3.2.77 \(\int \frac {\cos ^2(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx\) [177]

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

[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 = 0.47, 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 = {4105, 4107, 4005, 3859, 209, 3880} \begin {gather*} \frac {59 A \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a-a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac {167 A \text {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 {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}} \end {gather*}

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 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 3859

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 3880

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 4005

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 4105

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 4107

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) \text {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) \text {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] Result contains complex when optimal does not.
time = 2.08, size = 308, normalized size = 1.31 \begin {gather*} \frac {A \left (\sqrt {2} 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)}} \left (-118 i d x+118 \sinh ^{-1}\left (e^{i (c+d x)}\right )+167 \sqrt {2} \log \left (1-e^{i (c+d x)}\right )+118 \log \left (1+\sqrt {1+e^{2 i (c+d x)}}\right )-167 \sqrt {2} \log \left (1+e^{i (c+d x)}+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )+\frac {1}{4} \left (-19 \cos \left (\frac {1}{2} (c+d x)\right )+54 \cos \left (\frac {3}{2} (c+d x)\right )-62 \cos \left (\frac {5}{2} (c+d x)\right )+10 \cos \left (\frac {7}{2} (c+d x)\right )+\cos \left (\frac {9}{2} (c+d x)\right )\right ) \csc ^4\left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)}\right ) \sec ^{\frac {5}{2}}(c+d x) \sin ^5\left (\frac {1}{2} (c+d x)\right )}{4 d (a-a \sec (c+d x))^{5/2}} \end {gather*}

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[2]*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*((-118*I)*d*x + 118
*ArcSinh[E^(I*(c + d*x))] + 167*Sqrt[2]*Log[1 - E^(I*(c + d*x))] + 118*Log[1 + Sqrt[1 + E^((2*I)*(c + d*x))]]
- 167*Sqrt[2]*Log[1 + E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))/E^((I/2)*(c + d*x)) + ((-19*C
os[(c + d*x)/2] + 54*Cos[(3*(c + d*x))/2] - 62*Cos[(5*(c + d*x))/2] + 10*Cos[(7*(c + d*x))/2] + Cos[(9*(c + d*
x))/2])*Csc[(c + d*x)/2]^4*Sqrt[Sec[c + d*x]])/4)*Sec[c + d*x]^(5/2)*Sin[(c + d*x)/2]^5)/(4*d*(a - a*Sec[c + d
*x])^(5/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1474\) vs. \(2(201)=402\).
time = 9.09, size = 1475, normalized size = 6.25

method	result	size
default	\(\text {Expression too large to display}\)	\(1475\)

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,method=_RETURNVERBOSE)

[Out]

1/420*A/d*(-1+cos(d*x+c))^6*(17535*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+5145*(-2*cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)*2^(1/2)+24780*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))+5010*2^(1/2)*cos(
d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+5010*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-7
014*2^(1/2)*cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-2505*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(7/2)+11690*2^(1/2)*cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-35070*2^(1/2)*cos(d*x+c)^3*arctan
(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-5845*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)*cos(d*x+c)-1995*2^(
1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)-1322*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*cos(d
*x+c)^3-12768*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)*cos(d*x+c)^2+1015*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1
/2)*2^(1/2)*cos(d*x+c)+17535*2^(1/2)*cos(d*x+c)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-7014*2^(1/2)*co
s(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+420*2^(1/2)*cos(d*x+c)^7*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)-
2505*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+3507*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(5/2)-5845*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)+17535*2^(1/2)*cos(d*x+c)^5*arct
an(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))-49560*cos(d*x+c)^3*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*
2^(1/2))-2505*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)-11633*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)+3507*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)+15573*2^(1/2)*cos(d*x+c)^4*(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)+11690*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)-35070*arctan(1/(-2
*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*2^(1/2)*cos(d*x+c)^2+3507*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-4956
0*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^2+3570*2^(1/2)*cos(d*x+c)^6*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(1/2)-6405*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)-5670*2^(1/2)*cos(d*x+c
)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)-2505*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)+1470*2
^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)+4305*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c))
)^(9/2)+3507*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)-5845*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(3/2)+17535*2^(1/2)*cos(d*x+c)^4*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+1575*2^(1/2)
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)+24780*cos(d*x+c)^4*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/
2))-5845*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)+24780*cos(d*x+c)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(1/2)*2^(1/2))+24780*cos(d*x+c)^5*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)))/(-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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 1.61, size = 634, normalized size = 2.69 \begin {gather*} \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 ] \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 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 ) \end {gather*}

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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Giac [A]
time = 1.28, size = 224, normalized size = 0.95 \begin {gather*} \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}}} - \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}}} - \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}}}{16 \, d} \end {gather*}

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) - 236*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)*(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))/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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