3.511 \(\int \frac{\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt{a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=208 \[ \frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (35 A-7 B+31 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 a d}-\frac{4 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (7 B-C) \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}} \]

[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*(35*A
- 49*B + 37*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(7*B - C)*Sec[c + d*x]^2*Tan[c + d*x])/(35
*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A - 7
*B + 31*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(105*a*d)

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Rubi [A]  time = 0.637046, antiderivative size = 208, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{\sqrt{2} (A-B+C) \tan ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{\sqrt{a} d}+\frac{2 (35 A-7 B+31 C) \tan (c+d x) \sqrt{a \sec (c+d x)+a}}{105 a d}-\frac{4 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt{a \sec (c+d x)+a}}+\frac{2 (7 B-C) \tan (c+d x) \sec ^2(c+d x)}{35 d \sqrt{a \sec (c+d x)+a}}+\frac{2 C \tan (c+d x) \sec ^3(c+d x)}{7 d \sqrt{a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(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*(35*A
- 49*B + 37*C)*Tan[c + d*x])/(105*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(7*B - C)*Sec[c + d*x]^2*Tan[c + d*x])/(35
*d*Sqrt[a + a*Sec[c + d*x]]) + (2*C*Sec[c + d*x]^3*Tan[c + d*x])/(7*d*Sqrt[a + a*Sec[c + d*x]]) + (2*(35*A - 7
*B + 31*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(105*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 ^3(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 ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 \int \frac{\sec ^3(c+d x) \left (\frac{1}{2} a (7 A+6 C)+\frac{1}{2} a (7 B-C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{7 a}\\ &=\frac{2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{4 \int \frac{\sec ^2(c+d x) \left (a^2 (7 B-C)+\frac{1}{4} a^2 (35 A-7 B+31 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{35 a^2}\\ &=\frac{2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (35 A-7 B+31 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+\frac{8 \int \frac{\sec (c+d x) \left (\frac{1}{8} a^3 (35 A-7 B+31 C)-\frac{1}{4} a^3 (35 A-49 B+37 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{105 a^3}\\ &=-\frac{4 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (35 A-7 B+31 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}+(A-B+C) \int \frac{\sec (c+d x)}{\sqrt{a+a \sec (c+d x)}} \, dx\\ &=-\frac{4 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (35 A-7 B+31 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 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 (35 A-49 B+37 C) \tan (c+d x)}{105 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (7 B-C) \sec ^2(c+d x) \tan (c+d x)}{35 d \sqrt{a+a \sec (c+d x)}}+\frac{2 C \sec ^3(c+d x) \tan (c+d x)}{7 d \sqrt{a+a \sec (c+d x)}}+\frac{2 (35 A-7 B+31 C) \sqrt{a+a \sec (c+d x)} \tan (c+d x)}{105 a d}\\ \end{align*}

Mathematica [C]  time = 29.5461, size = 7134, normalized size = 34.3 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^3*(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 = 0.39, size = 1144, normalized size = 5.5 \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)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

