3.1.99 \(\int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}} \, dx\) [99]

Optimal. Leaf size=203 \[ -\frac {35 \text {ArcTan}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{64 \sqrt {2} a^2 c^{5/2} f}-\frac {35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac {7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac {35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}} \]

[Out]

-35/128*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))/a^2/c^(5/2)/f*2^(1/2)-35/48*tan(f*x+e)/a
^2/f/(c-c*sec(f*x+e))^(5/2)+1/3*tan(f*x+e)/f/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(5/2)+7/6*tan(f*x+e)/f/(a^2+a
^2*sec(f*x+e))/(c-c*sec(f*x+e))^(5/2)-35/64*tan(f*x+e)/a^2/c/f/(c-c*sec(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.27, antiderivative size = 203, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {4045, 3881, 3880, 209} \begin {gather*} -\frac {35 \text {ArcTan}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{64 \sqrt {2} a^2 c^{5/2} f}-\frac {35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}}-\frac {35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac {7 \tan (e+f x)}{6 f \left (a^2 \sec (e+f x)+a^2\right ) (c-c \sec (e+f x))^{5/2}}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2 (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])^2*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

(-35*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(64*Sqrt[2]*a^2*c^(5/2)*f) - (35*Tan[e
+ f*x])/(48*a^2*f*(c - c*Sec[e + f*x])^(5/2)) + Tan[e + f*x]/(3*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])
^(5/2)) + (7*Tan[e + f*x])/(6*f*(a^2 + a^2*Sec[e + f*x])*(c - c*Sec[e + f*x])^(5/2)) - (35*Tan[e + f*x])/(64*a
^2*c*f*(c - c*Sec[e + f*x])^(3/2))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3880

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

Rule 3881

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

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))^2 (c-c \sec (e+f x))^{5/2}} \, dx &=\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac {7 \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^{5/2}} \, dx}{6 a}\\ &=\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac {7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}+\frac {35 \int \frac {\sec (e+f x)}{(c-c \sec (e+f x))^{5/2}} \, dx}{12 a^2}\\ &=-\frac {35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac {7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}+\frac {35 \int \frac {\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{32 a^2 c}\\ &=-\frac {35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac {7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac {35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}}+\frac {35 \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}} \, dx}{128 a^2 c^2}\\ &=-\frac {35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac {7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac {35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}}-\frac {35 \text {Subst}\left (\int \frac {1}{2 c+x^2} \, dx,x,\frac {c \tan (e+f x)}{\sqrt {c-c \sec (e+f x)}}\right )}{64 a^2 c^2 f}\\ &=-\frac {35 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{64 \sqrt {2} a^2 c^{5/2} f}-\frac {35 \tan (e+f x)}{48 a^2 f (c-c \sec (e+f x))^{5/2}}+\frac {\tan (e+f x)}{3 f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{5/2}}+\frac {7 \tan (e+f x)}{6 f \left (a^2+a^2 \sec (e+f x)\right ) (c-c \sec (e+f x))^{5/2}}-\frac {35 \tan (e+f x)}{64 a^2 c f (c-c \sec (e+f x))^{3/2}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 2.41, size = 434, normalized size = 2.14 \begin {gather*} \frac {\cot ^4(e+f x) \left (\frac {105 i \sqrt {2} \left (-1+e^{i (e+f x)}\right )^5 \left (1+e^{i (e+f x)}\right )^4 \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right )}{\left (1+e^{2 i (e+f x)}\right )^{9/2}}+192 \cos ^4\left (\frac {1}{2} (e+f x)\right ) \csc \left (\frac {e}{2}\right ) \sec ^5(e+f x) \sin \left (\frac {f x}{2}\right ) \sin \left (\frac {1}{2} (e+f x)\right )-192 \cos ^4\left (\frac {1}{2} (e+f x)\right ) \cot \left (\frac {e}{2}\right ) \sec ^5(e+f x) \sin ^2\left (\frac {1}{2} (e+f x)\right )+256 \cos \left (\frac {1}{2} (e+f x)\right ) \sec ^5(e+f x) \sin ^5\left (\frac {1}{2} (e+f x)\right )+2752 \cos \left (\frac {e}{2}\right ) \cos \left (\frac {f x}{2}\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right ) \sec ^5(e+f x) \sin ^5\left (\frac {1}{2} (e+f x)\right )-13312 \csc ^5(2 (e+f x)) \sin ^2\left (\frac {1}{2} (e+f x)\right ) \sin ^8(e+f x)-3648 \csc \left (\frac {e}{2}\right ) \csc \left (\frac {1}{2} (e+f x)\right ) \csc ^5(2 (e+f x)) \sin \left (\frac {f x}{2}\right ) \sin ^9(e+f x)-5504 \csc ^5(2 (e+f x)) \sin \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \sin ^9(e+f x)+114 \cot \left (\frac {e}{2}\right ) \sec (e+f x) \tan ^4(e+f x)\right )}{384 a^2 c^2 f \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])^2*(c - c*Sec[e + f*x])^(5/2)),x]

