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

Optimal. Leaf size=236 \[ \frac {59 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac {167 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}+\frac {49 A \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}+\frac {23 A \sin (c+d x) \cos (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {15 A \sin (c+d x) \cos (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}-\frac {A \sin (c+d x) \cos (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}} \]

[Out]

59/4*A*arctan(a^(1/2)*tan(d*x+c)/(a-a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/2*A*cos(d*x+c)*sin(d*x+c)/d/(a-a*sec(d*x+
c))^(5/2)-15/8*A*cos(d*x+c)*sin(d*x+c)/a/d/(a-a*sec(d*x+c))^(3/2)-167/16*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)+49/8*A*sin(d*x+c)/a^2/d/(a-a*sec(d*x+c))^(1/2)+23/8*A*cos(d*x+c)*
sin(d*x+c)/a^2/d/(a-a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.73, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {4020, 4022, 3920, 3774, 203, 3795} \[ \frac {49 A \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}+\frac {59 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac {167 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a-a \sec (c+d x)}}\right )}{8 \sqrt {2} a^{5/2} d}+\frac {23 A \sin (c+d x) \cos (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {15 A \sin (c+d x) \cos (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}-\frac {A \sin (c+d x) \cos (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(59*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(4*a^(5/2)*d) - (167*A*ArcTan[(Sqrt[a]*Tan[c +
d*x])/(Sqrt[2]*Sqrt[a - a*Sec[c + d*x]])])/(8*Sqrt[2]*a^(5/2)*d) - (A*Cos[c + d*x]*Sin[c + d*x])/(2*d*(a - a*S
ec[c + d*x])^(5/2)) - (15*A*Cos[c + d*x]*Sin[c + d*x])/(8*a*d*(a - a*Sec[c + d*x])^(3/2)) + (49*A*Sin[c + d*x]
)/(8*a^2*d*Sqrt[a - a*Sec[c + d*x]]) + (23*A*Cos[c + d*x]*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 3774

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

Rule 3795

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2/f, Subst[Int[1/(2
*a + x^2), x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0
]

Rule 3920

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*
b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 4022

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1
/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*B*n - A*b*(m + n + 1)*Csc[e + f*x
], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx &=-\frac {A \cos (c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}+\frac {\int \frac {\cos ^2(c+d x) (8 a A+7 a A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac {A \cos (c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {15 A \cos (c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos ^2(c+d x) \left (46 a^2 A+\frac {75}{2} a^2 A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac {A \cos (c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {15 A \cos (c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac {23 A \cos (c+d x) \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}-\frac {\int \frac {\cos (c+d x) \left (-98 a^3 A-69 a^3 A \sec (c+d x)\right )}{\sqrt {a-a \sec (c+d x)}} \, dx}{16 a^5}\\ &=-\frac {A \cos (c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {15 A \cos (c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac {49 A \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}+\frac {23 A \cos (c+d x) \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}+\frac {\int \frac {118 a^4 A+49 a^4 A \sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{16 a^6}\\ &=-\frac {A \cos (c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {15 A \cos (c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac {49 A \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}+\frac {23 A \cos (c+d x) \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}+\frac {(59 A) \int \sqrt {a-a \sec (c+d x)} \, dx}{8 a^3}+\frac {(167 A) \int \frac {\sec (c+d x)}{\sqrt {a-a \sec (c+d x)}} \, dx}{16 a^2}\\ &=-\frac {A \cos (c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {15 A \cos (c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac {49 A \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}+\frac {23 A \cos (c+d x) \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}+\frac {(59 A) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^2 d}-\frac {(167 A) \operatorname {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,\frac {a \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{8 a^2 d}\\ &=\frac {59 A \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac {167 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 {A \cos (c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac {15 A \cos (c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac {49 A \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}+\frac {23 A \cos (c+d x) \sin (c+d x)}{8 a^2 d \sqrt {a-a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 6.82, size = 458, normalized size = 1.94 \[ A \left (\frac {\sin ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^3(c+d x) \left (\frac {12 \sin \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )}{d}-\frac {14 \sin \left (\frac {3 c}{2}\right ) \sin \left (\frac {3 d x}{2}\right )}{d}-\frac {\sin \left (\frac {5 c}{2}\right ) \sin \left (\frac {5 d x}{2}\right )}{d}-\frac {12 \cos \left (\frac {c}{2}\right ) \cos \left (\frac {d x}{2}\right )}{d}+\frac {14 \cos \left (\frac {3 c}{2}\right ) \cos \left (\frac {3 d x}{2}\right )}{d}+\frac {\cos \left (\frac {5 c}{2}\right ) \cos \left (\frac {5 d x}{2}\right )}{d}-\frac {\cot \left (\frac {c}{2}\right ) \csc ^3\left (\frac {c}{2}+\frac {d x}{2}\right )}{d}+\frac {31 \cot \left (\frac {c}{2}\right ) \csc \left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}+\frac {\csc \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \csc ^4\left (\frac {c}{2}+\frac {d x}{2}\right )}{d}-\frac {31 \csc \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right ) \csc ^2\left (\frac {c}{2}+\frac {d x}{2}\right )}{2 d}\right )}{(a-a \sec (c+d x))^{5/2}}+\frac {e^{-\frac {1}{2} i (c+d x)} \sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}} \sin ^5\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec ^{\frac {5}{2}}(c+d x) \left (59 \sinh ^{-1}\left (e^{i (c+d x)}\right )-\frac {167 \tanh ^{-1}\left (\frac {1+e^{i (c+d x)}}{\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}}\right )}{\sqrt {2}}+59 \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right )}{\sqrt {2} d (a-a \sec (c+d x))^{5/2}}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(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))]*(59*ArcSinh[E^(I*(c + d*x))]
- (167*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])/Sqrt[2] + 59*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*((-12*Cos[c/2]*Cos[(d*x)/2])/d + (14*Cos[(3*c)/2]*Cos[(3*d*x)/2])/d + (Cos[(5*c
)/2]*Cos[(5*d*x)/2])/d + (31*Cot[c/2]*Csc[c/2 + (d*x)/2])/(2*d) - (Cot[c/2]*Csc[c/2 + (d*x)/2]^3)/d - (31*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 + (12*Sin[c/2]*
Sin[(d*x)/2])/d - (14*Sin[(3*c)/2]*Sin[(3*d*x)/2])/d - (Sin[(5*c)/2]*Sin[(5*d*x)/2])/d)*Sin[c/2 + (d*x)/2]^5)/
(a - a*Sec[c + d*x])^(5/2))

