3.1.83 \(\int \frac {c-c \sec (e+f x)}{(a+a \sec (e+f x))^{5/2}} \, dx\) [83]

Optimal. Leaf size=148 \[ \frac {2 c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac {23 c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{8 \sqrt {2} a^{5/2} f}-\frac {c \tan (e+f x)}{2 f (a+a \sec (e+f x))^{5/2}}-\frac {7 c \tan (e+f x)}{8 a f (a+a \sec (e+f x))^{3/2}} \]

[Out]

2*c*arctan(a^(1/2)*tan(f*x+e)/(a+a*sec(f*x+e))^(1/2))/a^(5/2)/f-23/16*c*arctan(1/2*a^(1/2)*tan(f*x+e)*2^(1/2)/
(a+a*sec(f*x+e))^(1/2))/a^(5/2)/f*2^(1/2)-1/2*c*tan(f*x+e)/f/(a+a*sec(f*x+e))^(5/2)-7/8*c*tan(f*x+e)/a/f/(a+a*
sec(f*x+e))^(3/2)

________________________________________________________________________________________

Rubi [A]
time = 0.13, antiderivative size = 181, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3989, 3972, 482, 541, 536, 209} \begin {gather*} \frac {2 c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac {23 c \text {ArcTan}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a \sec (e+f x)+a}}\right )}{8 \sqrt {2} a^{5/2} f}-\frac {7 c \sin (e+f x) \sec ^2\left (\frac {1}{2} (e+f x)\right )}{16 a^2 f \sqrt {a \sec (e+f x)+a}}-\frac {c \sin (e+f x) \cos (e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right )}{8 a^2 f \sqrt {a \sec (e+f x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - (23*c*ArcTan[(Sqrt[a]*Tan[e + f*x]
)/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(8*Sqrt[2]*a^(5/2)*f) - (7*c*Sec[(e + f*x)/2]^2*Sin[e + f*x])/(16*a^2*f
*Sqrt[a + a*Sec[e + f*x]]) - (c*Cos[e + f*x]*Sec[(e + f*x)/2]^4*Sin[e + f*x])/(8*a^2*f*Sqrt[a + a*Sec[e + f*x]
])

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 482

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

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[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 {c-c \sec (e+f x)}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\left ((a c) \int \frac {\tan ^2(e+f x)}{(a+a \sec (e+f x))^{7/2}} \, dx\right )\\ &=\frac {(2 c) \text {Subst}\left (\int \frac {x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )^3} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a f}\\ &=-\frac {c \cos (e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x)}{8 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {c \text {Subst}\left (\int \frac {1-3 a x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{2 a^2 f}\\ &=-\frac {7 c \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x)}{16 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {c \cos (e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x)}{8 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {c \text {Subst}\left (\int \frac {9 a-7 a^2 x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{8 a^3 f}\\ &=-\frac {7 c \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x)}{16 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {c \cos (e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x)}{8 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {(2 c) \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 )}{a^2 f}+\frac {(23 c) \text {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{8 a^2 f}\\ &=\frac {2 c \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac {23 c \tan ^{-1}\left (\frac {\sqrt {a} \tan (e+f x)}{\sqrt {2} \sqrt {a+a \sec (e+f x)}}\right )}{8 \sqrt {2} a^{5/2} f}-\frac {7 c \sec ^2\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x)}{16 a^2 f \sqrt {a+a \sec (e+f x)}}-\frac {c \cos (e+f x) \sec ^4\left (\frac {1}{2} (e+f x)\right ) \sin (e+f x)}{8 a^2 f \sqrt {a+a \sec (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]
time = 1.49, size = 134, normalized size = 0.91 \begin {gather*} -\frac {c \cot \left (\frac {1}{2} (e+f x)\right ) \left ((3+8 \cos (e+f x)-11 \cos (2 (e+f x))) \sec ^4\left (\frac {1}{2} (e+f x)\right )-128 \text {ArcTan}\left (\sqrt {-1+\sec (e+f x)}\right ) \sqrt {-1+\sec (e+f x)}+92 \sqrt {2} \text {ArcTan}\left (\frac {\sqrt {-1+\sec (e+f x)}}{\sqrt {2}}\right ) \sqrt {-1+\sec (e+f x)}\right )}{64 a^2 f \sqrt {a (1+\sec (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/64*(c*Cot[(e + f*x)/2]*((3 + 8*Cos[e + f*x] - 11*Cos[2*(e + f*x)])*Sec[(e + f*x)/2]^4 - 128*ArcTan[Sqrt[-1
+ Sec[e + f*x]]]*Sqrt[-1 + Sec[e + f*x]] + 92*Sqrt[2]*ArcTan[Sqrt[-1 + Sec[e + f*x]]/Sqrt[2]]*Sqrt[-1 + Sec[e
+ f*x]]))/(a^2*f*Sqrt[a*(1 + Sec[e + f*x])])

