3.6.90 \(\int \frac {(a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\) [590]

Optimal. Leaf size=177 \[ \frac {a^{3/2} (2 B+3 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {a^2 (8 A+6 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}-\frac {a (2 A-3 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}} \]

[Out]

a^(3/2)*(2*B+3*C)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+2/3*A*(a+a*sec(d*x+c))^(3/2)*sin(d*x+c)
/d/sec(d*x+c)^(1/2)+1/3*a^2*(8*A+6*B-3*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)-1/3*a*(2*A-3*C)
*sin(d*x+c)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.35, antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4171, 4103, 4100, 3886, 221} \begin {gather*} \frac {a^{3/2} (2 B+3 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}+\frac {a^2 (8 A+6 B-3 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{3 d \sqrt {a \sec (c+d x)+a}}-\frac {a (2 A-3 C) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{3 d}+\frac {2 A \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{3 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(3/2),x]

[Out]

(a^(3/2)*(2*B + 3*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d + (a^2*(8*A + 6*B - 3*C)*Sqrt
[Sec[c + d*x]]*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]]) - (a*(2*A - 3*C)*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec
[c + d*x]]*Sin[c + d*x])/(3*d) + (2*A*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(3*d*Sqrt[Sec[c + d*x]])

Rule 221

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]

Rule 3886

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(a/(b
*f))*Sqrt[a*(d/b)], 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 4100

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

Rule 4103

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

Rule 4171

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

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx &=\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {3}{2} a (A+B)-\frac {1}{2} a (2 A-3 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{3 a}\\ &=-\frac {a (2 A-3 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (8 A+6 B-3 C)+\frac {3}{4} a^2 (2 B+3 C) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx}{3 a}\\ &=\frac {a^2 (8 A+6 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}-\frac {a (2 A-3 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {1}{2} (a (2 B+3 C)) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^2 (8 A+6 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}-\frac {a (2 A-3 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}-\frac {(a (2 B+3 C)) \text {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 )}{d}\\ &=\frac {a^{3/2} (2 B+3 C) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}+\frac {a^2 (8 A+6 B-3 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}-\frac {a (2 A-3 C) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{3 d}+\frac {2 A (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 1.14, size = 122, normalized size = 0.69 \begin {gather*} \frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \sqrt {a (1+\sec (c+d x))} \left (3 \sqrt {2} (2 B+3 C) \tanh ^{-1}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos (c+d x)+2 (A+3 C+2 (5 A+3 B) \cos (c+d x)+A \cos (2 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{6 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sec[c + d*x]^(3/2),x]

[Out]

(a*Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*(2*B + 3*C)*ArcTanh[Sqrt[2]*Sin[(
c + d*x)/2]]*Cos[c + d*x] + 2*(A + 3*C + 2*(5*A + 3*B)*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sin[(c + d*x)/2]))/(
6*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(382\) vs. \(2(153)=306\).
time = 0.23, size = 383, normalized size = 2.16

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (6 B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+6 B \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (-1-\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+9 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+9 C \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (-1-\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+8 A \left (\cos ^{3}\left (d x +c \right )\right )+32 A \left (\cos ^{2}\left (d x +c \right )\right )+24 B \left (\cos ^{2}\left (d x +c \right )\right )-40 A \cos \left (d x +c \right )-24 B \cos \left (d x +c \right )+12 C \cos \left (d x +c \right )-12 C \right ) \cos \left (d x +c \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} a}{12 d \sin \left (d x +c \right )}\)	\(383\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/12/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(6*B*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2
)*(1+cos(d*x+c)+sin(d*x+c))*2^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c)+6*B*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(-
1-cos(d*x+c)+sin(d*x+c))*2^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*2^(1/2)*sin(d*x+c)*cos(d*x+c)+9*C*(-2/(1+cos(d*x+c
)))^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(1+cos(d*x+c)+sin(d*x+c))*2^(1/2))*2^(1/2)*sin(d*x+c)*cos(d*x+c
)+9*C*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(-1-cos(d*x+c)+sin(d*x+c))*2^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*2^(1/
2)*sin(d*x+c)*cos(d*x+c)+8*A*cos(d*x+c)^3+32*A*cos(d*x+c)^2+24*B*cos(d*x+c)^2-40*A*cos(d*x+c)-24*B*cos(d*x+c)+
12*C*cos(d*x+c)-12*C)*cos(d*x+c)*(1/cos(d*x+c))^(3/2)/sin(d*x+c)*a

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 1455 vs. \(2 (153) = 306\).
time = 0.70, size = 1455, normalized size = 8.22 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*sqrt(2)*(sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1
/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^
2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a*log(2*cos(1/2*d*x + 1/2*c
)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2
)*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1
/2*d*x + 1/2*c) + 2) + 8*a*sin(1/2*d*x + 1/2*c))*B*sqrt(a) + 4*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*s
in(1/2*d*x + 1/2*c))*A*sqrt(a) + 3*(3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*c
os(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1
/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^
2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*
cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1
/2*c) + 2))*cos(2*d*x + 2*c)^2 + 3*(a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(
1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*
c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + a*log(2*cos(1/2*d*x + 1/2*c)^2 +
2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - a*log(2*cos
(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*
c) + 2))*sin(2*d*x + 2*c)^2 + 4*sqrt(2)*a*sin(3/2*d*x + 3/2*c) - 4*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 2*(2*sqrt(
2)*a*sin(3/2*d*x + 3/2*c) - 2*sqrt(2)*a*sin(1/2*d*x + 1/2*c) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*
x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x +
1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3
*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/
2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/
2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(2*d*x + 2*c) + 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*
x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x +
1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 3
*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/
2*d*x + 1/2*c) + 2) - 3*a*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/
2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 4*(sqrt(2)*a*cos(3/2*d*x + 3/2*c) - sqrt(2)*a*cos(1/2*d*x + 1/2*c
))*sin(2*d*x + 2*c))*C*sqrt(a)/(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1))/d

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Fricas [A]
time = 3.58, size = 404, normalized size = 2.28 \begin {gather*} \left [\frac {3 \, {\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) + {\left (2 \, B + 3 \, C\right )} a\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{12 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, \frac {3 \, {\left ({\left (2 \, B + 3 \, C\right )} a \cos \left (d x + c\right ) + {\left (2 \, B + 3 \, C\right )} a\right )} \sqrt {-a} \arctan \left (\frac {2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A + 3 \, B\right )} a \cos \left (d x + c\right ) + 3 \, C a\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}}{6 \, {\left (d \cos \left (d x + c\right ) + d\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/12*(3*((2*B + 3*C)*a*cos(d*x + c) + (2*B + 3*C)*a)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 - 4*(
cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c
)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(2*A*a*cos(d*x + c)^2 + 2*(5*A + 3*B)*a*cos(d*x + c) + 3*C*a)
*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c) + d), 1/6*(3*((2*B +
3*C)*a*cos(d*x + c) + (2*B + 3*C)*a)*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)) + 2*(2*A*a*cos(d*x + c)^2 + 2*(5*A + 3*B
)*a*cos(d*x + c) + 3*C*a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x
+ c) + d)]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3006 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(3/2)/sec(d*x + c)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(3/2),x)

[Out]

int(((a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(1/cos(c + d*x))^(3/2), x)

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