-c \sec (e+f x)}}{\sqrt {a+a \sec (e+f x)}} \, dx}{a^2}\\ &=-\frac {2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}+\frac {\left (c^3 \tan (e+f x)\right ) \text {Subst}\left (\int \frac {1}{a+a x} \, dx,x,\cos (e+f x)\right )}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {2 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{5/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^3 \log (1+\cos (e+f x)) \tan (e+f x)}{a^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 1.64, size = 154, normalized size = 1.57 \begin {gather*} \frac {i c^2 \cot \left (\frac {1}{2} (e+f x)\right ) \left (4 i+3 f x+\cos (2 (e+f x)) \left (f x+2 i \log \left (1+e^{i (e+f x)}\right )\right )+4 \cos (e+f x) \left (2 i+f x+2 i \log \left (1+e^{i (e+f x)}\right )\right )+6 i \log \left (1+e^{i (e+f x)}\right )\right ) \sqrt {c-c \sec (e+f x)}}{2 a^2 f (1+\cos (e+f x))^2 \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.25, size = 144, normalized size = 1.47

method	result	size
default	\(\frac {\left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+3 \left (\cos ^{2}\left (f x +e \right )\right )+4 \cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 \cos \left (f x +e \right )+2 \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-1\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}}{2 f \sin \left (f x +e \right )^{5} a^{3}}\)	\(144\)
risch	\(\frac {c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, x}{a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}-\frac {2 c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (f x +e \right )}{a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {8 i c^{2} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{3 i \left (f x +e \right )}+{\mathrm e}^{2 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {2 i c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(428\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Maxima [A]
time = 0.51, size = 108, normalized size = 1.10 \begin {gather*} \frac {\frac {2 \, c^{\frac {5}{2}} \log \left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}{\sqrt {-a} a^{2}} + \frac {\frac {2 \, \sqrt {-a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {\sqrt {-a} c^{\frac {5}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}}}{a^{3}}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Giac [A]
time = 1.76, size = 75, normalized size = 0.77 \begin {gather*} -\frac {\frac {2 \, \sqrt {-a c} c^{3} \log \left ({\left | c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + c \right |}\right )}{a^{3} {\left | c \right |}} + \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} \sqrt {-a c} {\left | c \right |}}{a^{3} c}}{2 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________