3.178 \(\int \frac{\cos ^3(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=280 \[ \frac{21 A \sin (c+d x)}{2 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{203 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a^{5/2} d}-\frac{287 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{2} a^{5/2} d}+\frac{77 A \sin (c+d x) \cos ^2(c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{119 A \sin (c+d x) \cos (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{19 A \sin (c+d x) \cos ^2(c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}-\frac{A \sin (c+d x) \cos ^2(c+d x)}{2 d (a-a \sec (c+d x))^{5/2}} \]
[Out]
(203*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(8*a^(5/2)*d) - (287*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]^2*Sin[c + d*x])/(2*d*(a -
a*Sec[c + d*x])^(5/2)) - (19*A*Cos[c + d*x]^2*Sin[c + d*x])/(8*a*d*(a - a*Sec[c + d*x])^(3/2)) + (21*A*Sin[c +
d*x])/(2*a^2*d*Sqrt[a - a*Sec[c + d*x]]) + (119*A*Cos[c + d*x]*Sin[c + d*x])/(24*a^2*d*Sqrt[a - a*Sec[c + d*x
]]) + (77*A*Cos[c + d*x]^2*Sin[c + d*x])/(24*a^2*d*Sqrt[a - a*Sec[c + d*x]])
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Rubi [A] time = 0.906291, antiderivative size = 280, normalized size of antiderivative = 1.,
number of steps used = 10, 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} \[ \frac{21 A \sin (c+d x)}{2 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{203 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a^{5/2} d}-\frac{287 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a-a \sec (c+d x)}}\right )}{8 \sqrt{2} a^{5/2} d}+\frac{77 A \sin (c+d x) \cos ^2(c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{119 A \sin (c+d x) \cos (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{19 A \sin (c+d x) \cos ^2(c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}-\frac{A \sin (c+d x) \cos ^2(c+d x)}{2 d (a-a \sec (c+d x))^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]^3*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(5/2),x]
[Out]
(203*A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a - a*Sec[c + d*x]]])/(8*a^(5/2)*d) - (287*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]^2*Sin[c + d*x])/(2*d*(a -
a*Sec[c + d*x])^(5/2)) - (19*A*Cos[c + d*x]^2*Sin[c + d*x])/(8*a*d*(a - a*Sec[c + d*x])^(3/2)) + (21*A*Sin[c +
d*x])/(2*a^2*d*Sqrt[a - a*Sec[c + d*x]]) + (119*A*Cos[c + d*x]*Sin[c + d*x])/(24*a^2*d*Sqrt[a - a*Sec[c + d*x
]]) + (77*A*Cos[c + d*x]^2*Sin[c + d*x])/(24*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 ^3(c+d x) (A+A \sec (c+d x))}{(a-a \sec (c+d x))^{5/2}} \, dx &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}+\frac{\int \frac{\cos ^3(c+d x) (10 a A+9 a A \sec (c+d x))}{(a-a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{19 A \cos ^2(c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{\int \frac{\cos ^3(c+d x) \left (77 a^2 A+\frac{133}{2} a^2 A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{19 A \cos ^2(c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{77 A \cos ^2(c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{\cos ^2(c+d x) \left (-238 a^3 A-\frac{385}{2} a^3 A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{24 a^5}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{19 A \cos ^2(c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{119 A \cos (c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{77 A \cos ^2(c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{\int \frac{\cos (c+d x) \left (504 a^4 A+357 a^4 A \sec (c+d x)\right )}{\sqrt{a-a \sec (c+d x)}} \, dx}{48 a^6}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{19 A \cos ^2(c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{21 A \sin (c+d x)}{2 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{119 A \cos (c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{77 A \cos ^2(c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}-\frac{\int \frac{-609 a^5 A-252 a^5 A \sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{48 a^7}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{19 A \cos ^2(c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{21 A \sin (c+d x)}{2 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{119 A \cos (c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{77 A \cos ^2(c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{(203 A) \int \sqrt{a-a \sec (c+d x)} \, dx}{16 a^3}+\frac{(287 A) \int \frac{\sec (c+d x)}{\sqrt{a-a \sec (c+d x)}} \, dx}{16 a^2}\\ &=-\frac{A \cos ^2(c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{19 A \cos ^2(c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{21 A \sin (c+d x)}{2 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{119 A \cos (c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{77 A \cos ^2(c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{(203 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 )}{8 a^2 d}-\frac{(287 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{203 A \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a-a \sec (c+d x)}}\right )}{8 a^{5/2} d}-\frac{287 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 ^2(c+d x) \sin (c+d x)}{2 d (a-a \sec (c+d x))^{5/2}}-\frac{19 A \cos ^2(c+d x) \sin (c+d x)}{8 a d (a-a \sec (c+d x))^{3/2}}+\frac{21 A \sin (c+d x)}{2 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{119 A \cos (c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}+\frac{77 A \cos ^2(c+d x) \sin (c+d x)}{24 a^2 d \sqrt{a-a \sec (c+d x)}}\\ \end{align*}