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(542\) vs. \(2(123)=246\).
time = 0.17, size = 543, normalized size = 3.67

method	result	size
default	\(-\frac {c \sqrt {\frac {a \left (\cos \left (f x +e \right )+1\right )}{\cos \left (f x +e \right )}}\, \left (16 \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}+23 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+32 \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}+46 \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right ) \cos \left (f x +e \right ) \sin \left (f x +e \right )+16 \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 )+23 \sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}\, \ln \left (\frac {\sin \left (f x +e \right ) \sqrt {-\frac {2 \cos \left (f x +e \right )}{\cos \left (f x +e \right )+1}}-\cos \left (f x +e \right )+1}{\sin \left (f x +e \right )}\right )-22 \left (\cos ^{3}\left (f x +e \right )\right )+8 \left (\cos ^{2}\left (f x +e \right )\right )+14 \cos \left (f x +e \right )\right )}{16 f \left (\cos \left (f x +e \right )+1\right )^{2} \sin \left (f x +e \right ) a^{3}}\)	\(543\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*c/f*(a*(cos(f*x+e)+1)/cos(f*x+e))^(1/2)*(16*(-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)+23*(-2*cos(f*x+e)/(co
s(f*x+e)+1))^(1/2)*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*cos(f*x+e)^2*
sin(f*x+e)+32*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)+46*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*ln((sin(f*x+e)*(-2*
cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x+e)+1)/sin(f*x+e))*cos(f*x+e)*sin(f*x+e)+16*2^(1/2)*arctanh(1/2*(-2*co
s(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)+
23*sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)*ln((sin(f*x+e)*(-2*cos(f*x+e)/(cos(f*x+e)+1))^(1/2)-cos(f*x
+e)+1)/sin(f*x+e))-22*cos(f*x+e)^3+8*cos(f*x+e)^2+14*cos(f*x+e))/(cos(f*x+e)+1)^2/sin(f*x+e)/a^3

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (131) = 262\).
time = 4.86, size = 656, normalized size = 4.43 \begin {gather*} \left [-\frac {23 \, \sqrt {2} {\left (c \cos \left (f x + e\right )^{3} + 3 \, c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) + c\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {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 ) - 3 \, a \cos \left (f x + e\right )^{2} - 2 \, a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1}\right ) + 32 \, {\left (c \cos \left (f x + e\right )^{3} + 3 \, c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) + c\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (f x + e\right )^{2} + 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 ) + a \cos \left (f x + e\right ) - a}{\cos \left (f x + e\right ) + 1}\right ) + 4 \, {\left (11 \, c \cos \left (f x + e\right )^{2} + 7 \, c \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{32 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}}, \frac {23 \, \sqrt {2} {\left (c \cos \left (f x + e\right )^{3} + 3 \, c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) + c\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - 32 \, {\left (c \cos \left (f x + e\right )^{3} + 3 \, c \cos \left (f x + e\right )^{2} + 3 \, c \cos \left (f x + e\right ) + c\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \cos \left (f x + e\right )}{\sqrt {a} \sin \left (f x + e\right )}\right ) - 2 \, {\left (11 \, c \cos \left (f x + e\right )^{2} + 7 \, c \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right ) + a}{\cos \left (f x + e\right )}} \sin \left (f x + e\right )}{16 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/32*(23*sqrt(2)*(c*cos(f*x + e)^3 + 3*c*cos(f*x + e)^2 + 3*c*cos(f*x + e) + c)*sqrt(-a)*log(-(2*sqrt(2)*sqr
t(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) - 3*a*cos(f*x + e)^2 - 2*a*cos(f*x + e
) + a)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) + 32*(c*cos(f*x + e)^3 + 3*c*cos(f*x + e)^2 + 3*c*cos(f*x + e) +
c)*sqrt(-a)*log((2*a*cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x
+ e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) + 4*(11*c*cos(f*x + e)^2 + 7*c*cos(f*x + e))*sqrt((a*cos(f*x +
e) + a)/cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a
^3*f), 1/16*(23*sqrt(2)*(c*cos(f*x + e)^3 + 3*c*cos(f*x + e)^2 + 3*c*cos(f*x + e) + c)*sqrt(a)*arctan(sqrt(2)*
sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - 32*(c*cos(f*x + e)^3 + 3*c*cos(
f*x + e)^2 + 3*c*cos(f*x + e) + c)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a
)*sin(f*x + e))) - 2*(11*c*cos(f*x + e)^2 + 7*c*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x
+ e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*f)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - c \left (\int \frac {\sec {\left (e + f x \right )}}{a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )} + a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}\, dx + \int \left (- \frac {1}{a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec ^{2}{\left (e + f x \right )} + 2 a^{2} \sqrt {a \sec {\left (e + f x \right )} + a} \sec {\left (e + f x \right )} + a^{2} \sqrt {a \sec {\left (e + f x \right )} + a}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(co

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________