3.1172 \(\int \frac{A+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=277 \[ \frac{(3 A+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{(3 A+115 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{5 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(A-15 C) \sin (c+d x)}{16 a d \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

[Out]

(-5*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(a^(5/2)
*d) + ((3*A + 115*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqr
t[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A + C)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(7/2)*(a
+ a*Sec[c + d*x])^(5/2)) + ((A - 15*C)*Sin[c + d*x])/(16*a*d*Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)) +
((3*A + 35*C)*Sin[c + d*x])/(16*a^2*d*Cos[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.919193, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.243, Rules used = {4265, 4085, 4019, 4021, 4023, 3808, 206, 3801, 215} \[ \frac{(3 A+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a \sec (c+d x)+a}}+\frac{(3 A+115 C) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{5 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)} \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}+\frac{(A-15 C) \sin (c+d x)}{16 a d \cos ^{\frac{5}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}-\frac{(A+C) \sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*C*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]]*Sqrt[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(a^(5/2)
*d) + ((3*A + 115*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqr
t[Cos[c + d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A + C)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(7/2)*(a
+ a*Sec[c + d*x])^(5/2)) + ((A - 15*C)*Sin[c + d*x])/(16*a*d*Cos[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)) +
((3*A + 35*C)*Sin[c + d*x])/(16*a^2*d*Cos[c + d*x]^(3/2)*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 4019

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

Rule 4021

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

Rule 4023

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

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])

Rule 3801

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

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \frac{A+C \sec ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{5}{2}}(c+d x) \left (-\frac{1}{2} a (3 A-5 C)-a (A+5 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac{(A-15 C) \sin (c+d x)}{16 a d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (-\frac{3}{4} a^2 (A-15 C)-\frac{1}{2} a^2 (3 A+35 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac{(A-15 C) \sin (c+d x)}{16 a d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(3 A+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)} \left (-\frac{1}{4} a^3 (3 A+35 C)+20 a^3 C \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^5}\\ &=-\frac{(A+C) \sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac{(A-15 C) \sin (c+d x)}{16 a d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(3 A+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}-\frac{\left (5 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx}{2 a^3}+\frac{\left ((3 A+115 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) \sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac{(A-15 C) \sin (c+d x)}{16 a d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(3 A+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}+\frac{\left (5 C \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{a}}} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^3 d}-\frac{\left ((3 A+115 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{5 C \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}}{a^{5/2} d}+\frac{(3 A+115 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) \sin (c+d x)}{4 d \cos ^{\frac{7}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac{(A-15 C) \sin (c+d x)}{16 a d \cos ^{\frac{5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac{(3 A+35 C) \sin (c+d x)}{16 a^2 d \cos ^{\frac{3}{2}}(c+d x) \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.57722, size = 187, normalized size = 0.68 \[ \frac{\cos ^5\left (\frac{1}{2} (c+d x)\right ) \left (A+C \sec ^2(c+d x)\right ) \left ((6 A+230 C) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+\frac{1}{2} \tan \left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sec ^3\left (\frac{1}{2} (c+d x)\right ) (2 (7 A+55 C) \cos (c+d x)+(3 A+35 C) \cos (2 (c+d x))+3 A+67 C)-160 \sqrt{2} C \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{4 d \sqrt{\cos (c+d x)} (a (\sec (c+d x)+1))^{5/2} (A \cos (2 (c+d x))+A+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^5*(A + C*Sec[c + d*x]^2)*((6*A + 230*C)*ArcTanh[Sin[(c + d*x)/2]] - 160*Sqrt[2]*C*ArcTanh[Sq
rt[2]*Sin[(c + d*x)/2]] + ((3*A + 67*C + 2*(7*A + 55*C)*Cos[c + d*x] + (3*A + 35*C)*Cos[2*(c + d*x)])*Sec[(c +
d*x)/2]^3*Sec[c + d*x]*Tan[(c + d*x)/2])/2))/(4*d*Sqrt[Cos[c + d*x]]*(A + 2*C + A*Cos[2*(c + d*x)])*(a*(1 + S
ec[c + d*x]))^(5/2))

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Maple [B]  time = 0.305, size = 605, normalized size = 2.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/d*(-1+cos(d*x+c))^2*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)*(40*C*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2
)*(cos(d*x+c)+1-sin(d*x+c)))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)-40*C*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)
*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)+3*A*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2
))*cos(d*x+c)^2*sin(d*x+c)-3*A*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3+40*C*sin(d*x+c)*2^(1/2)*cos(d*x+c)*arcta
n(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))-40*C*sin(d*x+c)*2^(1/2)*cos(d*x+c)*arctan(1
/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+115*C*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1)
)^(1/2))*cos(d*x+c)^2*sin(d*x+c)-35*C*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^3+3*A*sin(d*x+c)*cos(d*x+c)*arctan(
1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-4*A*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)+115*C*sin(d*x+c)*cos(d*x+
c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-20*C*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)+7*A*cos(d*x+c)
*(-2/(cos(d*x+c)+1))^(1/2)+39*C*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)+16*C*(-2/(cos(d*x+c)+1))^(1/2))/a^3/(-2/(
cos(d*x+c)+1))^(1/2)/sin(d*x+c)^5/cos(d*x+c)^(1/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: RuntimeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: RuntimeError

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Fricas [A]  time = 0.717105, size = 2107, normalized size = 7.61 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/64*(sqrt(2)*((3*A + 115*C)*cos(d*x + c)^4 + 3*(3*A + 115*C)*cos(d*x + c)^3 + 3*(3*A + 115*C)*cos(d*x + c)^2
+ (3*A + 115*C)*cos(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/co
s(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*((3*A + 35*C)*cos(d*x + c)^2 + (7*A + 55*C)*cos(d*x + c) + 16*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sq
rt(cos(d*x + c))*sin(d*x + c) + 80*(C*cos(d*x + c)^4 + 3*C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + C*cos(d*x + c
))*sqrt(a)*log((a*cos(d*x + c)^3 + 4*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*(cos(d*x + c) - 2)*sqrt(c
os(d*x + c))*sin(d*x + c) - 7*a*cos(d*x + c)^2 + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)))/(a^3*d*cos(d*x + c)^
4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3*d*cos(d*x + c)), -1/32*(sqrt(2)*((3*A + 115*C)*cos(d
*x + c)^4 + 3*(3*A + 115*C)*cos(d*x + c)^3 + 3*(3*A + 115*C)*cos(d*x + c)^2 + (3*A + 115*C)*cos(d*x + 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*
((3*A + 35*C)*cos(d*x + c)^2 + (7*A + 55*C)*cos(d*x + c) + 16*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(
cos(d*x + c))*sin(d*x + c) + 80*(C*cos(d*x + c)^4 + 3*C*cos(d*x + c)^3 + 3*C*cos(d*x + c)^2 + C*cos(d*x + c))*
sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x
+ c)^2 - a*cos(d*x + c) - 2*a)))/(a^3*d*cos(d*x + c)^4 + 3*a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + a^3
*d*cos(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((A+C*sec(d*x+c)**2)/cos(d*x+c)**(5/2)/(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{C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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