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

Optimal. Leaf size=212 \[ -\frac {7 \text {ArcTan}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}-\frac {7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{12 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}} \]

[Out]

-7/32*arctan(1/2*c^(1/2)*tan(f*x+e)*2^(1/2)/(c-c*sec(f*x+e))^(1/2))/a^3/c^(3/2)/f*2^(1/2)-7/16*tan(f*x+e)/a^3/
f/(c-c*sec(f*x+e))^(3/2)+1/5*tan(f*x+e)/f/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(3/2)+7/30*tan(f*x+e)/a/f/(a+a*s
ec(f*x+e))^2/(c-c*sec(f*x+e))^(3/2)+7/12*tan(f*x+e)/f/(a^3+a^3*sec(f*x+e))/(c-c*sec(f*x+e))^(3/2)

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Rubi [A]
time = 0.33, antiderivative size = 212, 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 {7 \text {ArcTan}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}-\frac {7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{12 f \left (a^3 \sec (e+f x)+a^3\right ) (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{30 a f (a \sec (e+f x)+a)^2 (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{5 f (a \sec (e+f x)+a)^3 (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])^3*(c - c*Sec[e + f*x])^(3/2)),x]

[Out]

(-7*ArcTan[(Sqrt[c]*Tan[e + f*x])/(Sqrt[2]*Sqrt[c - c*Sec[e + f*x]])])/(16*Sqrt[2]*a^3*c^(3/2)*f) - (7*Tan[e +
f*x])/(16*a^3*f*(c - c*Sec[e + f*x])^(3/2)) + Tan[e + f*x]/(5*f*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(
3/2)) + (7*Tan[e + f*x])/(30*a*f*(a + a*Sec[e + f*x])^2*(c - c*Sec[e + f*x])^(3/2)) + (7*Tan[e + f*x])/(12*f*(
a^3 + a^3*Sec[e + f*x])*(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))^3 (c-c \sec (e+f x))^{3/2}} \, dx &=\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac {7 \int \frac {\sec (e+f x)}{(a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}} \, dx}{10 a}\\ &=\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {7 \int \frac {\sec (e+f x)}{(a+a \sec (e+f x)) (c-c \sec (e+f x))^{3/2}} \, dx}{12 a^2}\\ &=\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{12 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}+\frac {7 \int \frac {\sec (e+f x)}{(c-c \sec (e+f x))^{3/2}} \, dx}{8 a^3}\\ &=-\frac {7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{12 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}+\frac {7 \int \frac {\sec (e+f x)}{\sqrt {c-c \sec (e+f x)}} \, dx}{32 a^3 c}\\ &=-\frac {7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{12 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}-\frac {7 \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 )}{16 a^3 c f}\\ &=-\frac {7 \tan ^{-1}\left (\frac {\sqrt {c} \tan (e+f x)}{\sqrt {2} \sqrt {c-c \sec (e+f x)}}\right )}{16 \sqrt {2} a^3 c^{3/2} f}-\frac {7 \tan (e+f x)}{16 a^3 f (c-c \sec (e+f x))^{3/2}}+\frac {\tan (e+f x)}{5 f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{30 a f (a+a \sec (e+f x))^2 (c-c \sec (e+f x))^{3/2}}+\frac {7 \tan (e+f x)}{12 f \left (a^3+a^3 \sec (e+f x)\right ) (c-c \sec (e+f x))^{3/2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 6.44, size = 398, normalized size = 1.88 \begin {gather*} \frac {7 e^{-\frac {1}{2} i (e+f x)} \sqrt {\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}} \sqrt {1+e^{2 i (e+f x)}} \tanh ^{-1}\left (\frac {1+e^{i (e+f x)}}{\sqrt {2} \sqrt {1+e^{2 i (e+f x)}}}\right ) \cos ^6\left (\frac {e}{2}+\frac {f x}{2}\right ) \sec ^{\frac {9}{2}}(e+f x) \sin ^3\left (\frac {e}{2}+\frac {f x}{2}\right )}{f (a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}}+\frac {\cos ^6\left (\frac {e}{2}+\frac {f x}{2}\right ) \sec ^5(e+f x) \left (-\frac {278 \cos \left (\frac {e}{2}\right ) \cos \left (\frac {f x}{2}\right )}{15 f}-\frac {\cot \left (\frac {e}{2}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}\right )}{f}+\frac {242 \sec \left (\frac {e}{2}+\frac {f x}{2}\right )}{15 f}-\frac {56 \sec ^3\left (\frac {e}{2}+\frac {f x}{2}\right )}{15 f}+\frac {2 \sec ^5\left (\frac {e}{2}+\frac {f x}{2}\right )}{5 f}+\frac {\csc \left (\frac {e}{2}\right ) \csc ^2\left (\frac {e}{2}+\frac {f x}{2}\right ) \sin \left (\frac {f x}{2}\right )}{f}+\frac {278 \sin \left (\frac {e}{2}\right ) \sin \left (\frac {f x}{2}\right )}{15 f}\right ) \sin ^3\left (\frac {e}{2}+\frac {f x}{2}\right )}{(a+a \sec (e+f x))^3 (c-c \sec (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(7*Sqrt[E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)))]*Sqrt[1 + E^((2*I)*(e + f*x))]*ArcTanh[(1 + E^(I*(e + f*x)))
/(Sqrt[2]*Sqrt[1 + E^((2*I)*(e + f*x))])]*Cos[e/2 + (f*x)/2]^6*Sec[e + f*x]^(9/2)*Sin[e/2 + (f*x)/2]^3)/(E^((I
/2)*(e + f*x))*f*(a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(3/2)) + (Cos[e/2 + (f*x)/2]^6*Sec[e + f*x]^5*((-
278*Cos[e/2]*Cos[(f*x)/2])/(15*f) - (Cot[e/2]*Csc[e/2 + (f*x)/2])/f + (242*Sec[e/2 + (f*x)/2])/(15*f) - (56*Se
c[e/2 + (f*x)/2]^3)/(15*f) + (2*Sec[e/2 + (f*x)/2]^5)/(5*f) + (Csc[e/2]*Csc[e/2 + (f*x)/2]^2*Sin[(f*x)/2])/f +
(278*Sin[e/2]*Sin[(f*x)/2])/(15*f))*Sin[e/2 + (f*x)/2]^3)/((a + a*Sec[e + f*x])^3*(c - c*Sec[e + f*x])^(3/2))