[Out]

(Cot[e + f*x]^4*(((105*I)*Sqrt[2]*(-1 + E^(I*(e + f*x)))^5*(1 + E^(I*(e + f*x)))^4*ArcTanh[(1 + E^(I*(e + f*x)
))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))])])/(1 + E^((2*I)*(e + f*x)))^(9/2) + 192*Cos[(e + f*x)/2]^4*Csc[e/2]
*Sec[e + f*x]^5*Sin[(f*x)/2]*Sin[(e + f*x)/2] - 192*Cos[(e + f*x)/2]^4*Cot[e/2]*Sec[e + f*x]^5*Sin[(e + f*x)/2
]^2 + 256*Cos[(e + f*x)/2]*Sec[e + f*x]^5*Sin[(e + f*x)/2]^5 + 2752*Cos[e/2]*Cos[(f*x)/2]*Cos[(e + f*x)/2]^4*S
ec[e + f*x]^5*Sin[(e + f*x)/2]^5 - 13312*Csc[2*(e + f*x)]^5*Sin[(e + f*x)/2]^2*Sin[e + f*x]^8 - 3648*Csc[e/2]*
Csc[(e + f*x)/2]*Csc[2*(e + f*x)]^5*Sin[(f*x)/2]*Sin[e + f*x]^9 - 5504*Csc[2*(e + f*x)]^5*Sin[e/2]*Sin[(f*x)/2
]*Sin[(e + f*x)/2]*Sin[e + f*x]^9 + 114*Cot[e/2]*Sec[e + f*x]*Tan[e + f*x]^4))/(384*a^2*c^2*f*Sqrt[c - c*Sec[e
+ f*x]])

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(550\) vs. \(2(176)=352\).
time = 2.42, size = 551, normalized size = 2.71

method	result	size
default	\(\frac {\left (-1+\cos \left (f x +e \right )\right )^{3} \left (21 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {9}{2}} \left (\cos ^{2}\left (f x +e \right )\right )+12 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {9}{2}} \cos \left (f x +e \right )-9 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {9}{2}}+15 \left (\cos ^{2}\left (f x +e \right )\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {7}{2}}-30 \cos \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {7}{2}}+15 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {7}{2}}-21 \left (\cos ^{2}\left (f x +e \right )\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}}+42 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \cos \left (f x +e \right )-21 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}}+35 \left (\cos ^{2}\left (f x +e \right )\right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}}-70 \cos \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}}+35 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}}-105 \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-105 \left (\cos ^{2}\left (f x +e \right )\right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )+210 \cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+210 \cos \left (f x +e \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )-105 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-105 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )\right )}{48 a^{2} f \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 )^{5} \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}}}\)	\(551\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(a+a*sec(f*x+e))^2/(c-c*sec(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/48/a^2/f*(-1+cos(f*x+e))^3*(21*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(9/2)*cos(f*x+e)^2+12*(-2*cos(f*x+e)/(cos(f*x+
e)+1))^(9/2)*cos(f*x+e)-9*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(9/2)+15*cos(f*x+e)^2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^
(7/2)-30*cos(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(7/2)+15*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(7/2)-21*cos(f*x+e)
^2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)+42*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(5/2)*cos(f*x+e)-21*(-2*cos(f*x+e)/(
cos(f*x+e)+1))^(5/2)+35*cos(f*x+e)^2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)-70*cos(f*x+e)*(-2*cos(f*x+e)/(cos(f*
x+e)+1))^(3/2)+35*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(3/2)-105*cos(f*x+e)^2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-1
05*cos(f*x+e)^2*arctan(1/(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2))+210*cos(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(
1/2)+210*cos(f*x+e)*arctan(1/(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2))-105*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-10
5*arctan(1/(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)))/(c*(-1+cos(f*x+e))/cos(f*x+e))^(5/2)/sin(f*x+e)^5/(-2*cos(f*
x+e)/(cos(f*x+e)+1))^(5/2)

