3.1.56 \(\int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^4} \, dx\) [56]

Optimal. Leaf size=172 \[ \frac {2 a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^4 f}+\frac {2 a \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^4 f}-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^4 f}+\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a c^4 f}-\frac {4 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^2 c^4 f} \]

[Out]

2*a^(3/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c^4/f-2/3*cot(f*x+e)^3*(a+a*sec(f*x+e))^(3/2)/c^4/
f+2/5*cot(f*x+e)^5*(a+a*sec(f*x+e))^(5/2)/a/c^4/f-4/7*cot(f*x+e)^7*(a+a*sec(f*x+e))^(7/2)/a^2/c^4/f+2*a*cot(f*
x+e)*(a+a*sec(f*x+e))^(1/2)/c^4/f

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Rubi [A]
time = 0.13, antiderivative size = 172, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3989, 3972, 464, 331, 209} \begin {gather*} \frac {2 a^{3/2} \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^4 f}-\frac {4 \cot ^7(e+f x) (a \sec (e+f x)+a)^{7/2}}{7 a^2 c^4 f}+\frac {2 \cot ^5(e+f x) (a \sec (e+f x)+a)^{5/2}}{5 a c^4 f}-\frac {2 \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 c^4 f}+\frac {2 a \cot (e+f x) \sqrt {a \sec (e+f x)+a}}{c^4 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(3/2)/(c - c*Sec[e + f*x])^4,x]

[Out]

(2*a^(3/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c^4*f) + (2*a*Cot[e + f*x]*Sqrt[a + a*Sec
[e + f*x]])/(c^4*f) - (2*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(3*c^4*f) + (2*Cot[e + f*x]^5*(a + a*Sec[e
+ f*x])^(5/2))/(5*a*c^4*f) - (4*Cot[e + f*x]^7*(a + a*Sec[e + f*x])^(7/2))/(7*a^2*c^4*f)

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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 464

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[c*(e*x)^(m +
1)*((a + b*x^n)^(p + 1)/(a*e*(m + 1))), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 3972

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[-2*(a^(m/2 +
n + 1/2)/d), Subst[Int[x^m*((2 + a*x^2)^(m/2 + n - 1/2)/(1 + a*x^2)), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rule 3989

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

Rubi steps

\begin {align*} \int \frac {(a+a \sec (e+f x))^{3/2}}{(c-c \sec (e+f x))^4} \, dx &=\frac {\int \cot ^8(e+f x) (a+a \sec (e+f x))^{11/2} \, dx}{a^4 c^4}\\ &=-\frac {2 \text {Subst}\left (\int \frac {2+a x^2}{x^8 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^2 c^4 f}\\ &=-\frac {4 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^2 c^4 f}+\frac {2 \text {Subst}\left (\int \frac {1}{x^6 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a c^4 f}\\ &=\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a c^4 f}-\frac {4 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^2 c^4 f}-\frac {2 \text {Subst}\left (\int \frac {1}{x^4 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^4 f}\\ &=-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^4 f}+\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a c^4 f}-\frac {4 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^2 c^4 f}+\frac {(2 a) \text {Subst}\left (\int \frac {1}{x^2 \left (1+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^4 f}\\ &=\frac {2 a \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^4 f}-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^4 f}+\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a c^4 f}-\frac {4 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^2 c^4 f}-\frac {\left (2 a^2\right ) \text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^4 f}\\ &=\frac {2 a^{3/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^4 f}+\frac {2 a \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^4 f}-\frac {2 \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^4 f}+\frac {2 \cot ^5(e+f x) (a+a \sec (e+f x))^{5/2}}{5 a c^4 f}-\frac {4 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a^2 c^4 f}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
time = 1.32, size = 102, normalized size = 0.59 \begin {gather*} -\frac {2 a \sqrt {\cos (e+f x)} \left (10 \cos ^{\frac {7}{2}}(e+f x)+7 (-1+\cos (e+f x)) \, _2F_1\left (-\frac {5}{2},-\frac {5}{2};-\frac {3}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )\right ) \sqrt {a (1+\sec (e+f x))} \tan \left (\frac {1}{2} (e+f x)\right )}{35 c^4 f (-1+\cos (e+f x))^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[e + f*x])^(3/2)/(c - c*Sec[e + f*x])^4,x]

