3.2.18 \(\int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx\) [118]

Optimal. Leaf size=96 \[ -\frac {4 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^3 \log (\cos (e+f x)) \tan (e+f x)}{a f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 96, 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, 3990, 3556} \begin {gather*} \frac {c^3 \tan (e+f x) \log (\cos (e+f x))}{a f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {4 c^3 \tan (e+f x)}{f (a \sec (e+f x)+a)^{3/2} \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])^(3/2),x]

[Out]

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

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3990

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

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)]

Rubi steps

\begin {align*} \int \frac {(c-c \sec (e+f x))^{5/2}}{(a+a \sec (e+f x))^{3/2}} \, dx &=-\frac {4 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^2 \int \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)} \, dx}{a^2}\\ &=-\frac {4 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}-\frac {\left (c^3 \tan (e+f x)\right ) \int \tan (e+f x) \, dx}{a \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {4 c^3 \tan (e+f x)}{f (a+a \sec (e+f x))^{3/2} \sqrt {c-c \sec (e+f x)}}+\frac {c^3 \log (\cos (e+f x)) \tan (e+f x)}{a 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 = 0.78, size = 116, normalized size = 1.21 \begin {gather*} \frac {i c^2 \cot \left (\frac {1}{2} (e+f x)\right ) \left (4 i+f x+\cos (e+f x) \left (f x+i \log \left (1+e^{2 i (e+f x)}\right )\right )+i \log \left (1+e^{2 i (e+f x)}\right )\right ) \sqrt {c-c \sec (e+f x)}}{a f (1+\cos (e+f x)) \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])^(3/2),x]

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(235\) vs. \(2(88)=176\).
time = 0.25, size = 236, normalized size = 2.46

method	result	size
default	\(\frac {\left (\cos \left (f x +e \right ) \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\cos \left (f x +e \right ) \ln \left (\frac {-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2 \cos \left (f x +e \right )+\ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\ln \left (\frac {-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )+2\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (f x +e \right )\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}}{f \sin \left (f x +e \right )^{3} \left (-1+\cos \left (f x +e \right )\right ) a^{2}}\)	\(236\)
risch	\(\frac {c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, x}{a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}-\frac {2 c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (f x +e \right )}{a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {8 i c^{2} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{i \left (f x +e \right )}}{a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}-\frac {i c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{a \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) f}\)	\(409\)

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))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

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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))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 233 vs. \(2 (95) = 190\).
time = 4.20, size = 489, normalized size = 5.09 \begin {gather*} \left [-\frac {4 \, c^{2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2} \cos \left (f x + e\right ) + a c^{2}\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c \cos \left (f x + e\right )^{4} - {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt {-\frac {c}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) + c}{2 \, \cos \left (f x + e\right )^{2}}\right )}{2 \, {\left (a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f\right )}}, -\frac {2 \, c^{2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (a c^{2} \cos \left (f x + e\right )^{2} + 2 \, a c^{2} \cos \left (f x + e\right ) + a c^{2}\right )} \sqrt {\frac {c}{a}} \arctan \left (\frac {\sqrt {\frac {c}{a}} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{c \cos \left (f x + e\right )^{2} + c}\right )}{a^{2} f \cos \left (f x + e\right )^{2} + 2 \, a^{2} f \cos \left (f x + e\right ) + a^{2} f}\right ] \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))^(3/2),x, algorithm="fricas")

[Out]

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

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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))**(3/2),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))^(3/2),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}}{{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{3/2}} \,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))^(3/2),x)

[Out]

int((c - c/cos(e + f*x))^(5/2)/(a + a/cos(e + f*x))^(3/2), x)

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