3.1167 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=315 \[ \frac{(157 A+45 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{80 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(787 A+195 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{240 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(2671 A+735 C) \sin (c+d x)}{240 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{(283 A+75 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(21 A+5 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

-((283*A + 75*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A + C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(4*d*(a + a
*Sec[c + d*x])^(5/2)) - ((21*A + 5*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) + (
(2671*A + 735*C)*Sin[c + d*x])/(240*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((787*A + 195*C)*Sqrt
[Cos[c + d*x]]*Sin[c + d*x])/(240*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + ((157*A + 45*C)*Cos[c + d*x]^(3/2)*Sin[c +
d*x])/(80*a^2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 1.07039, antiderivative size = 315, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.189, Rules used = {4265, 4085, 4020, 4022, 4013, 3808, 206} \[ \frac{(157 A+45 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{80 a^2 d \sqrt{a \sec (c+d x)+a}}-\frac{(787 A+195 C) \sin (c+d x) \sqrt{\cos (c+d x)}}{240 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(2671 A+735 C) \sin (c+d x)}{240 a^2 d \sqrt{\cos (c+d x)} \sqrt{a \sec (c+d x)+a}}-\frac{(283 A+75 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(21 A+5 C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((283*A + 75*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Co
s[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A + C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(4*d*(a + a
*Sec[c + d*x])^(5/2)) - ((21*A + 5*C)*Cos[c + d*x]^(3/2)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) + (
(2671*A + 735*C)*Sin[c + d*x])/(240*a^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) - ((787*A + 195*C)*Sqrt
[Cos[c + d*x]]*Sin[c + d*x])/(240*a^2*d*Sqrt[a + a*Sec[c + d*x]]) + ((157*A + 45*C)*Cos[c + d*x]^(3/2)*Sin[c +
d*x])/(80*a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 4265

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 4085

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

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 4013

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[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 3808

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

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{A+C \sec ^2(c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{2} a (13 A+5 C)+4 a A \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(21 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{4} a^2 (157 A+45 C)+\frac{3}{2} a^2 (21 A+5 C) \sec (c+d x)}{\sec ^{\frac{5}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(21 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(157 A+45 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{8} a^3 (787 A+195 C)-\frac{1}{2} a^3 (157 A+45 C) \sec (c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}} \, dx}{20 a^5}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(21 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac{(787 A+195 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(157 A+45 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{16} a^4 (2671 A+735 C)+\frac{1}{8} a^4 (787 A+195 C) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)}} \, dx}{30 a^6}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(21 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(2671 A+735 C) \sin (c+d x)}{240 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(787 A+195 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(157 A+45 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{\left ((283 A+75 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(21 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(2671 A+735 C) \sin (c+d x)}{240 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(787 A+195 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(157 A+45 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\left ((283 A+75 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{16 a^2 d}\\ &=-\frac{(283 A+75 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{16 \sqrt{2} a^{5/2} d}-\frac{(A+C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(21 A+5 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(2671 A+735 C) \sin (c+d x)}{240 a^2 d \sqrt{\cos (c+d x)} \sqrt{a+a \sec (c+d x)}}-\frac{(787 A+195 C) \sqrt{\cos (c+d x)} \sin (c+d x)}{240 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(157 A+45 C) \cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{80 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.97109, size = 150, normalized size = 0.48 \[ \frac{\sec \left (\frac{1}{2} (c+d x)\right ) \left (4 \sin \left (\frac{1}{2} (c+d x)\right ) (5 (887 A+255 C) \cos (c+d x)+16 (52 A+15 C) \cos (2 (c+d x))-40 A \cos (3 (c+d x))+12 A \cos (4 (c+d x))+3491 A+975 C)-120 (283 A+75 C) \cos ^4\left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{960 a d \cos ^{\frac{3}{2}}(c+d x) (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[(c + d*x)/2]*(-120*(283*A + 75*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^4 + 4*(3491*A + 975*C + 5*(8
87*A + 255*C)*Cos[c + d*x] + 16*(52*A + 15*C)*Cos[2*(c + d*x)] - 40*A*Cos[3*(c + d*x)] + 12*A*Cos[4*(c + d*x)]
)*Sin[(c + d*x)/2]))/(960*a*d*Cos[c + d*x]^(3/2)*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [A]  time = 0.39, size = 450, normalized size = 1.4 \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)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

-1/480/d*cos(d*x+c)^(1/2)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(192*A*cos(d*x+c)^5-4245*A*sin
(d*x+c)*cos(d*x+c)^2*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)-1125*C*cos(d*x
+c)^2*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)-512*A*cos(d*x+c)^4
-8490*A*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)-2250*
C*cos(d*x+c)*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)+3456*A*cos(
d*x+c)^3-4245*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*(-2/(cos(d*x+c)+1))^(1/2)*A*sin(d*x+c)+960*C*co
s(d*x+c)^3-1125*C*(-2/(cos(d*x+c)+1))^(1/2)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+5974*A
*cos(d*x+c)^2+1590*C*cos(d*x+c)^2-3768*A*cos(d*x+c)-1080*C*cos(d*x+c)-5342*A-1470*C)/a^3/sin(d*x+c)^5

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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)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.580757, size = 1534, normalized size = 4.87 \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)^(5/2)*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/960*(15*sqrt(2)*((283*A + 75*C)*cos(d*x + c)^3 + 3*(283*A + 75*C)*cos(d*x + c)^2 + 3*(283*A + 75*C)*cos(d*x
+ c) + 283*A + 75*C)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c
))*sqrt(cos(d*x + c))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(96*A*
cos(d*x + c)^4 - 160*A*cos(d*x + c)^3 + 32*(49*A + 15*C)*cos(d*x + c)^2 + 5*(911*A + 255*C)*cos(d*x + c) + 267
1*A + 735*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 +
3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/480*(15*sqrt(2)*((283*A + 75*C)*cos(d*x + c)^3 + 3*(
283*A + 75*C)*cos(d*x + c)^2 + 3*(283*A + 75*C)*cos(d*x + c) + 283*A + 75*C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*
sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) + 2*(96*A*cos(d*x + c)^4 - 160*A*
cos(d*x + c)^3 + 32*(49*A + 15*C)*cos(d*x + c)^2 + 5*(911*A + 255*C)*cos(d*x + c) + 2671*A + 735*C)*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2
+ 3*a^3*d*cos(d*x + c) + a^3*d)]

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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)**(5/2)*(A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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