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Maple [A]
time = 2.60, size = 370, normalized size = 1.75

method	result	size
default	\(\frac {\left (-1+\cos \left (f x +e \right )\right )^{2} \left (15 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {9}{2}} \cos \left (f x +e \right )+15 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {9}{2}}+15 \cos \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {7}{2}}-15 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {7}{2}}-21 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}} \cos \left (f x +e \right )+21 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {5}{2}}+35 \cos \left (f x +e \right ) \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}}-35 \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}}-105 \cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-105 \cos \left (f x +e \right ) \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )+105 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}+105 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}}\right )\right )}{120 a^{3} f \left (\frac {c \left (-1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}\right )^{\frac {3}{2}} \sin \left (f x +e \right )^{3} \left (-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}\right )^{\frac {3}{2}}}\)	\(370\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)/(a+a*sec(f*x+e))^3/(c-c*sec(f*x+e))^(3/2),x,method=_RETURNVERBOSE)

[Out]

1/120/a^3/f*(-1+cos(f*x+e))^2*(15*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(9/2)*cos(f*x+e)+15*(-2*cos(f*x+e)/(cos(f*x+e
)+1))^(9/2)+15*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*(-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*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*cos(f*x+e)/(cos(f*x+e)+1))
^(1/2)-105*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)+
105*arctan(1/(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)))/(c*(-1+cos(f*x+e))/cos(f*x+e))^(3/2)/sin(f*x+e)^3/(-2*cos(
f*x+e)/(cos(f*x+e)+1))^(3/2)

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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))^3/(c-c*sec(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)/((a*sec(f*x + e) + a)^3*(-c*sec(f*x + e) + c)^(3/2)), x)

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Fricas [A]
time = 4.31, size = 523, normalized size = 2.47 \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 (139 \, \cos \left (f x + e\right )^{4} + 21 \, \cos \left (f x + e\right )^{3} - 175 \, \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 )}}}{960 \, {\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 {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 (139 \, \cos \left (f x + e\right )^{4} + 21 \, \cos \left (f x + e\right )^{3} - 175 \, \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 )}}}{480 \, {\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))^3/(c-c*sec(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

[-1/960*(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*(139*cos(f*x + e)^4 + 21*cos(f*x + e)^3 - 175*cos(f*x + e)^2
- 105*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)), 1/480*(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*(139*cos(f*x + e)^4 + 21*cos(f*x + e)^3 - 175*cos(f*x + e)^2 - 105*cos(f*x + e))*s
qrt((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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sec {\left (e + f x \right )}}{- c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{4}{\left (e + f x \right )} - 2 c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec ^{3}{\left (e + f x \right )} + 2 c \sqrt {- c \sec {\left (e + f x \right )} + c} \sec {\left (e + f x \right )} + c \sqrt {- c \sec {\left (e + f x \right )} + c}}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)/(a+a*sec(f*x+e))**3/(c-c*sec(f*x+e))**(3/2),x)

[Out]

Integral(sec(e + f*x)/(-c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x)**4 - 2*c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*
x)**3 + 2*c*sqrt(-c*sec(e + f*x) + c)*sec(e + f*x) + c*sqrt(-c*sec(e + f*x) + c)), x)/a**3

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Giac [A]
time = 0.76, size = 154, normalized size = 0.73 \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 {15 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c}}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2}} - \frac {2 \, {\left (3 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {5}{2}} c^{8} - 10 \, {\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c\right )}^{\frac {3}{2}} c^{9} + 45 \, \sqrt {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c} c^{10}\right )}}{c^{10}}\right )}}{480 \, a^{3} c^{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))^3/(c-c*sec(f*x+e))^(3/2),x, algorithm="giac")

[Out]

1/480*sqrt(2)*(105*sqrt(c)*arctan(sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)/sqrt(c)) - 15*sqrt(c*tan(1/2*f*x + 1/2*e)
^2 - c)/tan(1/2*f*x + 1/2*e)^2 - 2*(3*(c*tan(1/2*f*x + 1/2*e)^2 - c)^(5/2)*c^8 - 10*(c*tan(1/2*f*x + 1/2*e)^2
- c)^(3/2)*c^9 + 45*sqrt(c*tan(1/2*f*x + 1/2*e)^2 - c)*c^10)/c^10)/(a^3*c^2*f)

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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 )}^3\,{\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))^3*(c - c/cos(e + f*x))^(3/2)),x)

[Out]

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

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