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fricas [A]  time = 0.49, size = 634, normalized size = 2.69 \[ \left [-\frac {167 \, \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 ) + 236 \, {\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 (4 \, A \cos \left (d x + c\right )^{5} + 22 \, A \cos \left (d x + c\right )^{4} - 57 \, A \cos \left (d x + c\right )^{3} - 26 \, A \cos \left (d x + c\right )^{2} + 49 \, 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 {167 \, \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 ) - 236 \, {\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 (4 \, A \cos \left (d x + c\right )^{5} + 22 \, A \cos \left (d x + c\right )^{4} - 57 \, A \cos \left (d x + c\right )^{3} - 26 \, A \cos \left (d x + c\right )^{2} + 49 \, 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(cos(d*x+c)^2*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/32*(167*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) + 236*(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))/si
n(d*x + c))*sin(d*x + c) + 4*(4*A*cos(d*x + c)^5 + 22*A*cos(d*x + c)^4 - 57*A*cos(d*x + c)^3 - 26*A*cos(d*x +
c)^2 + 49*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*(167*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) - 236*(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*(4*A*cos(d*x + c)^5 + 22*A*cos(d*x + c)^4 - 57*A*cos(d*x + c)^3 - 26*A*cos(d*x + c)
^2 + 49*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 = 2.92, size = 308, normalized size = 1.31 \[ -\frac {\frac {167 \, \sqrt {2} A \arctan \left (\frac {\sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {236 \, A \arctan \left (\frac {\sqrt {2} \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a}}{2 \, \sqrt {a}}\right )}{a^{\frac {5}{2}} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} - \frac {\sqrt {2} {\left (69 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {7}{2}} A + 315 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {5}{2}} A a + 444 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{\frac {3}{2}} A a^{2} + 196 \, \sqrt {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a} A a^{3}\right )}}{{\left ({\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} + 3 \, {\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )} a + 2 \, a^{2}\right )}^{2} a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right ) \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(167*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))) - 236*A*arctan(1/2*sqrt(2)*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))) - sqrt(2)*(69*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(7/2)*
A + 315*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(5/2)*A*a + 444*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*A*a^2 + 196*sqrt(a
*tan(1/2*d*x + 1/2*c)^2 - a)*A*a^3)/(((a*tan(1/2*d*x + 1/2*c)^2 - a)^2 + 3*(a*tan(1/2*d*x + 1/2*c)^2 - a)*a +
2*a^2)^2*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))))/d

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maple [B]  time = 1.89, size = 1475, normalized size = 6.25 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/420*A/d*(-1+cos(d*x+c))^6*(5845*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)*2^(1/2)-17535*2^(1/2)*arcta
n(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^5+1995*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*cos(d
*x+c)^5+2505*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)^5-3507*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(5/2)*cos(d*x+c)^5+2505*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)^4-4305*2^(1/2)*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(9/2)*cos(d*x+c)-3507*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^4+5845*2^(1
/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^4-17535*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2))*cos(d*x+c)^4-420*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^7-3570*2^(1/2)*(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2)*cos(d*x+c)^6-24780*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))-5010*2^(1/2)*(
-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)^3-5010*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)^
2+2505*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*cos(d*x+c)+7014*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/
2)*cos(d*x+c)^3-11690*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^3+35070*2^(1/2)*arctan(1/(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^3-5145*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)+7014*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)^2*2^(1/2)-17535*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*2^(1/2)+58
45*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^5+1322*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x
+c)^3*2^(1/2)+12768*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2*2^(1/2)-17535*arctan(1/(-2*cos(d*x+c)/(1
+cos(d*x+c)))^(1/2))*cos(d*x+c)*2^(1/2)-1015*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)*2^(1/2)-24780*arc
tan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^5+6405*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(9/2)*cos(d*x+c)^4+5670*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*cos(d*x+c)^3-1470*2^(1/2)*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(9/2)*cos(d*x+c)^2-3507*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*cos(d*x+c)*2^(1/2)+11633*(-2*cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^5*2^(1/2)-11690*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)^2*2^(
1/2)-15573*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^4*2^(1/2)+35070*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(1/2))*cos(d*x+c)^2*2^(1/2)-1575*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)-24780*arctan(1/2*(-2*cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^4-24780*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)
)*cos(d*x+c)-3507*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*2^(1/2)+49560*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*2^(1/2))*cos(d*x+c)^2+5845*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)+49560*arctan(1/2*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^3+2505*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2))/(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(5/2)/(a*(-1+cos(d*x+c))/cos(d*x+c))^(5/2)/sin(d*x+c)^11*2^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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