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

Optimal. Leaf size=152 \[ \frac {2 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {23 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}-\frac {7 A \tan (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}-\frac {A \tan (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}} \]

[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)

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Rubi [A]  time = 0.21, antiderivative size = 185, 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 = {3904, 3887, 471, 527, 522, 203} \[ \frac {2 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {23 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}+\frac {7 A \sin (c+d x) \csc ^2\left (\frac {1}{2} (c+d x)\right )}{16 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {A \sin (c+d x) \cos (c+d x) \csc ^4\left (\frac {1}{2} (c+d x)\right )}{8 a^2 d \sqrt {a-a \sec (c+d x)}} \]

Antiderivative was successfully verified.

[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 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 471

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 522

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 527

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 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 (c+d x)}{(a-a \sec (c+d x))^{5/2}} \, dx &=-\left ((a A) \int \frac {\tan ^2(c+d x)}{(a-a \sec (c+d x))^{7/2}} \, dx\right )\\ &=\frac {(2 A) \operatorname {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 \operatorname {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 \operatorname {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) \operatorname {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) \operatorname {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 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {23 A \tan ^{-1}\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*}

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Mathematica [C]  time = 6.81, size = 387, normalized size = 2.55 \[ A \left (\frac {\sin ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^3(c+d x) \left (\frac {11 \sin \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}-\frac {11 \cos \left (\frac {c}{2}\right ) \cos \left (\frac {d x}{2}\right )}{d}-\frac {\cot \left (\frac {c}{2}\right ) \csc ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d}+\frac {15 \cot \left (\frac {c}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}+\frac {\csc \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \csc ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {15 \csc \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \csc ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}\right )}{(a-a \sec (c+d x))^{5/2}}+\frac {e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \sin ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^{\frac {5}{2}}(c+d x) \left (8 \sinh ^{-1}\left (e^{i (c+d x)}\right )-\frac {23 \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{\sqrt {2}}+8 \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{\sqrt {2} d (a-a \sec (c+d x))^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

A*((Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(8*ArcSinh[E^(I*(c + d*x))]
- (23*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[2] + 8*ArcTanh[Sqrt[1 + E^(
(2*I)*(c + d*x))]])*Sec[c + d*x]^(5/2)*Sin[c/2 + (d*x)/2]^5)/(Sqrt[2]*d*E^((I/2)*(c + d*x))*(a - a*Sec[c + d*x
])^(5/2)) + (Sec[c + d*x]^3*((-11*Cos[c/2]*Cos[(d*x)/2])/d + (15*Cot[c/2]*Csc[c/2 + (d*x)/2])/(2*d) - (Cot[c/2
]*Csc[c/2 + (d*x)/2]^3)/d - (15*Csc[c/2]*Csc[c/2 + (d*x)/2]^2*Sin[(d*x)/2])/(2*d) + (Csc[c/2]*Csc[c/2 + (d*x)/
2]^4*Sin[(d*x)/2])/d + (11*Sin[c/2]*Sin[(d*x)/2])/d)*Sin[c/2 + (d*x)/2]^5)/(a - a*Sec[c + d*x])^(5/2))

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fricas [B]  time = 0.47, size = 590, normalized size = 3.88 \[ \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 ] \]

Verification of antiderivative is not currently implemented for this CAS.

[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))]

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giac [A]  time = 1.37, size = 222, normalized size = 1.46 \[ -\frac {\frac {23 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {32 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (9 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A + 7 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a\right )}}{a^{4} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1
)*sgn(tan(1/2*d*x + 1/2*c))) - 32*A*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(5/2)*sg
n(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - sqrt(2)*(9*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A +
7*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*A*a)/(a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))*tan(
1/2*d*x + 1/2*c)^4))/d

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maple [B]  time = 1.55, size = 695, normalized size = 4.57 \[ \frac {A \left (-1+\cos \left (d x +c \right )\right )^{4} \left (-21 \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (d x +c \right )\right )-33 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-23 \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{3}\left (d x +c \right )\right )-3 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right ) \sqrt {2}+23 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+9 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sqrt {2}+5 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+69 \sqrt {2}\, \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{3}\left (d x +c \right )\right )+23 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right ) \sqrt {2}+96 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (\cos ^{3}\left (d x +c \right )\right )+11 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-69 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}-23 \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}-96 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-37 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \sqrt {2}-69 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \cos \left (d x +c \right ) \sqrt {2}-96 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \cos \left (d x +c \right )+21 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}+69 \arctan \left (\frac {1}{\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}}\right ) \sqrt {2}+96 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right )\right ) \sqrt {2}}{12 d \left (\frac {a \left (-1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )^{7} \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*A/d*(-1+cos(d*x+c))^4*(-21*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^3-33*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(5/2)*cos(d*x+c)^2*2^(1/2)-23*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^3-3*(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)*2^(1/2)+23*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2*2^(1/2)
+9*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)+5*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^3*2^(1/2)+69
*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^3+23*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*c
os(d*x+c)*2^(1/2)+96*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^3+11*(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*2^(1/2)-69*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2*2^(1/
2)-23*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)-96*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))
*cos(d*x+c)^2-37*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*2^(1/2)-69*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(1/2))*cos(d*x+c)*2^(1/2)-96*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)+21*(-2*
cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)+69*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*2^(1/2)+96*arctan(1
/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)))/(a*(-1+cos(d*x+c))/cos(d*x+c))^(5/2)/sin(d*x+c)^7/(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A \sec \left (d x + c\right ) + A}{{\left (-a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ A \left (\int \frac {\sec {\left (c + d x \right )}}{a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx + \int \frac {1}{a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec ^{2}{\left (c + d x \right )} - 2 a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a} \sec {\left (c + d x \right )} + a^{2} \sqrt {- a \sec {\left (c + d x \right )} + a}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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))

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