Mathematica [C] time = 6.8404, size = 514, normalized size = 1.84 \[ A \left (\frac{\sin ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^3(c+d x) \left (\frac{7 \sin \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right )}{6 d}-\frac{92 \sin \left (\frac{3 c}{2}\right ) \sin \left (\frac{3 d x}{2}\right )}{3 d}-\frac{7 \sin \left (\frac{5 c}{2}\right ) \sin \left (\frac{5 d x}{2}\right )}{2 d}-\frac{\sin \left (\frac{7 c}{2}\right ) \sin \left (\frac{7 d x}{2}\right )}{3 d}-\frac{7 \cos \left (\frac{c}{2}\right ) \cos \left (\frac{d x}{2}\right )}{6 d}+\frac{92 \cos \left (\frac{3 c}{2}\right ) \cos \left (\frac{3 d x}{2}\right )}{3 d}+\frac{7 \cos \left (\frac{5 c}{2}\right ) \cos \left (\frac{5 d x}{2}\right )}{2 d}+\frac{\cos \left (\frac{7 c}{2}\right ) \cos \left (\frac{7 d x}{2}\right )}{3 d}-\frac{\cot \left (\frac{c}{2}\right ) \csc ^3\left (\frac{c}{2}+\frac{d x}{2}\right )}{d}+\frac{39 \cot \left (\frac{c}{2}\right ) \csc \left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}+\frac{\csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc ^4\left (\frac{c}{2}+\frac{d x}{2}\right )}{d}-\frac{39 \csc \left (\frac{c}{2}\right ) \sin \left (\frac{d x}{2}\right ) \csc ^2\left (\frac{c}{2}+\frac{d x}{2}\right )}{2 d}\right )}{(a-a \sec (c+d x))^{5/2}}+\frac{7 e^{-\frac{1}{2} i (c+d x)} \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sin ^5\left (\frac{c}{2}+\frac{d x}{2}\right ) \sec ^{\frac{5}{2}}(c+d x) \left (29 \sinh ^{-1}\left (e^{i (c+d x)}\right )-41 \sqrt{2} \tanh ^{-1}\left (\frac{1+e^{i (c+d x)}}{\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}}\right )+29 \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{2 \sqrt{2} d (a-a \sec (c+d x))^{5/2}}\right ) \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]^3*(A + A*Sec[c + d*x]))/(a - a*Sec[c + d*x])^(5/2),x]
[Out]
A*((7*Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(29*ArcSinh[E^(I*(c + d*x)
)] - 41*Sqrt[2]*ArcTanh[(1 + E^(I*(c + d*x)))/(Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] + 29*ArcTanh[Sqrt[1 + E
^((2*I)*(c + d*x))]])*Sec[c + d*x]^(5/2)*Sin[c/2 + (d*x)/2]^5)/(2*Sqrt[2]*d*E^((I/2)*(c + d*x))*(a - a*Sec[c +
d*x])^(5/2)) + (Sec[c + d*x]^3*((-7*Cos[c/2]*Cos[(d*x)/2])/(6*d) + (92*Cos[(3*c)/2]*Cos[(3*d*x)/2])/(3*d) + (
7*Cos[(5*c)/2]*Cos[(5*d*x)/2])/(2*d) + (Cos[(7*c)/2]*Cos[(7*d*x)/2])/(3*d) + (39*Cot[c/2]*Csc[c/2 + (d*x)/2])/
(2*d) - (Cot[c/2]*Csc[c/2 + (d*x)/2]^3)/d - (39*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 + (7*Sin[c/2]*Sin[(d*x)/2])/(6*d) - (92*Sin[(3*c)/2]*Sin[(3*d*x)/2])/(3*d
) - (7*Sin[(5*c)/2]*Sin[(5*d*x)/2])/(2*d) - (Sin[(7*c)/2]*Sin[(7*d*x)/2])/(3*d))*Sin[c/2 + (d*x)/2]^5)/(a - a*
Sec[c + d*x])^(5/2))
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Maple [B] time = 0.44, size = 1964, normalized size = 7. \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)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x)
[Out]
-1/180*A/d*2^(1/2)*(-1+cos(d*x+c))^7*(-10335*cos(d*x+c)^3*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-945*2^(
1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)+18270*cos(d*x+c)^6*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1)
)^(1/2))+18270*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-2583*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5
/2)*cos(d*x+c)^2*2^(1/2)+1845*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(d*x+c)^2*2^(1/2)-3690*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(7/2)*cos(d*x+c)*2^(1/2)-10332*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)^3*2^(1/2)+17220*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^3*2^(1/2)-51660*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*
cos(d*x+c)^3*2^(1/2)+7380*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(d*x+c)^3*2^(1/2)+2583*(-2*cos(d*x+c)/(cos(d
*x+c)+1))^(5/2)*2^(1/2)-18270*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2+3780*2^(1/
2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+12915*2^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+5166*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*2^(1/2)+14285*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^5*2^
(1/2)+4305*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*cos(d*x+c)^2*2^(1/2)+6254*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
*cos(d*x+c)^4*2^(1/2)-12915*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*2^(1/2)+1435*2^(1/2)*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)-18270*cos(d*x+c)^4*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
)-8610*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)-8652*cos(d*x+c)^2*2^(1/2)*(-2*cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)+4515*cos(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+25830*cos(d*x+c)*2^(1/2)*arctan(1
/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+36540*cos(d*x+c)^5*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2))+2870*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)-3690*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(
cos(d*x+c)+1))^(7/2)+5166*2^(1/2)*cos(d*x+c)^5*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)-8610*2^(1/2)*cos(d*x+c)^5*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+25830*2^(1/2)*cos(d*x+c)^5*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))
+120*2^(1/2)*cos(d*x+c)^9*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+930*2^(1/2)*cos(d*x+c)^8*(-2*cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)+1125*2^(1/2)*cos(d*x+c)^6*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)+4680*2^(1/2)*cos(d*x+c)^5*(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(11/2)+1435*2^(1/2)*cos(d*x+c)^6*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)+6525*2^(1/2)*c
os(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)+1800*2^(1/2)*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11
/2)-1845*2^(1/2)*cos(d*x+c)^6*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)-3825*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(11/2)-3600*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)+2583*2^(1/2)*cos(d*x+c)^6*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)-4305*2^(1/2)*cos(d*x+c)^6*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)+12915*2^(1/
2)*cos(d*x+c)^6*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-4305*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/
2)+36540*cos(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-1845*(-2*cos(d*x+c)/(cos(d*x+c)+1
))^(7/2)*2^(1/2)-73080*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3+2870*2^(1/2)*cos(
d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)-2583*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)+430
5*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)-12915*2^(1/2)*cos(d*x+c)^4*arctan(1/(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2))+4215*2^(1/2)*cos(d*x+c)^7*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-15112*2^(1/2)*cos(d*x+c
)^6*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-1435*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)-5740*2
^(1/2)*cos(d*x+c)^3*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)+1845*2^(1/2)*cos(d*x+c)^4*(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(7/2)-1435*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2))/(-2*cos(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)^13
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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)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
Fricas [A] time = 0.612642, size = 1744, normalized size = 6.23 \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)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
[-1/96*(861*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) + 1218*(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))/s
in(d*x + c))*sin(d*x + c) + 4*(8*A*cos(d*x + c)^6 + 30*A*cos(d*x + c)^5 + 113*A*cos(d*x + c)^4 - 294*A*cos(d*x
+ c)^3 - 133*A*cos(d*x + c)^2 + 252*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)), 1/48*(861*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) - 1218*(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*(8*A*cos(d*x + c)^6 + 30*A*cos(d*x + c)^5 + 113*A*cos(d
*x + c)^4 - 294*A*cos(d*x + c)^3 - 133*A*cos(d*x + c)^2 + 252*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)**3*(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 = 3.04878, size = 459, normalized size = 1.64 \begin{align*} -\frac{A{\left (\frac{861 \, \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{1218 \, \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{2 \, \sqrt{2}{\left (129 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{5}{2}} + 560 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} a + 636 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a} a^{2}\right )}}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a\right )}^{3} 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 )} - \frac{3 \, \sqrt{2}{\left (33 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{\frac{3}{2}} + 31 \, \sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 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}}\right )}}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(A+A*sec(d*x+c))/(a-a*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
-1/48*A*(861*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))) - 1218*arctan(1/2*sqrt(2)*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)/sqrt(a))/(a^(5/2)*s
gn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - 2*sqrt(2)*(129*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(5/2
) + 560*(a*tan(1/2*d*x + 1/2*c)^2 - a)^(3/2)*a + 636*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - a)*a^2)/((a*tan(1/2*d*x +
1/2*c)^2 + a)^3*a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)*sgn(tan(1/2*d*x + 1/2*c))) - 3*sqrt(2)*(33*(a*tan(1/2*d*x
+ 1/2*c)^2 - a)^(3/2) + 31*sqrt(a*tan(1/2*d*x + 1/2*c)^2 - 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