3.2.29 \(\int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx\) [129]

Optimal. Leaf size=145 \[ -\frac {a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a^2 \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{c f (c-c \sec (e+f x))^{3/2}}+\frac {a^3 \log (1-\sec (e+f x)) \tan (e+f x)}{c^2 f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.29, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.056, Rules used = {4039, 4037} \begin {gather*} \frac {a^3 \tan (e+f x) \log (1-\sec (e+f x))}{c^2 f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {a^2 \tan (e+f x) \sqrt {a \sec (e+f x)+a}}{c f (c-c \sec (e+f x))^{3/2}}-\frac {a \tan (e+f x) (a \sec (e+f x)+a)^{3/2}}{2 f (c-c \sec (e+f x))^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*(a*(a + a*Sec[e + f*x])^(3/2)*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^(5/2)) + (a^2*Sqrt[a + a*Sec[e + f*x]
]*Tan[e + f*x])/(c*f*(c - c*Sec[e + f*x])^(3/2)) + (a^3*Log[1 - Sec[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 4037

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

Rule 4039

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

Rubi steps

\begin {align*} \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^{5/2}} \, dx &=-\frac {a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}-\frac {a \int \frac {\sec (e+f x) (a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^{3/2}} \, dx}{c}\\ &=-\frac {a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a^2 \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{c f (c-c \sec (e+f x))^{3/2}}+\frac {a^2 \int \frac {\sec (e+f x) \sqrt {a+a \sec (e+f x)}}{\sqrt {c-c \sec (e+f x)}} \, dx}{c^2}\\ &=-\frac {a (a+a \sec (e+f x))^{3/2} \tan (e+f x)}{2 f (c-c \sec (e+f x))^{5/2}}+\frac {a^2 \sqrt {a+a \sec (e+f x)} \tan (e+f x)}{c f (c-c \sec (e+f x))^{3/2}}+\frac {a^3 \log (1-\sec (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 = 1.32, size = 182, normalized size = 1.26 \begin {gather*} -\frac {a^2 \left (4-6 \log \left (1-e^{i (e+f x)}\right )+\cos (e+f x) \left (8 \log \left (1-e^{i (e+f x)}\right )-4 \log \left (1+e^{2 i (e+f x)}\right )\right )+3 \log \left (1+e^{2 i (e+f x)}\right )+\cos (2 (e+f x)) \left (-2 \log \left (1-e^{i (e+f x)}\right )+\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 )}{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[(Sec[e + f*x]*(a + a*Sec[e + f*x])^(5/2))/(c - c*Sec[e + f*x])^(5/2),x]

[Out]

-1/2*(a^2*(4 - 6*Log[1 - E^(I*(e + f*x))] + Cos[e + f*x]*(8*Log[1 - E^(I*(e + f*x))] - 4*Log[1 + E^((2*I)*(e +
f*x))]) + 3*Log[1 + E^((2*I)*(e + f*x))] + Cos[2*(e + f*x)]*(-2*Log[1 - E^(I*(e + f*x))] + Log[1 + E^((2*I)*(
e + f*x))]))*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/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. \(365\) vs. \(2(131)=262\).
time = 2.99, size = 366, normalized size = 2.52

method	result	size
risch	\(\frac {8 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{3} \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}-\frac {2 i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}+\frac {i a^{2} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{c^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(325\)
default	\(\frac {\left (-1+\cos \left (f x +e \right )\right ) \left (2 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-4 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right )-4 \cos \left (f x +e \right ) \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+8 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-4 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \cos \left (f x +e \right )+2 \ln \left (-\frac {\cos \left (f x +e \right )-1+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-4 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+2 \ln \left (-\frac {-1+\cos \left (f x +e \right )-\sin \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+3\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, a^{2}}{2 f \cos \left (f x +e \right )^{2} \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {5}{2}} \sin \left (f x +e \right )}\)	\(366\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Maxima [A]
time = 0.49, size = 179, normalized size = 1.23 \begin {gather*} -\frac {\frac {2 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{c^{\frac {5}{2}}} + \frac {2 \, \sqrt {-a} a^{2} \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{c^{\frac {5}{2}}} - \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 {{\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(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

-1/2*(2*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/c^(5/2) + 2*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f
*x + e) + 1) - 1)/c^(5/2) - 4*sqrt(-a)*a^2*log(sin(f*x + e)/(cos(f*x + e) + 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(sec(f*x+e)*(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)^3 + 2*a^2*sec(f*x + e)^2 + a^2*sec(f*x + e))*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(sec(f*x+e)*(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 4370 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(sec(f*x+e)*(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(5/2),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}}{\cos \left (e+f\,x\right )\,{\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)/(cos(e + f*x)*(c - c/cos(e + f*x))^(5/2)),x)

[Out]

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

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