3.63 \(\int \frac {(a+a \sec (e+f x))^{5/2}}{(c-c \sec (e+f x))^4} \, dx\)

Optimal. Leaf size=140 \[ \frac {2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^4 f}+\frac {2 a^2 \cot (e+f x) \sqrt {a \sec (e+f x)+a}}{c^4 f}-\frac {8 \cot ^7(e+f x) (a \sec (e+f x)+a)^{7/2}}{7 a c^4 f}-\frac {2 a \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 c^4 f} \]

[Out]

2*a^(5/2)*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/c^4/f-2/3*a*cot(f*x+e)^3*(a+a*sec(f*x+e))^(3/2)/c^
4/f-8/7*cot(f*x+e)^7*(a+a*sec(f*x+e))^(7/2)/a/c^4/f+2*a^2*cot(f*x+e)*(a+a*sec(f*x+e))^(1/2)/c^4/f

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Rubi [A]  time = 0.18, antiderivative size = 140, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3904, 3887, 461, 203} \[ \frac {2 a^2 \cot (e+f x) \sqrt {a \sec (e+f x)+a}}{c^4 f}+\frac {2 a^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{c^4 f}-\frac {8 \cot ^7(e+f x) (a \sec (e+f x)+a)^{7/2}}{7 a c^4 f}-\frac {2 a \cot ^3(e+f x) (a \sec (e+f x)+a)^{3/2}}{3 c^4 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^4,x]

[Out]

(2*a^(5/2)*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(c^4*f) + (2*a^2*Cot[e + f*x]*Sqrt[a + a*S
ec[e + f*x]])/(c^4*f) - (2*a*Cot[e + f*x]^3*(a + a*Sec[e + f*x])^(3/2))/(3*c^4*f) - (8*Cot[e + f*x]^7*(a + a*S
ec[e + f*x])^(7/2))/(7*a*c^4*f)

Rule 203

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

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3887

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 3904

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))^{5/2}}{(c-c \sec (e+f x))^4} \, dx &=\frac {\int \cot ^8(e+f x) (a+a \sec (e+f x))^{13/2} \, dx}{a^4 c^4}\\ &=-\frac {2 \operatorname {Subst}\left (\int \frac {\left (2+a x^2\right )^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 c^4 f}\\ &=-\frac {2 \operatorname {Subst}\left (\int \left (\frac {4}{x^8}+\frac {a^2}{x^4}-\frac {a^3}{x^2}+\frac {a^4}{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 a^2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^4 f}-\frac {2 a \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^4 f}-\frac {8 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a c^4 f}-\frac {\left (2 a^3\right ) \operatorname {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^{5/2} \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{c^4 f}+\frac {2 a^2 \cot (e+f x) \sqrt {a+a \sec (e+f x)}}{c^4 f}-\frac {2 a \cot ^3(e+f x) (a+a \sec (e+f x))^{3/2}}{3 c^4 f}-\frac {8 \cot ^7(e+f x) (a+a \sec (e+f x))^{7/2}}{7 a c^4 f}\\ \end {align*}

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Mathematica [C]  time = 7.98, size = 361, normalized size = 2.58 \[ -\frac {\sin ^8\left (\frac {e}{2}+\frac {f x}{2}\right ) \sqrt {\frac {1}{1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )}} \sqrt {1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )} \csc ^7\left (\frac {1}{2} (e+f x)\right ) \sec ^5\left (\frac {1}{2} (e+f x)\right ) \sec ^{\frac {3}{2}}(e+f x) (a (\sec (e+f x)+1))^{5/2} \left (336 \sin ^2\left (\frac {1}{2} (e+f x)\right ) \left (5 \sin ^4\left (\frac {1}{2} (e+f x)\right )-8 \sin ^2\left (\frac {1}{2} (e+f x)\right )+3\right ) \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )+4 \left (35 \sin ^4\left (\frac {1}{2} (e+f x)\right )-42 \sin ^2\left (\frac {1}{2} (e+f x)\right )+15\right ) \, _2F_1\left (-\frac {7}{2},-\frac {3}{2};-\frac {1}{2};2 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right )-105 \left (3 \sqrt {2} \sin ^{-1}\left (\sqrt {2} \sqrt {\sin ^2\left (\frac {1}{2} (e+f x)\right )}\right ) \sin ^2\left (\frac {1}{2} (e+f x)\right )^{3/2}+2 \left (5-4 \sin ^2\left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {1-2 \sin ^2\left (\frac {1}{2} (e+f x)\right )} \sin ^4\left (\frac {1}{2} (e+f x)\right )\right ) \cos ^4\left (\frac {1}{2} (e+f x)\right )\right )}{210 f (c-c \sec (e+f x))^4} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[e + f*x])^(5/2)/(c - c*Sec[e + f*x])^4,x]

[Out]

-1/210*(Csc[(e + f*x)/2]^7*Sec[(e + f*x)/2]^5*Sec[e + f*x]^(3/2)*(a*(1 + Sec[e + f*x]))^(5/2)*Sin[e/2 + (f*x)/
2]^8*Sqrt[(1 - 2*Sin[(e + f*x)/2]^2)^(-1)]*Sqrt[1 - 2*Sin[(e + f*x)/2]^2]*(336*Hypergeometric2F1[-5/2, -1/2, 1
/2, 2*Sin[(e + f*x)/2]^2]*Sin[(e + f*x)/2]^2*(3 - 8*Sin[(e + f*x)/2]^2 + 5*Sin[(e + f*x)/2]^4) + 4*Hypergeomet
ric2F1[-7/2, -3/2, -1/2, 2*Sin[(e + f*x)/2]^2]*(15 - 42*Sin[(e + f*x)/2]^2 + 35*Sin[(e + f*x)/2]^4) - 105*Cos[
(e + f*x)/2]^4*(3*Sqrt[2]*ArcSin[Sqrt[2]*Sqrt[Sin[(e + f*x)/2]^2]]*(Sin[(e + f*x)/2]^2)^(3/2) + 2*Sin[(e + f*x
)/2]^4*(5 - 4*Sin[(e + f*x)/2]^2)*Sqrt[1 - 2*Sin[(e + f*x)/2]^2])))/(f*(c - c*Sec[e + f*x])^4)

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fricas [B]  time = 0.57, size = 527, normalized size = 3.76 \[ \left [\frac {21 \, {\left (a^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {-a} \log \left (-\frac {8 \, a \cos \left (f x + e\right )^{3} - 4 \, {\left (2 \, \cos \left (f x + e\right )^{2} - \cos \left (f x + e\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right ) - 7 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right ) + 1}\right ) \sin \left (f x + e\right ) + 4 \, {\left (40 \, a^{2} \cos \left (f x + e\right )^{4} - 77 \, a^{2} \cos \left (f x + e\right )^{3} + 70 \, a^{2} \cos \left (f x + e\right )^{2} - 21 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{42 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )}, \frac {21 \, {\left (a^{2} \cos \left (f x + e\right )^{3} - 3 \, a^{2} \cos \left (f x + e\right )^{2} + 3 \, a^{2} \cos \left (f x + e\right ) - a^{2}\right )} \sqrt {a} \arctan \left (\frac {2 \, \sqrt {a} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}{2 \, a \cos \left (f x + e\right )^{2} + a \cos \left (f x + e\right ) - a}\right ) \sin \left (f x + e\right ) + 2 \, {\left (40 \, a^{2} \cos \left (f x + e\right )^{4} - 77 \, a^{2} \cos \left (f x + e\right )^{3} + 70 \, a^{2} \cos \left (f x + e\right )^{2} - 21 \, a^{2} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}}}{21 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^4,x, algorithm="fricas")

[Out]