________________________________________________________________________________________

Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to 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))^2/(c-c*sec(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]
time = 3.96, size = 527, normalized size = 2.60 \begin {gather*} \left [-\frac {105 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) + 1\right )} \sqrt {-c} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{2} + \cos \left (f x + e\right )\right )} \sqrt {-c} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} + {\left (3 \, c \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right )}{{\left (\cos \left (f x + e\right ) - 1\right )} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 4 \, {\left (43 \, \cos \left (f x + e\right )^{4} - 161 \, \cos \left (f x + e\right )^{3} - 35 \, \cos \left (f x + e\right )^{2} + 105 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{768 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} - a^{2} c^{3} f \cos \left (f x + e\right )^{2} - a^{2} c^{3} f \cos \left (f x + e\right ) + a^{2} c^{3} f\right )} \sin \left (f x + e\right )}, \frac {105 \, \sqrt {2} {\left (\cos \left (f x + e\right )^{3} - \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right ) + 1\right )} \sqrt {c} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {c} \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, {\left (43 \, \cos \left (f x + e\right )^{4} - 161 \, \cos \left (f x + e\right )^{3} - 35 \, \cos \left (f x + e\right )^{2} + 105 \, \cos \left (f x + e\right )\right )} \sqrt {\frac {c \cos \left (f x + e\right ) - c}{\cos \left (f x + e\right )}}}{384 \, {\left (a^{2} c^{3} f \cos \left (f x + e\right )^{3} - a^{2} c^{3} f \cos \left (f x + e\right )^{2} - a^{2} c^{3} f \cos \left (f x + e\right ) + a^{2} c^{3} 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))^2/(c-c*sec(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

[-1/768*(105*sqrt(2)*(cos(f*x + e)^3 - cos(f*x + e)^2 - cos(f*x + e) + 1)*sqrt(-c)*log((2*sqrt(2)*(cos(f*x + e
)^2 + cos(f*x + e))*sqrt(-c)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)) + (3*c*cos(f*x + e) + c)*sin(f*x + e))/((
cos(f*x + e) - 1)*sin(f*x + e)))*sin(f*x + e) + 4*(43*cos(f*x + e)^4 - 161*cos(f*x + e)^3 - 35*cos(f*x + e)^2
+ 105*cos(f*x + e))*sqrt((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c^3*f*cos(f*x + e)^3 - a^2*c^3*f*cos(f*x +
e)^2 - a^2*c^3*f*cos(f*x + e) + a^2*c^3*f)*sin(f*x + e)), 1/384*(105*sqrt(2)*(cos(f*x + e)^3 - cos(f*x + e)^2
- cos(f*x + e) + 1)*sqrt(c)*arctan(sqrt(2)*sqrt((c*cos(f*x + e) - c)/cos(f*x + e))*cos(f*x + e)/(sqrt(c)*sin(f
*x + e)))*sin(f*x + e) - 2*(43*cos(f*x + e)^4 - 161*cos(f*x + e)^3 - 35*cos(f*x + e)^2 + 105*cos(f*x + e))*sqr
t((c*cos(f*x + e) - c)/cos(f*x + e)))/((a^2*c^3*f*cos(f*x + e)^3 - a^2*c^3*f*cos(f*x + e)^2 - a^2*c^3*f*cos(f*
x + e) + a^2*c^3*f)*sin(f*x + e))]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sec {\left (e + f x \right )}}{c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{4}{\left (e + f x \right )} - 2 c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{2}{\left (e + f x \right )} + c^{2} \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx}{a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))**2/(c-c*sec(f*x+e))**(5/2),x)

[Out]

Integral(sec(e + f*x)/(c**2*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**4 - 2*c**2*sqrt(-c*sec(e + f*x) + c)*sec(e
+ f*x)**2 + c**2*sqrt(-c*sec(e + f*x) + c)), x)/a**2

________________________________________________________________________________________

Giac [A]
time = 0.69, size = 160, normalized size = 0.79 \begin {gather*} \frac {\sqrt {2} {\left (105 \, \sqrt {c} \arctan \left (\frac {\sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\sqrt {c}}\right ) + \frac {8 \, {\left ({\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} c^{2} - 9 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c^{3}\right )}}{c^{3}} - \frac {3 \, {\left (13 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} c + 11 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c^{2}\right )}}{c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4}}\right )}}{384 \, a^{2} c^{3} 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))^2/(c-c*sec(f*x+e))^(5/2),x, algorithm="giac")

[Out]

1/384*sqrt(2)*(105*sqrt(c)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c)) + 8*((c*tan(1/2*f*x + 1/2*e)^2 -
c)^(3/2)*c^2 - 9*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^3)/c^3 - 3*(13*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(3/2)*c +
11*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^2)/(c^2*tan(1/2*f*x + 1/2*e)^4))/(a^2*c^3*f)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________