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 \[ \frac{49 A \sin (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{59 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{4 a^{5/2} d}-\frac{167 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{2} a^{5/2} d}+\frac{23 A \sin (c+d x) \cos (c+d x)}{8 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{15 A \sin (c+d x) \cos (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}-\frac{A \sin (c+d x) \cos (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}} \]
[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]])
________________________________________________________________________________________
Rubi [A] time = 0.730891, antiderivative size = 236, normalized size of antiderivative = 1.,
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 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]
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 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 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 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
]
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*}
Mathematica [C] time = 6.81463, 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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Maple [B] time = 0.34, size = 1475, normalized size = 6.3 \begin{align*} \text{result too large to display} \end{align*}
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*2^(1/2)*(-1+cos(d*x+c))^6*(-1322*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+24780*arc
tan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-7014*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^2*2
^(1/2)+5010*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(d*x+c)^2*2^(1/2)-2505*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2
)*cos(d*x+c)*2^(1/2)-7014*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^3*2^(1/2)+11690*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(3/2)*cos(d*x+c)^3*2^(1/2)-35070*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3*2^(1/2
)+5010*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(d*x+c)^3*2^(1/2)+3507*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*2^(
1/2)-49560*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2+5145*2^(1/2)*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(1/2)+17535*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+3507*(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(5/2)*cos(d*x+c)*2^(1/2)-11633*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^5*2^(1/2)+11690*(-2*cos
(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^2*2^(1/2)+15573*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^4*2^(
1/2)-35070*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*2^(1/2)+1575*2^(1/2)*(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(9/2)+24780*cos(d*x+c)^4*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-5845*cos(d*x+c)
*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)-12768*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+
1015*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+17535*cos(d*x+c)*2^(1/2)*arctan(1/(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(1/2))+24780*cos(d*x+c)^5*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-1995*2^(1/2
)*cos(d*x+c)^5*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)-2505*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(
7/2)+3507*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)-5845*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(3/2)+17535*2^(1/2)*cos(d*x+c)^5*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-5845*2^(1/2)*(-
2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+24780*cos(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2
505*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*2^(1/2)-49560*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
)*cos(d*x+c)^3+4305*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)+3507*2^(1/2)*cos(d*x+c)^4*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(5/2)-5845*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+17535*2^(1/2)*cos(
d*x+c)^4*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+420*2^(1/2)*cos(d*x+c)^7*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2)+3570*2^(1/2)*cos(d*x+c)^6*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-6405*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(9/2)-5670*2^(1/2)*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)-2505*2^(1/2)*cos(d*x+c)
^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)+1470*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2))/(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(5/2)/(a*(-1+cos(d*x+c))/cos(d*x+c))^(5/2)/sin(d*x+c)^11
________________________________________________________________________________________
Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out
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Fricas [A] time = 0.574989, size = 1671, normalized size = 7.08 \begin{align*} \text{result too large to display} \end{align*}
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(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
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]
Timed out
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Giac [A] time = 3.15859, size = 409, normalized size = 1.73 \begin{align*} -\frac{A{\left (\frac{167 \, \sqrt{2} \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 \, \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}} + 315 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{5}{2}} a + 444 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} a^{2} + 196 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 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 )}\right )}}{16 \, d} \end{align*}
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*A*(167*sqrt(2)*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*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)*(69*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(7/2) +
315*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(5/2)*a + 444*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*a^2 + 196*sqrt(a*tan(1/2
*d*x + 1/2*c)^2 - 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