[1/42*(21*(a^2*cos(f*x + e)^3 - 3*a^2*cos(f*x + e)^2 + 3*a^2*cos(f*x + e) - a^2)*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*(40*a^2*cos(f*x + e)^4 - 77*a^2*cos(f*x + e)^3 + 70*a^
2*cos(f*x + e)^2 - 21*a^2*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)), 1/21*(21*(a^2*cos(f*x + e)^3 - 3*a^2*cos(f*x
+ e)^2 + 3*a^2*cos(f*x + e) - a^2)*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*(40*a^2*cos(f*x + e)^4 - 77*a^2*c
os(f*x + e)^3 + 70*a^2*cos(f*x + e)^2 - 21*a^2*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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)4*(-1/90602450860736855426226665952476127623724728371125153965
332993965887984762880000*(56087231485218053359092697970580459957543879467839381026158520074121133424640000*sqr
t(2)*a^9*sqrt(-a)*sign(cos(f*x+exp(1)))+9060245086073685542622666595247612762372472837112515396533299396588798
4762880000*sqrt(2)*a^3*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^12*sign(co
s(f*x+exp(1)))-362409803442947421704906663809904510494898913484500615861331975863551939051520000*sqrt(2)*a^4*s
qrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^10*sign(cos(f*x+exp(1)))[phone]
560947506071008881508919985445155526501626590975107604314175842549760000*sqrt(2)*a^5*sqrt(-a)*(sqrt(-a*tan(1/2
*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^8*sign(cos(f*x+exp(1)))-1026827776421684361497235547461396
113068880254872751744940440598280063827312640000*sqrt(2)*a^6*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt
(-a)*tan(1/2*(f*x+exp(1))))^6*sign(cos(f*x+exp(1)))+8154220577466316988360399935722851486135225553401263856879
96945692991862865920000*sqrt(2)*a^7*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))
))^4*sign(cos(f*x+exp(1)))-302008169535789518087422219841587092079082427903750513217776646552959949209600000*s
qrt(2)*a^8*sqrt(-a)*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2*sign(cos(f*x+exp(1))
))/c^4/((sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2-a)^7-1/4*a^3*sqrt(-a)*sign(cos(f
*x+exp(1)))*ln(abs(2*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2-4*sqrt(2)*abs(a)-6*
a)/abs(2*(sqrt(-a*tan(1/2*(f*x+exp(1)))^2+a)-sqrt(-a)*tan(1/2*(f*x+exp(1))))^2+4*sqrt(2)*abs(a)-6*a))/c^4/abs(
a))/f

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maple [B]  time = 1.63, size = 395, normalized size = 2.82 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (f x +e \right )\right )}{\cos \left (f x +e \right )}}\, \left (-21 \left (\cos ^{3}\left (f x +e \right )\right ) \sin \left (f x +e \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )+63 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )-63 \cos \left (f x +e \right ) \sin \left (f x +e \right ) \sqrt {2}\, \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right )+80 \left (\cos ^{4}\left (f x +e \right )\right )+21 \sqrt {2}\, \sin \left (f x +e \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}\, \sin \left (f x +e \right ) \sqrt {2}}{2 \cos \left (f x +e \right )}\right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{1+\cos \left (f x +e \right )}}-154 \left (\cos ^{3}\left (f x +e \right )\right )+140 \left (\cos ^{2}\left (f x +e \right )\right )-42 \cos \left (f x +e \right )\right ) a^{2}}{21 c^{4} f \sin \left (f x +e \right ) \left (-1+\cos \left (f x +e \right )\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^4,x)

[Out]

1/21/c^4/f*(a*(1+cos(f*x+e))/cos(f*x+e))^(1/2)*(-21*cos(f*x+e)^3*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+
e)))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))+63*sin(f*x+e)*cos(f
*x+e)^2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+
e)/cos(f*x+e)*2^(1/2))-63*cos(f*x+e)*sin(f*x+e)*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*(-2*c
os(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))+80*cos(f*x+e)^4+21*2^(1/2)*sin(f*x+e)*arctanh(1
/2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e)*2^(1/2))*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)-15
4*cos(f*x+e)^3+140*cos(f*x+e)^2-42*cos(f*x+e))/sin(f*x+e)/(-1+cos(f*x+e))^3*a^2

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))^(5/2)/(c-c*sec(f*x+e))^4,x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {a}{\cos \left (e+f\,x\right )}\right )}^{5/2}}{{\left (c-\frac {c}{\cos \left (e+f\,x\right )}\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(e + f*x))^(5/2)/(c - c/cos(e + f*x))^4,x)

[Out]

int((a + a/cos(e + f*x))^(5/2)/(c - c/cos(e + f*x))^4, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(f*x+e))**(5/2)/(c-c*sec(f*x+e))**4,x)

[Out]

Timed out

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