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

Optimal. Leaf size=217 \[ \frac{(35 A-11 B+3 C) \sin (c+d x)}{16 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(115 A-43 B+3 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(5 A-2 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(15 A-7 B-C) \sin (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[Out]

-(((5*A - 2*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d)) + ((115*A - 43*B + 3*C)*A
rcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sin[c
+ d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((15*A - 7*B - C)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))
+ ((35*A - 11*B + 3*C)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Sec[c + d*x]])

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Rubi [A]  time = 0.610402, antiderivative size = 217, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.171, Rules used = {4084, 4020, 4022, 3920, 3774, 203, 3795} \[ \frac{(35 A-11 B+3 C) \sin (c+d x)}{16 a^2 d \sqrt{a \sec (c+d x)+a}}+\frac{(115 A-43 B+3 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(5 A-2 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{5/2} d}-\frac{(15 A-7 B-C) \sin (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}-\frac{(A-B+C) \sin (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((5*A - 2*B)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(5/2)*d)) + ((115*A - 43*B + 3*C)*A
rcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B + C)*Sin[c
+ d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) - ((15*A - 7*B - C)*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2))
+ ((35*A - 11*B + 3*C)*Sin[c + d*x])/(16*a^2*d*Sqrt[a + a*Sec[c + d*x]])

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, 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 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 (c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{5/2}} \, dx &=-\frac{(A-B+C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac{\int \frac{\cos (c+d x) \left (a (5 A-B+C)-\frac{1}{2} a (5 A-5 B-3 C) \sec (c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\cos (c+d x) \left (\frac{1}{2} a^2 (35 A-11 B+3 C)-\frac{3}{4} a^2 (15 A-7 B-C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(35 A-11 B+3 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{-4 a^3 (5 A-2 B)+\frac{1}{4} a^3 (35 A-11 B+3 C) \sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{8 a^5}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(35 A-11 B+3 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}-\frac{(5 A-2 B) \int \sqrt{a+a \sec (c+d x)} \, dx}{2 a^3}+\frac{(115 A-43 B+3 C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac{(A-B+C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(35 A-11 B+3 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}+\frac{(5 A-2 B) \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 )}{a^2 d}-\frac{(115 A-43 B+3 C) \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 )}{16 a^2 d}\\ &=-\frac{(5 A-2 B) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{5/2} d}+\frac{(115 A-43 B+3 C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{16 \sqrt{2} a^{5/2} d}-\frac{(A-B+C) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac{(15 A-7 B-C) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac{(35 A-11 B+3 C) \sin (c+d x)}{16 a^2 d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 5.28304, size = 181, normalized size = 0.83 \[ \frac{-2 \tan ^3\left (\frac{1}{2} (c+d x)\right ) ((55 A-15 B+7 C) \cos (c+d x)+8 A \cos (2 (c+d x))+43 A-11 B+3 C)-\sqrt{2} (115 A-43 B+3 C) \sin (c+d x) \sqrt{\sec (c+d x)-1} \tan ^{-1}\left (\frac{\sqrt{\sec (c+d x)-1}}{\sqrt{2}}\right )+32 (5 A-2 B) \sin (c+d x) \sqrt{\sec (c+d x)-1} \tan ^{-1}\left (\sqrt{\sec (c+d x)-1}\right )}{32 a^2 d (\cos (c+d x)-1) \sqrt{a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(32*(5*A - 2*B)*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Sqrt[-1 + Sec[c + d*x]]*Sin[c + d*x] - Sqrt[2]*(115*A - 43*B +
3*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Sqrt[-1 + Sec[c + d*x]]*Sin[c + d*x] - 2*(43*A - 11*B + 3*C + (5
5*A - 15*B + 7*C)*Cos[c + d*x] + 8*A*Cos[2*(c + d*x)])*Tan[(c + d*x)/2]^3)/(32*a^2*d*(-1 + Cos[c + d*x])*Sqrt[
a*(1 + Sec[c + d*x])])

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Maple [B]  time = 0.39, size = 1338, normalized size = 6.2 \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)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)

[Out]

1/32/d/a^3*(-1+cos(d*x+c))^2*(115*A*cos(d*x+c)^2*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x
+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-43*B*cos(d*x+c)^2*sin(d*x+c)*ln(-(-(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+3*C*cos
(d*x+c)^2*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)+230*A*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+
cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-86*B*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+115*A*ln(-(-(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
sin(d*x+c)-43*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+3*C*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(
d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+30*B*cos(d*x+c)^3+70*A*cos(d*x+c)+160*A*sin(d*x+c)*(-2
*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+
c))*2^(1/2)*cos(d*x+c)+80*A*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)^2*2^(1/2)-32*A*cos(d*x+c)^4-14*C*cos(d*x+c)^3+8*C*cos
(d*x+c)^2+6*C*cos(d*x+c)+6*C*cos(d*x+c)*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d
*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-78*A*cos(d*x+c)^3-8*B*cos(d*x+c)^2-64*B*cos(d*x+c)*2
^(1/2)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*sin(d*x+c)/cos(d*x+c))+80*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)-32*B*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*2^(1/2)+40*A*cos(d
*x+c)^2-32*B*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d
*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-22*B*cos(d*x+c))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/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)*(A+B*sec(d*x+c)+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 = 129.546, size = 2043, normalized size = 9.41 \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)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/64*(sqrt(2)*((115*A - 43*B + 3*C)*cos(d*x + c)^3 + 3*(115*A - 43*B + 3*C)*cos(d*x + c)^2 + 3*(115*A - 43*B
+ 3*C)*cos(d*x + c) + 115*A - 43*B + 3*C)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x
+ c))*cos(d*x + c)*sin(d*x + c) + 3*a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c)
+ 1)) - 32*((5*A - 2*B)*cos(d*x + c)^3 + 3*(5*A - 2*B)*cos(d*x + c)^2 + 3*(5*A - 2*B)*cos(d*x + c) + 5*A - 2*B
)*sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x +
c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 4*(16*A*cos(d*x + c)^3 + (55*A - 15*B + 7*C)*cos(d*x + c)^2 +
(35*A - 11*B + 3*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/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), -1/32*(sqrt(2)*((115*A - 43*B + 3*C)*cos(d*x + c)^3
+ 3*(115*A - 43*B + 3*C)*cos(d*x + c)^2 + 3*(115*A - 43*B + 3*C)*cos(d*x + c) + 115*A - 43*B + 3*C)*sqrt(a)*ar
ctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 32*((5*A - 2*B)*co
s(d*x + c)^3 + 3*(5*A - 2*B)*cos(d*x + c)^2 + 3*(5*A - 2*B)*cos(d*x + c) + 5*A - 2*B)*sqrt(a)*arctan(sqrt((a*c
os(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*(16*A*cos(d*x + c)^3 + (55*A - 15*B +
7*C)*cos(d*x + c)^2 + (35*A - 11*B + 3*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/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)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError