3.510 \(\int \frac{\sec ^4(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=254 \[ \frac{2 (21 A-3 B+19 C) \tan (c+d x) \sec ^2(c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (21 A-93 B+29 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 a d}+\frac{4 (147 A-111 B+143 C) \tan (c+d x)}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt{a \sec (c+d x)+a}} \]

[Out]

-((Sqrt[2]*(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (4*(1
47*A - 111*B + 143*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(21*A - 3*B + 19*C)*Sec[c + d*x]^2*T
an[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(9*B - C)*Sec[c + d*x]^3*Tan[c + d*x])/(63*d*Sqrt[a + a*Sec
[c + d*x]]) + (2*C*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) - (2*(21*A - 93*B + 29*C)*Sqrt[
a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*a*d)

________________________________________________________________________________________

Rubi [A]  time = 0.856623, antiderivative size = 254, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 43, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used = {4088, 4021, 4010, 4001, 3795, 203} \[ \frac{2 (21 A-3 B+19 C) \tan (c+d x) \sec ^2(c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}-\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}-\frac{2 (21 A-93 B+29 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{315 a d}+\frac{4 (147 A-111 B+143 C) \tan (c+d x)}{315 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (9 B-C) \tan (c+d x) \sec ^3(c+d x)}{63 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^4(c+d x)}{9 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

-((Sqrt[2]*(A - B + C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d)) + (4*(1
47*A - 111*B + 143*C)*Tan[c + d*x])/(315*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(21*A - 3*B + 19*C)*Sec[c + d*x]^2*T
an[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(9*B - C)*Sec[c + d*x]^3*Tan[c + d*x])/(63*d*Sqrt[a + a*Sec
[c + d*x]]) + (2*C*Sec[c + d*x]^4*Tan[c + d*x])/(9*d*Sqrt[a + a*Sec[c + d*x]]) - (2*(21*A - 93*B + 29*C)*Sqrt[
a + a*Sec[c + d*x]]*Tan[c + d*x])/(315*a*d)

Rule 4088

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[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d
*Csc[e + f*x])^n)/(f*(m + n + 1)), x] + Dist[1/(b*(m + n + 1)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n*
Simp[A*b*(m + n + 1) + b*C*n + (a*C*m + b*B*(m + n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A,
B, C, m, n}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)] &&  !LtQ[n, -2^(-1)] && NeQ[m + n + 1, 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 4010

Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_
)), x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), I
nt[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*B*(m + 1) + (A*b*(m + 2) - a*B)*Csc[e + f*x], x], x], x] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4001

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

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
]

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])

Rubi steps

\begin{align*} \int \frac{\sec ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx &=\frac{2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{\sec ^4(c+d x) \left (\frac{1}{2} a (9 A+8 C)+\frac{1}{2} a (9 B-C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{9 a}\\ &=\frac{2 (9 B-C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{4 \int \frac{\sec ^3(c+d x) \left (\frac{3}{2} a^2 (9 B-C)+\frac{3}{4} a^2 (21 A-3 B+19 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{63 a^2}\\ &=\frac{2 (21 A-3 B+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (9 B-C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}+\frac{8 \int \frac{\sec ^2(c+d x) \left (\frac{3}{2} a^3 (21 A-3 B+19 C)-\frac{3}{8} a^3 (21 A-93 B+29 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{315 a^3}\\ &=\frac{2 (21 A-3 B+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (9 B-C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}-\frac{2 (21 A-93 B+29 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 a d}+\frac{16 \int \frac{\sec (c+d x) \left (-\frac{3}{16} a^4 (21 A-93 B+29 C)+\frac{3}{8} a^4 (147 A-111 B+143 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{945 a^4}\\ &=\frac{4 (147 A-111 B+143 C) \tan (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (21 A-3 B+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (9 B-C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}-\frac{2 (21 A-93 B+29 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 a d}+(-A+B-C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=\frac{4 (147 A-111 B+143 C) \tan (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (21 A-3 B+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (9 B-C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}-\frac{2 (21 A-93 B+29 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 a d}+\frac{(2 (A-B+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 )}{d}\\ &=-\frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{\sqrt{a} d}+\frac{4 (147 A-111 B+143 C) \tan (c+d x)}{315 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (21 A-3 B+19 C) \sec ^2(c+d x) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (9 B-C) \sec ^3(c+d x) \tan (c+d x)}{63 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^4(c+d x) \tan (c+d x)}{9 d \sqrt{a+a \sec (c+d x)}}-\frac{2 (21 A-93 B+29 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{315 a d}\\ \end{align*}

Mathematica [C]  time = 29.6254, size = 7186, normalized size = 28.29 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^4*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

