\(\int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx\) [175]

Optimal result

Integrand size = 26, antiderivative size = 152 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {23 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}-\frac {A \tan (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {7 A \tan (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.22, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3989, 3972, 482, 541, 536, 209} \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {23 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}+\frac {7 A \sin (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{16 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {A \sin (c+d x) \cos (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d \sqrt {a-a \sec (c+d x)}} \]

[In]

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

[Out]

(2*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(a^(5/2)*d) - (23*A*ArcTan[(Sqrt[a]*Tan[c + d*x]
)/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(8*Sqrt[2]*a^(5/2)*d) + (7*A*Csc[(c + d*x)/2]^2*Sin[c + d*x])/(16*a^2*d
*Sqrt[a - a*Sec[c + d*x]]) - (A*Cos[c + d*x]*Csc[(c + d*x)/2]^4*Sin[c + d*x])/(8*a^2*d*Sqrt[a - a*Sec[c + d*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*} \text {integral}& = -\left ((a A) \int \frac {\tan ^2(c+d x)}{(a-a \sec (c+d x))^{7/2}} \, dx\right ) \\ & = \frac {(2 A) \text {Subst}\left (\int \frac {x^2}{\left (1+a x^2\right ) \left (2+a x^2\right )^3} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a d} \\ & = -\frac {A \cos (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {A \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 (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{2 a^2 d} \\ & = \frac {7 A \csc ^2\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{16 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {A \cos (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {A \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 (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 a^3 d} \\ & = \frac {7 A \csc ^2\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{16 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {A \cos (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {(2 A) \text {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^2 d}+\frac {(23 A) \text {Subst}\left (\int \frac {1}{2+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 a^2 d} \\ & = \frac {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {23 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}+\frac {7 A \csc ^2\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{16 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {A \cos (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.43 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.98 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {A \left (16 \text {arctanh}\left (\sqrt {1+\sec (c+d x)}\right ) (-1+\sec (c+d x))^2+\sqrt {1+\sec (c+d x)} (-11+7 \sec (c+d x))-46 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {1+\sec (c+d x)}}{\sqrt {2}}\right ) \sec ^2(c+d x) \sin ^4\left (\frac {1}{2} (c+d x)\right )\right ) \tan (c+d x)}{8 a^2 d (-1+\sec (c+d x))^2 \sqrt {1+\sec (c+d x)} \sqrt {a-a \sec (c+d x)}} \]

[In]

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

[Out]

(A*(16*ArcTanh[Sqrt[1 + Sec[c + d*x]]]*(-1 + Sec[c + d*x])^2 + Sqrt[1 + Sec[c + d*x]]*(-11 + 7*Sec[c + d*x]) -
46*Sqrt[2]*ArcTanh[Sqrt[1 + Sec[c + d*x]]/Sqrt[2]]*Sec[c + d*x]^2*Sin[(c + d*x)/2]^4)*Tan[c + d*x])/(8*a^2*d*
(-1 + Sec[c + d*x])^2*Sqrt[1 + Sec[c + d*x]]*Sqrt[a - a*Sec[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(416\) vs. \(2(127)=254\).

Time = 4.25 (sec) , antiderivative size = 417, normalized size of antiderivative = 2.74

method	result	size
default	\(\frac {A \sqrt {2}\, \left (15 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \left (1-\cos \left (d x +c \right )\right )^{2} \sin \left (d x +c \right )-23 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (1-\cos \left (d x +c \right )\right )^{4} \csc \left (d x +c \right )+8 \left (1-\cos \left (d x +c \right )\right )^{6} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{3}+48 \sqrt {2}\, \arctan \left (\frac {\sqrt {2}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}{2}\right ) \left (1-\cos \left (d x +c \right )\right )^{4} \csc \left (d x +c \right )-6 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \sin \left (d x +c \right )^{3}+13 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (1-\cos \left (d x +c \right )\right )^{4} \csc \left (d x +c \right )+69 \arctan \left (\frac {1}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \left (1-\cos \left (d x +c \right )\right )^{4} \csc \left (d x +c \right )\right )}{48 a^{2} d \sqrt {\frac {a \left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (1-\cos \left (d x +c \right )\right )^{3} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\)	\(417\)
parts	\(-\frac {A \sqrt {2}\, \left (15 \cos \left (d x +c \right )^{2} \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-32 \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sqrt {2}\, \cos \left (d x +c \right )^{2}+4 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {2}\, \cos \left (d x +c \right )-43 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \cos \left (d x +c \right )^{2}+64 \sqrt {2}\, \cos \left (d x +c \right ) \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-11 \sqrt {2}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}+86 \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )-32 \sqrt {2}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right )-43 \arctan \left (\frac {\sqrt {2}}{2 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )\right ) \sec \left (d x +c \right ) \tan \left (d x +c \right )}{32 d \left (\cos \left (d x +c \right )+1\right ) a^{2} \left (\sec \left (d x +c \right )-1\right )^{2} \sqrt {-a \left (\sec \left (d x +c \right )-1\right )}\, \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}+\frac {A \sqrt {2}\, \left (1-\cos \left (d x +c \right )\right ) \left (\left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-\left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (1-\cos \left (d x +c \right )\right )^{4} \csc \left (d x +c \right )^{4}+3 \arctan \left (\frac {1}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right ) \left (1-\cos \left (d x +c \right )\right )^{4} \csc \left (d x +c \right )^{4}-2 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}+3 \left (1-\cos \left (d x +c \right )\right )^{4} \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \csc \left (d x +c \right )^{4}\right ) \csc \left (d x +c \right )}{32 d \left (\frac {a \left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )^{\frac {5}{2}} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}}\)	\(674\)

