3.2.5 \(\int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{3/2}} \, dx\) [105]

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

[Out]

-4*a^3*tan(f*x+e)/f/(c-c*sec(f*x+e))^(3/2)/(a+a*sec(f*x+e))^(1/2)+a^3*ln(cos(f*x+e))*tan(f*x+e)/c/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 {a^3 \tan (e+f x) \log (\cos (e+f x))}{c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {4 a^3 \tan (e+f x)}{f \sqrt {a \sec (e+f x)+a} (c-c \sec (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^(3/2),x]

[Out]

(-4*a^3*Tan[e + f*x])/(f*Sqrt[a + a*Sec[e + f*x]]*(c - c*Sec[e + f*x])^(3/2)) + (a^3*Log[Cos[e + f*x]]*Tan[e +
f*x])/(c*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 {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{3/2}} \, dx &=-\frac {4 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a^2 \int \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)} \, dx}{c^2}\\ &=-\frac {4 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}-\frac {\left (a^3 \tan (e+f x)\right ) \int \tan (e+f x) \, dx}{c \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=-\frac {4 a^3 \tan (e+f x)}{f \sqrt {a+a \sec (e+f x)} (c-c \sec (e+f x))^{3/2}}+\frac {a^3 \log (\cos (e+f x)) \tan (e+f x)}{c 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 = 1.28, size = 111, normalized size = 1.16 \begin {gather*} \frac {a^2 \left (-4+i f x-\log \left (1+e^{2 i (e+f x)}\right )+\cos (e+f x) \left (-i f x+\log \left (1+e^{2 i (e+f x)}\right )\right )\right ) \sqrt {a (1+\sec (e+f x))} \tan \left (\frac {1}{2} (e+f x)\right )}{c f (-1+\cos (e+f x)) \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])^(3/2),x]

[Out]

(a^2*(-4 + I*f*x - Log[1 + E^((2*I)*(e + f*x))] + Cos[e + f*x]*((-I)*f*x + Log[1 + E^((2*I)*(e + f*x))]))*Sqrt
[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(c*f*(-1 + Cos[e + f*x])*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. \(236\) vs. \(2(88)=176\).
time = 0.25, size = 237, normalized size = 2.47

method	result	size
default	\(-\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (\cos \left (f x +e \right ) \ln \left (\frac {-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+\cos \left (f x +e \right ) \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\cos \left (f x +e \right ) \ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-\ln \left (\frac {-\cos \left (f x +e \right )+1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-\ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-2 \cos \left (f x +e \right )+\ln \left (\frac {2}{\cos \left (f x +e \right )+1}\right )-2\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, a^{2}}{f \cos \left (f x +e \right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )}\)	\(237\)
risch	\(\frac {a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) x}{c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}}-\frac {2 a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \left (f x +e \right )}{c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}+\frac {8 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{i \left (f x +e \right )}}{c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}-\frac {i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{c \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(409\)

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

[Out]

-1/f*(-1+cos(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(-(cos(f*x+e)-1+sin(f*
x+e))/sin(f*x+e))-cos(f*x+e)*ln(2/(cos(f*x+e)+1))-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))-2*cos(f*x+e)+ln(2/(cos(f*x+e)+1))-2)*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)/cos(f*x+e)/(c
*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(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((a+a*sec(f*x+e))^(5/2)/(c-c*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. 227 vs. \(2 (95) = 190\).
time = 3.46, size = 478, normalized size = 4.98 \begin {gather*} \left [\frac {{\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c\right )} \sqrt {-\frac {a}{c}} \log \left (\frac {a \cos \left (f x + e\right )^{4} - {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )\right )} \sqrt {-\frac {a}{c}} \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 ) + a}{2 \, \cos \left (f x + e\right )^{2}}\right ) \sin \left (f x + e\right ) + 4 \, {\left (a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right )\right )} \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 )}}}{2 \, {\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}, \frac {{\left (a^{2} c \cos \left (f x + e\right ) - a^{2} c\right )} \sqrt {\frac {a}{c}} \arctan \left (\frac {\sqrt {\frac {a}{c}} \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 )}{a \cos \left (f x + e\right )^{2} + a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (a^{2} \cos \left (f x + e\right )^{2} + a^{2} \cos \left (f x + e\right )\right )} \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 )}}}{{\left (c^{2} f \cos \left (f x + e\right ) - c^{2} f\right )} \sin \left (f x + e\right )}\right ] \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))^(3/2),x, algorithm="fricas")

[Out]

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

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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))**(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((a+a*sec(f*x+e))^(5/2)/(c-c*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 (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (c-\frac {c}{\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((a + a/cos(e + f*x))^(5/2)/(c - c/cos(e + f*x))^(3/2),x)

[Out]

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

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