3.2.49 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx\) [149]

Optimal. Leaf size=146 \[ \frac {3 \csc (e+f x)}{8 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}}-\frac {3 \tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{8 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}} \]

[Out]

3/8*csc(f*x+e)/a^2/c/f/(a+a*sec(f*x+e))^(1/2)/(c-c*sec(f*x+e))^(1/2)+1/4*tan(f*x+e)/f/(a+a*sec(f*x+e))^(5/2)/(
c-c*sec(f*x+e))^(3/2)-3/8*arctanh(cos(f*x+e))*tan(f*x+e)/a^2/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 = 0.23, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4045, 4044, 2691, 3855} \begin {gather*} \frac {3 \csc (e+f x)}{8 a^2 c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}-\frac {3 \tan (e+f x) \tanh ^{-1}(\cos (e+f x))}{8 a^2 c f \sqrt {a \sec (e+f x)+a} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{4 f (a \sec (e+f x)+a)^{5/2} (c-c \sec (e+f x))^{3/2}} \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])^(3/2)),x]

[Out]

(3*Csc[e + f*x])/(8*a^2*c*f*Sqrt[a + a*Sec[e + f*x]]*Sqrt[c - c*Sec[e + f*x]]) + Tan[e + f*x]/(4*f*(a + a*Sec[
e + f*x])^(5/2)*(c - c*Sec[e + f*x])^(3/2)) - (3*ArcTanh[Cos[e + f*x]]*Tan[e + f*x])/(8*a^2*c*f*Sqrt[a + a*Sec
[e + f*x]]*Sqrt[c - c*Sec[e + f*x]])

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3855

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

Rule 4044

Int[csc[(e_.) + (f_.)*(x_)]*(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[Csc[e + f*x]*Cot[e + f*x]^(2*m), 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]

Rule 4045

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

Rubi steps

\begin {align*} \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}} \, dx &=\frac {\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}}+\frac {3 \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^{3/2} (c-c \sec (e+f x))^{3/2}} \, dx}{4 a}\\ &=\frac {\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}}-\frac {(3 \tan (e+f x)) \int \cot ^2(e+f x) \csc (e+f x) \, dx}{4 a^2 c \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=\frac {3 \csc (e+f x)}{8 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}}+\frac {(3 \tan (e+f x)) \int \csc (e+f x) \, dx}{8 a^2 c \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}\\ &=\frac {3 \csc (e+f x)}{8 a^2 c f \sqrt {a+a \sec (e+f x)} \sqrt {c-c \sec (e+f x)}}+\frac {\tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2} (c-c \sec (e+f x))^{3/2}}-\frac {3 \tanh ^{-1}(\cos (e+f x)) \tan (e+f x)}{8 a^2 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 = 1.42, size = 130, normalized size = 0.89 \begin {gather*} -\frac {\left (1+2 \cos (e+f x)+5 \cos (2 (e+f x))-3 \tanh ^{-1}\left (e^{i (e+f x)}\right ) (2+\cos (e+f x)-2 \cos (2 (e+f x))-\cos (3 (e+f x)))\right ) \tan (e+f x)}{16 a^2 c f (-1+\cos (e+f x)) (1+\cos (e+f x))^2 \sqrt {a (1+\sec (e+f x))} \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])^(3/2)),x]

[Out]

-1/16*((1 + 2*Cos[e + f*x] + 5*Cos[2*(e + f*x)] - 3*ArcTanh[E^(I*(e + f*x))]*(2 + Cos[e + f*x] - 2*Cos[2*(e +
f*x)] - Cos[3*(e + f*x)]))*Tan[e + f*x])/(a^2*c*f*(-1 + Cos[e + f*x])*(1 + Cos[e + f*x])^2*Sqrt[a*(1 + Sec[e +
f*x])]*Sqrt[c - c*Sec[e + f*x]])

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Maple [A]
time = 3.55, size = 204, normalized size = 1.40

method	result	size
default	\(-\frac {\left (12 \left (\cos ^{3}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+12 \left (\cos ^{2}\left (f x +e \right )\right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )+5 \left (\cos ^{3}\left (f x +e \right )\right )-12 \cos \left (f x +e \right ) \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-15 \left (\cos ^{2}\left (f x +e \right )\right )-12 \ln \left (-\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}\right )-9 \cos \left (f x +e \right )+3\right ) \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (\cos ^{2}\left (f x +e \right )\right ) \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}}}{32 f \,c^{3} \sin \left (f x +e \right )^{5} a^{3}}\)	\(204\)
risch	\(\frac {i \left (5 \,{\mathrm e}^{5 i \left (f x +e \right )}+2 \,{\mathrm e}^{4 i \left (f x +e \right )}+2 \,{\mathrm e}^{3 i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )}+5 \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{2} c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}-\frac {3 i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{8 a^{2} c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}+\frac {3 i \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{8 a^{2} c \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) \sqrt {\frac {a \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \sqrt {\frac {c \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{2}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(407\)

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

[Out]

-1/32/f*(12*cos(f*x+e)^3*ln(-(-1+cos(f*x+e))/sin(f*x+e))+12*cos(f*x+e)^2*ln(-(-1+cos(f*x+e))/sin(f*x+e))+5*cos
(f*x+e)^3-12*cos(f*x+e)*ln(-(-1+cos(f*x+e))/sin(f*x+e))-15*cos(f*x+e)^2-12*ln(-(-1+cos(f*x+e))/sin(f*x+e))-9*c
os(f*x+e)+3)*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*cos(f*x+e)^2*(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/c^3/sin(f*x
+e)^5/a^3

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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(sec(f*x+e)/(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 [A]
time = 2.50, size = 580, normalized size = 3.97 \begin {gather*} \left [-\frac {3 \, {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) - 1\right )} \sqrt {-a c} \log \left (-\frac {4 \, {\left (2 \, \sqrt {-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 )^{2} + {\left (a c \cos \left (f x + e\right )^{2} + a c\right )} \sin \left (f x + e\right )\right )}}{{\left (\cos \left (f x + e\right )^{2} - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (5 \, \cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - 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 )}}}{16 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} c^{2} f\right )} \sin \left (f x + e\right )}, \frac {3 \, {\left (\cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) - 1\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {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 )}}}{a c \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + {\left (5 \, \cos \left (f x + e\right )^{3} + \cos \left (f x + e\right )^{2} - 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 )}}}{8 \, {\left (a^{3} c^{2} f \cos \left (f x + e\right )^{3} + a^{3} c^{2} f \cos \left (f x + e\right )^{2} - a^{3} c^{2} f \cos \left (f x + e\right ) - a^{3} 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(sec(f*x+e)/(a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(3*(cos(f*x + e)^3 + cos(f*x + e)^2 - cos(f*x + e) - 1)*sqrt(-a*c)*log(-4*(2*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)^2 + (a*c*cos(f*x + e)^2 + a*c)*s
in(f*x + e))/((cos(f*x + e)^2 - 1)*sin(f*x + e)))*sin(f*x + e) - 2*(5*cos(f*x + e)^3 + cos(f*x + e)^2 - 2*cos(
f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^3*c^2*f*cos(f*x
+ e)^3 + a^3*c^2*f*cos(f*x + e)^2 - a^3*c^2*f*cos(f*x + e) - a^3*c^2*f)*sin(f*x + e)), 1/8*(3*(cos(f*x + e)^3
+ cos(f*x + e)^2 - cos(f*x + e) - 1)*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))/(a*c*sin(f*x + e)))*sin(f*x + e) + (5*cos(f*x + e)^3 + cos(f*x + e)^2 - 2*c
os(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^3*c^2*f*cos(
f*x + e)^3 + a^3*c^2*f*cos(f*x + e)^2 - a^3*c^2*f*cos(f*x + e) - a^3*c^2*f)*sin(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(sec(f*x+e)/(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 3006 deep

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Giac [A]
time = 1.96, size = 151, normalized size = 1.03 \begin {gather*} \frac {\frac {2 \, {\left (3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}}{c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}} - \frac {{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{2} c^{2} - 4 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )} c^{3}}{c^{4}} - 6 \, \log \left ({\left | c \right |} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}\right ) + 6 \, \log \left ({\left | c \right |}\right ) - 4}{32 \, \sqrt {-a c} a^{2} f {\left | c \right |} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )} \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))^(3/2),x, algorithm="giac")

[Out]

1/32*(2*(3*c*tan(1/2*f*x + 1/2*e)^2 - c)/(c*tan(1/2*f*x + 1/2*e)^2) - ((c*tan(1/2*f*x + 1/2*e)^2 - c)^2*c^2 -
4*(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^3)/c^4 - 6*log(abs(c)*tan(1/2*f*x + 1/2*e)^2) + 6*log(abs(c)) - 4)/(sqrt(-a
*c)*a^2*f*abs(c)*sgn(tan(1/2*f*x + 1/2*e)^3 + tan(1/2*f*x + 1/2*e)))

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{\cos \left (e+f\,x\right )\,{\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(1/(cos(e + f*x)*(a + a/cos(e + f*x))^(5/2)*(c - c/cos(e + f*x))^(3/2)),x)

[Out]

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

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