[In]

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

[Out]

1/48*A/a^2/d*2^(1/2)/(a*(1-cos(d*x+c))^2/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)*csc(d*x+c)^2)^(1/2)/(1-cos(d*x+c))^
3/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(15*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(5/2)*(1-cos(d*x+c))^2*sin(d*x
+c)-23*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(1-cos(d*x+c))^4*csc(d*x+c)+8*(1-cos(d*x+c))^6*((1-cos(d*x+c))^
2*csc(d*x+c)^2-1)^(1/2)*csc(d*x+c)^3+48*2^(1/2)*arctan(1/2*2^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*(1
-cos(d*x+c))^4*csc(d*x+c)-6*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(5/2)*sin(d*x+c)^3+13*((1-cos(d*x+c))^2*csc(d*x+
c)^2-1)^(1/2)*(1-cos(d*x+c))^4*csc(d*x+c)+69*arctan(1/((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2))*(1-cos(d*x+c))^
4*csc(d*x+c))

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 255 vs. \(2 (127) = 254\).

Time = 0.30 (sec) , antiderivative size = 590, normalized size of antiderivative = 3.88 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {23 \, \sqrt {2} {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} + {\left (3 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{{\left (\cos \left (d x + c\right ) - 1\right )} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 32 \, {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {-a} \log \left (\frac {2 \, {\left (\cos \left (d x + c\right )^{2} + \cos \left (d x + c\right )\right )} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} - {\left (2 \, a \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right )}{\sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 4 \, {\left (11 \, A \cos \left (d x + c\right )^{3} + 4 \, A \cos \left (d x + c\right )^{2} - 7 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )} \sin \left (d x + c\right )}, \frac {23 \, \sqrt {2} {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) - 32 \, {\left (A \cos \left (d x + c\right )^{2} - 2 \, A \cos \left (d x + c\right ) + A\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) \sin \left (d x + c\right ) + 2 \, {\left (11 \, A \cos \left (d x + c\right )^{3} + 4 \, A \cos \left (d x + c\right )^{2} - 7 \, A \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )}}}{16 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} - 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )} \sin \left (d x + c\right )}\right ] \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=A \left (\int \frac {\sec {\left (c + d x \right )}}{a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {1}{a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \]

[In]

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))**(5/2),x)

[Out]

A*(Integral(sec(c + d*x)/(a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x)**2 - 2*a**2*sqrt(-a*sec(c + d*x) + a)*se
c(c + d*x) + a**2*sqrt(-a*sec(c + d*x) + a)), x) + Integral(1/(a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x)**2
- 2*a**2*sqrt(-a*sec(c + d*x) + a)*sec(c + d*x) + a**2*sqrt(-a*sec(c + d*x) + a)), x))

Maxima [F]

\[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\int { \frac {A \sec \left (d x + c\right ) + A}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((A*sec(d*x + c) + A)/(-a*sec(d*x + c) + a)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 1.20 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.91 \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\frac {\frac {23 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {32 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {5}{2}}} - \frac {\sqrt {2} {\left (9 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 7 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{16 \, d} \]

[In]

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

[Out]

1/16*(23*sqrt(2)*A*arctan(sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a^(5/2) - 32*A*arctan(1/2*sqrt(2)*sqrt(a
*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/a^(5/2) - sqrt(2)*(9*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A + 7*sqrt(a*t
an(1/2*d*x + 1/2*c)^2 - a)*A*a)/(a^4*tan(1/2*d*x + 1/2*c)^4))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {A+A \sec (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {A}{\cos \left (c+d\,x\right )}}{{\left (a-\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int((A + A/cos(c + d*x))/(a - a/cos(c + d*x))^(5/2),x)

[Out]

int((A + A/cos(c + d*x))/(a - a/cos(c + d*x))^(5/2), x)