3.2.11 \(\int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx\) [111]

Optimal. Leaf size=151 \[ \frac {c^3 \log (\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {4 c^3 \log (1+\sec (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {c^3 \sec (e+f x) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.08, antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {3997, 84} \begin {gather*} -\frac {c^3 \tan (e+f x) \sec (e+f x)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {4 c^3 \tan (e+f x) \log (\sec (e+f x)+1)}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {c^3 \tan (e+f x) \log (\cos (e+f x))}{f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(c^3*Log[Cos[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + (4*c^3*Log[1 + Se
c[e + f*x]]*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) - (c^3*Sec[e + f*x]*Tan[e + f*
x])/(f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 84

Int[((e_.) + (f_.)*(x_))^(p_.)/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Int[ExpandIntegrand[(
e + f*x)^p/((a + b*x)*(c + d*x)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IntegerQ[p]

Rule 3997

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

Rubi steps

\begin {align*} \int \frac {(c-c \sec (e+f x))^{5/2}}{\sqrt {a+a \sec (e+f x)}} \, dx &=-\frac {(a c \tan (e+f x)) \text {Subst}\left (\int \frac {(c-c x)^2}{x (a+a x)} \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {(a c \tan (e+f x)) \text {Subst}\left (\int \left (\frac {c^2}{a}+\frac {c^2}{a x}-\frac {4 c^2}{a (1+x)}\right ) \, dx,x,\sec (e+f x)\right )}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=\frac {c^3 \log (\cos (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {4 c^3 \log (1+\sec (e+f x)) \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}-\frac {c^3 \sec (e+f x) \tan (e+f x)}{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 = 4.01, size = 181, normalized size = 1.20 \begin {gather*} \frac {c^2 e^{-3 i (e+f x)} \left (1+e^{2 i (e+f x)}\right )^3 \cos \left (\frac {1}{2} (e+f x)\right ) \cot \left (\frac {1}{2} (e+f x)\right ) \left (1+\cos (e+f x) \left (i f x-8 \log \left (1+e^{i (e+f x)}\right )+3 \log \left (1+e^{2 i (e+f x)}\right )\right )\right ) \sec ^4(e+f x) \sqrt {c-c \sec (e+f x)} \left (\cos \left (\frac {1}{2} (e+f x)\right )+i \sin \left (\frac {1}{2} (e+f x)\right )\right )}{4 \left (1+e^{i (e+f x)}\right ) f \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.26, size = 169, normalized size = 1.12

method	result	size
default	\(\frac {\left (3 \cos \left (f x +e \right ) \ln \left (\frac {-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+3 \cos \left (f x +e \right ) \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+\cos \left (f x +e \right )+1\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}}{f \sin \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )^{2} a}\)	\(169\)
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}{\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 )}{\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} \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}^{2 i \left (f x +e \right )}+{\mathrm e}^{i \left (f x +e \right )}\right )}{\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 \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}-\frac {8 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 )}{\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 {3 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}^{2 i \left (f x +e \right )}+1\right )}{\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}\)	\(512\)

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))^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \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))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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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((c-c*sec(f*x+e))^(5/2)/(a+a*sec(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral((c^2*sec(f*x + e)^2 - 2*c^2*sec(f*x + e) + c^2)*sqrt(-c*sec(f*x + e) + c)/sqrt(a*sec(f*x + e) + a), 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((c-c*sec(f*x+e))**(5/2)/(a+a*sec(f*x+e))**(1/2),x)

[Out]

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

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

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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}}{\sqrt {a+\frac {a}{\cos \left (e+f\,x\right )}}} \,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))^(1/2),x)

[Out]

int((c - c/cos(e + f*x))^(5/2)/(a + a/cos(e + f*x))^(1/2), x)

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