-1/840/d/a*(105*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+c
os(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)*cos(d*x+c)^3-105*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+c
os(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(d*x+c)^3*sin(d*x+c)+105*C*(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*
cos(d*x+c)^3+315*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*cos(d*x+c)^2-315*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))^(7/2)*cos(d*x+c)^2*sin(d*x+c)+315*C*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(7/2)*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))*sin(d*x+c)
*cos(d*x+c)^2+315*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)*cos(d*x+c)-315*B*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+c
os(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*cos(d*x+c)*sin(d*x+c)+315*C*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(7/2)*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))*sin(d*x+c)*co
s(d*x+c)+105*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)-105*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))^(7/2)*sin(d*x+c)+105*C*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(7/2)*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))*sin(d*x+c)-560*A*cos(d*x+c)^4+1456*B*cos
(d*x+c)^4-688*C*cos(d*x+c)^4+1120*A*cos(d*x+c)^3-1568*B*cos(d*x+c)^3+1184*C*cos(d*x+c)^3-560*A*cos(d*x+c)^2+44
8*B*cos(d*x+c)^2-544*C*cos(d*x+c)^2-336*B*cos(d*x+c)+288*C*cos(d*x+c)-240*C)*(a*(cos(d*x+c)+1)/cos(d*x+c))^(1/
2)/cos(d*x+c)^3/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)^3*(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 = 0.633937, size = 1176, normalized size = 5.65 \begin{align*} \left [\frac{105 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{4} +{\left (A - B + C\right )} a \cos \left (d x + c\right )^{3}\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 (35 \, A - 91 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{3} -{\left (35 \, A - 7 \, B + 31 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (7 \, B - C\right )} \cos \left (d x + c\right ) - 15 \, C\right )} \sqrt{\frac{a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{210 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}, -\frac{2 \,{\left ({\left (35 \, A - 91 \, B + 43 \, C\right )} \cos \left (d x + c\right )^{3} -{\left (35 \, A - 7 \, B + 31 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (7 \, B - C\right )} \cos \left (d x + c\right ) - 15 \, 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{105 \, \sqrt{2}{\left ({\left (A - B + C\right )} a \cos \left (d x + c\right )^{4} +{\left (A - B + C\right )} a \cos \left (d x + c\right )^{3}\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}}}{105 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[1/210*(105*sqrt(2)*((A - B + C)*a*cos(d*x + c)^4 + (A - B + C)*a*cos(d*x + c)^3)*sqrt(-1/a)*log(-(2*sqrt(2)*s
qrt((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*((35*A - 91*B + 43*C)*cos(d*x + c)^3 - (35*A - 7*B + 31*C)*c
os(d*x + c)^2 - 3*(7*B - C)*cos(d*x + c) - 15*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a*d*co
s(d*x + c)^4 + a*d*cos(d*x + c)^3), -1/105*(2*((35*A - 91*B + 43*C)*cos(d*x + c)^3 - (35*A - 7*B + 31*C)*cos(d
*x + c)^2 - 3*(7*B - C)*cos(d*x + c) - 15*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) + 105*sqrt(2
)*((A - B + C)*a*cos(d*x + c)^4 + (A - B + C)*a*cos(d*x + c)^3)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d
*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c)))/sqrt(a))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)]

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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 ^{3}{\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)**3*(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)**3/sqrt(a*(sec(c + d*x) + 1)), x)

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Giac [A]  time = 9.26495, size = 412, normalized size = 1.98 \begin{align*} \frac{\frac{105 \, \sqrt{2}{\left (A - B + C\right )} \log \left ({\left | -\sqrt{-a} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt{-a} \mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} + \frac{2 \,{\left (\frac{105 \, \sqrt{2} B a^{3}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} +{\left ({\left (\frac{\sqrt{2}{\left (70 \, A a^{3} - 119 \, B a^{3} + 92 \, C a^{3}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )} - \frac{7 \, \sqrt{2}{\left (20 \, A a^{3} - 37 \, B a^{3} + 16 \, C a^{3}\right )}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \frac{35 \, \sqrt{2}{\left (2 \, A a^{3} - 7 \, B a^{3} + 4 \, C a^{3}\right )}}{\mathrm{sgn}\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a\right )}^{3} \sqrt{-a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}}}{105 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/105*(105*sqrt(2)*(A - B + 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*(105*sqrt(2)*B*a^3/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + ((sqrt(2)*
(70*A*a^3 - 119*B*a^3 + 92*C*a^3)*tan(1/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 7*sqrt(2)*(20*A*a^3
- 37*B*a^3 + 16*C*a^3)/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/2*c)^2 + 35*sqrt(2)*(2*A*a^3 - 7*B*a^
3 + 4*C*a^3)/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)/((a*tan(1/2*d*x + 1
/2*c)^2 - a)^3*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d