Result too large to show

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Maple [B]  time = 0.45, size = 1429, normalized size = 5.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-1/5040/d/a*(-1120*C-3264*B*cos(d*x+c)^3-9152*C*cos(d*x+c)^4-9408*A*cos(d*x+c)^4+2752*C*cos(d*x+c)^3-1984*C*co
s(d*x+c)^2+1280*C*cos(d*x+c)+315*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))^(9/2)*sin(d*x+c)-315*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))^(9/2)*sin(d*x+c)+315*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))^(9/2)*sin(d*x+c)+8736*A*co
s(d*x+c)^5-4128*B*cos(d*x+c)^5+1260*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))^(9/2)-1260*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))^(9/2)+8224*C*
cos(d*x+c)^5+7104*B*cos(d*x+c)^4+2688*A*cos(d*x+c)^3+1728*B*cos(d*x+c)^2-2016*A*cos(d*x+c)^2+1260*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)/(co
s(d*x+c)+1))^(9/2)-315*B*cos(d*x+c)^4*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))^(9/2)+315*C*cos(d*x+c)^4*sin(d*x+c)*ln(-(-(-2*cos(d*x+c)/(co
s(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))^(9/2)+1890*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))^(9/2)-1890*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))^(9/2)+1890*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))^(9/2)+1260*A*cos(d*
x+c)^3*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))^(9/2)-1260*B*cos(d*x+c)^3*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))^(9/2)+1260*C*cos(d*x+c)^3*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))^(9/2)+315*A*cos
(d*x+c)^4*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))^(9/2)-1440*B*cos(d*x+c))*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/2)/cos(d*x+c)^4/sin(d*x+c)

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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(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 0.648656, size = 1289, normalized size = 5.07 \begin{align*} \left [\frac{315 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{5} +{\left (A - B + C\right )} a \cos \left (d x + c\right )^{4}\right )} \sqrt{-\frac{1}{a}} \log \left (\frac{2 \, \sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt{-\frac{1}{a}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, \cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \,{\left ({\left (273 \, A - 129 \, B + 257 \, C\right )} \cos \left (d x + c\right )^{4} -{\left (21 \, A - 93 \, B + 29 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (21 \, A - 3 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \,{\left (9 \, B - C\right )} \cos \left (d x + c\right ) + 35 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{630 \,{\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}, \frac{2 \,{\left ({\left (273 \, A - 129 \, B + 257 \, C\right )} \cos \left (d x + c\right )^{4} -{\left (21 \, A - 93 \, B + 29 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \,{\left (21 \, A - 3 \, B + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 5 \,{\left (9 \, B - C\right )} \cos \left (d x + c\right ) + 35 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right ) + \frac{315 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{5} +{\left (A - B + C\right )} a \cos \left (d x + c\right )^{4}\right )} \arctan \left (\frac{\sqrt{2} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt{a} \sin \left (d x + c\right )}\right )}{\sqrt{a}}}{315 \,{\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/630*(315*sqrt(2)*((A - B + C)*a*cos(d*x + c)^5 + (A - B + C)*a*cos(d*x + c)^4)*sqrt(-1/a)*log((2*sqrt(2)*sq
rt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(-1/a)*cos(d*x + c)*sin(d*x + c) + 3*cos(d*x + c)^2 + 2*cos(d*x + c)
- 1)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*((273*A - 129*B + 257*C)*cos(d*x + c)^4 - (21*A - 93*B + 29*C
)*cos(d*x + c)^3 + 3*(21*A - 3*B + 19*C)*cos(d*x + c)^2 + 5*(9*B - C)*cos(d*x + c) + 35*C)*sqrt((a*cos(d*x + c
) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4), 1/315*(2*((273*A - 129*B + 257*C
)*cos(d*x + c)^4 - (21*A - 93*B + 29*C)*cos(d*x + c)^3 + 3*(21*A - 3*B + 19*C)*cos(d*x + c)^2 + 5*(9*B - C)*co
s(d*x + c) + 35*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) + 315*sqrt(2)*((A - B + C)*a*cos(d*x +
c)^5 + (A - B + C)*a*cos(d*x + c)^4)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqr
t(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\sqrt{a \left (\sec{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**4/sqrt(a*(sec(c + d*x) + 1)), x)

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Giac [B]  time = 9.52989, size = 689, normalized size = 2.71 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/315*(315*(sqrt(2)*A - sqrt(2)*B + sqrt(2)*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x +
1/2*c)^2 + a)))/(sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 2*(315*sqrt(2)*A*a^4*sgn(tan(1/2*d*x + 1/2*c)^2
- 1) + 315*sqrt(2)*C*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (1050*sqrt(2)*A*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)
- 420*sqrt(2)*B*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 840*sqrt(2)*C*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (15
12*sqrt(2)*A*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 756*sqrt(2)*B*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 1638*sq
rt(2)*C*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (1134*sqrt(2)*A*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 612*sqrt(2
)*B*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 936*sqrt(2)*C*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (357*sqrt(2)*A*a
^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 276*sqrt(2)*B*a^4*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + 383*sqrt(2)*C*a^4*sgn
(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*
d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d