[Out]

(-2*a*Sqrt[Cos[e + f*x]]*(10*Cos[e + f*x]^(7/2) + 7*(-1 + Cos[e + f*x])*Hypergeometric2F1[-5/2, -5/2, -3/2, 2*
Sin[(e + f*x)/2]^2])*Sqrt[a*(1 + Sec[e + f*x])]*Tan[(e + f*x)/2])/(35*c^4*f*(-1 + Cos[e + f*x])^4)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(400\) vs. \(2(152)=304\).
time = 0.26, size = 401, normalized size = 2.33

method	result	size
default	\(\frac {\sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (\cos \left (f x +e \right )+1\right ) \left (105 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {2}-315 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {2}+315 \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {2}-105 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \sin \left (f x +e \right )-382 \left (\cos ^{4}\left (f x +e \right )\right )+812 \left (\cos ^{3}\left (f x +e \right )\right )-700 \left (\cos ^{2}\left (f x +e \right )\right )+210 \cos \left (f x +e \right )\right ) a}{105 c^{4} f \sin \left (f x +e \right )^{3} \left (-1+\cos \left (f x +e \right )\right )^{2}}\)	\(401\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^4,x,method=_RETURNVERBOSE)

[Out]

1/105/c^4/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(cos(f*x+e)+1)*(105*cos(f*x+e)^3*sin(f*x+e)*(-2*cos(f*x+e)/(co
s(f*x+e)+1))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*2^(1/2)-315
*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2
^(1/2))*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)+315*sin(f*x+e)*cos(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*arctanh
(1/2*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*2^(1/2)-105*2^(1/2)*arctanh(1/2*(-2*c
os(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*sin(f*x+e)
-382*cos(f*x+e)^4+812*cos(f*x+e)^3-700*cos(f*x+e)^2+210*cos(f*x+e))/sin(f*x+e)^3/(-1+cos(f*x+e))^2*a

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 3.09, size = 537, normalized size = 3.12 \begin {gather*} \left [\frac {105 \, {\left (a \cos \left (f x + e\right )^{3} - 3 \, a \cos \left (f x + e\right )^{2} + 3 \, a \cos \left (f x + e\right ) - a\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (191 \, a \cos \left (f x + e\right )^{4} - 406 \, a \cos \left (f x + e\right )^{3} + 350 \, a \cos \left (f x + e\right )^{2} - 105 \, a \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{210 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )}, \frac {105 \, {\left (a \cos \left (f x + e\right )^{3} - 3 \, a \cos \left (f x + e\right )^{2} + 3 \, a \cos \left (f x + e\right ) - a\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (191 \, a \cos \left (f x + e\right )^{4} - 406 \, a \cos \left (f x + e\right )^{3} + 350 \, a \cos \left (f x + e\right )^{2} - 105 \, a \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{105 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} 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))^(3/2)/(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

[1/210*(105*(a*cos(f*x + e)^3 - 3*a*cos(f*x + e)^2 + 3*a*cos(f*x + e) - a)*sqrt(-a)*log(-(8*a*cos(f*x + e)^3 -
4*(2*cos(f*x + e)^2 - cos(f*x + e))*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e) - 7*a*cos(f
*x + e) + a)/(cos(f*x + e) + 1))*sin(f*x + e) + 4*(191*a*cos(f*x + e)^4 - 406*a*cos(f*x + e)^3 + 350*a*cos(f*x
+ e)^2 - 105*a*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^4*f*cos(f*x + e)^3 - 3*c^4*f*cos(f*
x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*f)*sin(f*x + e)), 1/105*(105*(a*cos(f*x + e)^3 - 3*a*cos(f*x + e)^2 + 3*
a*cos(f*x + e) - a)*sqrt(a)*arctan(2*sqrt(a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e)
/(2*a*cos(f*x + e)^2 + a*cos(f*x + e) - a))*sin(f*x + e) + 2*(191*a*cos(f*x + e)^4 - 406*a*cos(f*x + e)^3 + 35
0*a*cos(f*x + e)^2 - 105*a*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e)))/((c^4*f*cos(f*x + e)^3 - 3*c
^4*f*cos(f*x + e)^2 + 3*c^4*f*cos(f*x + e) - c^4*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 {a \sqrt {a \sec {\left (e + f x \right )} + a}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {a \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec {\left (e + f x \right )} + 1}\, dx}{c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*sqrt(a*sec(e + f*x) + a)/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x)
+ 1), x) + Integral(a*sqrt(a*sec(e + f*x) + a)*sec(e + f*x)/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e +
f*x)**2 - 4*sec(e + f*x) + 1), x))/c**4

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 555 vs. \(2 (152) = 304\).
time = 1.60, size = 555, normalized size = 3.23 \begin {gather*} -\frac {\frac {105 \, \sqrt {-a} a^{2} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right ) \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{c^{4} {\left | a \right |}} + \frac {420 \, \sqrt {2} {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{12} \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - 1785 \, \sqrt {2} {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{10} \sqrt {-a} a^{3} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 4235 \, \sqrt {2} {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{8} \sqrt {-a} a^{4} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - 4970 \, \sqrt {2} {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{6} \sqrt {-a} a^{5} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 3738 \, \sqrt {2} {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{4} \sqrt {-a} a^{6} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) - 1421 \, \sqrt {2} {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} \sqrt {-a} a^{7} \mathrm {sgn}\left (\cos \left (f x + e\right )\right ) + 263 \, \sqrt {2} \sqrt {-a} a^{8} \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{{\left ({\left (\sqrt {-a} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + a}\right )}^{2} - a\right )}^{7} c^{4}}}{105 \, f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(3/2)/(c-c*sec(f*x+e))^4,x, algorithm="giac")

[Out]

-1/105*(105*sqrt(-a)*a^2*log(abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 - 4
*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 + 4*sqrt(
2)*abs(a) - 6*a))*sgn(cos(f*x + e))/(c^4*abs(a)) + (420*sqrt(2)*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1
/2*f*x + 1/2*e)^2 + a))^12*sqrt(-a)*a^2*sgn(cos(f*x + e)) - 1785*sqrt(2)*(sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt
(-a*tan(1/2*f*x + 1/2*e)^2 + a))^10*sqrt(-a)*a^3*sgn(cos(f*x + e)) + 4235*sqrt(2)*(sqrt(-a)*tan(1/2*f*x + 1/2*
e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^8*sqrt(-a)*a^4*sgn(cos(f*x + e)) - 4970*sqrt(2)*(sqrt(-a)*tan(1/2*f*
x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^6*sqrt(-a)*a^5*sgn(cos(f*x + e)) + 3738*sqrt(2)*(sqrt(-a)*ta
n(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^4*sqrt(-a)*a^6*sgn(cos(f*x + e)) - 1421*sqrt(2)*(sqr
t(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(-a)*a^7*sgn(cos(f*x + e)) + 263*sqrt(
2)*sqrt(-a)*a^8*sgn(cos(f*x + e)))/(((sqrt(-a)*tan(1/2*f*x + 1/2*e) - sqrt(-a*tan(1/2*f*x + 1/2*e)^2 + a))^2 -
a)^7*c^4))/f

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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 )}^{3/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(3/2)/(c - c/cos(e + f*x))^4,x)

[Out]

int((a + a/cos(e + f*x))^(3/2)/(c - c/cos(e + f*x))^4, x)

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