3.2.94 \(\int \frac {\sec ^2(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [194]

Optimal. Leaf size=169 \[ \frac {(3 A+11 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(3 A+13 C) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}+\frac {(3 A+7 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d} \]

[Out]

1/4*(3*A+11*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(A+C)*sec(d
*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-1/3*(3*A+13*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/6*(3*A+7*C)
*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d

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Rubi [A]
time = 0.30, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4170, 4095, 4086, 3880, 209} \begin {gather*} \frac {(3 A+11 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(3 A+7 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{6 a^2 d}-\frac {(A+C) \tan (c+d x) \sec ^2(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {(3 A+13 C) \tan (c+d x)}{3 a d \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((3*A + 11*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A +
C)*Sec[c + d*x]^2*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((3*A + 13*C)*Tan[c + d*x])/(3*a*d*Sqrt[a
+ a*Sec[c + d*x]]) + ((3*A + 7*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(6*a^2*d)

Rule 209

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

Rule 3880

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 4086

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 4095

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)),
Int[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] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4170

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)]

Rubi steps

\begin {align*} \int \frac {\sec ^2(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {\int \frac {\sec ^2(c+d x) \left (2 a C-\frac {1}{2} a (3 A+7 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(3 A+7 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}-\frac {\int \frac {\sec (c+d x) \left (-\frac {1}{4} a^2 (3 A+7 C)+\frac {1}{2} a^2 (3 A+13 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{3 a^3}\\ &=-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(3 A+13 C) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}+\frac {(3 A+7 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}+\frac {(3 A+11 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(3 A+13 C) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}+\frac {(3 A+7 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}-\frac {(3 A+11 C) \text {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 )}{2 a d}\\ &=\frac {(3 A+11 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sec ^2(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(3 A+13 C) \tan (c+d x)}{3 a d \sqrt {a+a \sec (c+d x)}}+\frac {(3 A+7 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{6 a^2 d}\\ \end {align*}

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Mathematica [A]
time = 3.78, size = 162, normalized size = 0.96 \begin {gather*} \frac {\cos (c+d x) \left (A+C \sec ^2(c+d x)\right ) \left (3 \sqrt {2} (3 A+11 C) \text {ArcTan}\left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \cot ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {-1+\sec (c+d x)}-(3 A+11 C+24 C \cos (c+d x)+(3 A+19 C) \cos (2 (c+d x))) \sec ^2(c+d x)\right ) \sin (c+d x)}{6 d (A+2 C+A \cos (2 (c+d x))) (a (1+\sec (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[c + d*x]*(A + C*Sec[c + d*x]^2)*(3*Sqrt[2]*(3*A + 11*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Cot[(c +
d*x)/2]^2*Sqrt[-1 + Sec[c + d*x]] - (3*A + 11*C + 24*C*Cos[c + d*x] + (3*A + 19*C)*Cos[2*(c + d*x)])*Sec[c + d
*x]^2)*Sin[c + d*x])/(6*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(593\) vs. \(2(146)=292\).
time = 12.13, size = 594, normalized size = 3.51

method	result	size
default	\(-\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (-9 A \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )-33 C \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )-18 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )-66 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right )-9 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )-33 C \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \ln \left (-\frac {-\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )+12 A \left (\cos ^{3}\left (d x +c \right )\right )+76 C \left (\cos ^{3}\left (d x +c \right )\right )-12 A \left (\cos ^{2}\left (d x +c \right )\right )-28 C \left (\cos ^{2}\left (d x +c \right )\right )-64 C \cos \left (d x +c \right )+16 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{24 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) a^{2}}\)	\(594\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24/d*(-1+cos(d*x+c))*(-9*A*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(-(-(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-33*C*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(3/2)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-18*A*cos(d*x+
c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d
*x+c)-1)/sin(d*x+c))-66*C*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(-(-(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-9*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*ln(-(-(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)-33*C*(-2*cos(d*x+c)/(1+cos(d*x+c)
))^(3/2)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*sin(d*x+c)+12*A*cos(d
*x+c)^3+76*C*cos(d*x+c)^3-12*A*cos(d*x+c)^2-28*C*cos(d*x+c)^2-64*C*cos(d*x+c)+16*C)*(a*(1+cos(d*x+c))/cos(d*x+
c))^(1/2)/sin(d*x+c)^3/cos(d*x+c)/a^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 2.94, size = 451, normalized size = 2.67 \begin {gather*} \left [-\frac {3 \, \sqrt {2} {\left ({\left (3 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A + 11 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + 3 \, a \cos \left (d x + c\right )^{2} + 2 \, a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + 4 \, {\left ({\left (3 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, C \cos \left (d x + c\right ) - 4 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{24 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}, -\frac {3 \, \sqrt {2} {\left ({\left (3 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{3} + 2 \, {\left (3 \, A + 11 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, A + 11 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {a} \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 ) + 2 \, {\left ({\left (3 \, A + 19 \, C\right )} \cos \left (d x + c\right )^{2} + 12 \, C \cos \left (d x + c\right ) - 4 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{12 \, {\left (a^{2} d \cos \left (d x + c\right )^{3} + 2 \, a^{2} d \cos \left (d x + c\right )^{2} + a^{2} d \cos \left (d x + c\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[-1/24*(3*sqrt(2)*((3*A + 11*C)*cos(d*x + c)^3 + 2*(3*A + 11*C)*cos(d*x + c)^2 + (3*A + 11*C)*cos(d*x + c))*sq
rt(-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)) + 4*((3*A + 19*C)*cos(d*x + c)^2 + 12*C
*cos(d*x + c) - 4*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos
(d*x + c)^2 + a^2*d*cos(d*x + c)), -1/12*(3*sqrt(2)*((3*A + 11*C)*cos(d*x + c)^3 + 2*(3*A + 11*C)*cos(d*x + c)
^2 + (3*A + 11*C)*cos(d*x + c))*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(s
qrt(a)*sin(d*x + c))) + 2*((3*A + 19*C)*cos(d*x + c)^2 + 12*C*cos(d*x + c) - 4*C)*sqrt((a*cos(d*x + c) + a)/co
s(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^3 + 2*a^2*d*cos(d*x + c)^2 + a^2*d*cos(d*x + c))]

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**2/(a*(sec(c + d*x) + 1))**(3/2), x)

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Giac [A]
time = 1.98, size = 246, normalized size = 1.46 \begin {gather*} \frac {\frac {{\left ({\left (\frac {3 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a} - \frac {2 \, {\left (3 \, \sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 23 \, \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {3 \, {\left (\sqrt {2} A a \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 9 \, \sqrt {2} C a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{a}\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 )} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}} - \frac {3 \, {\left (3 \, \sqrt {2} A + 11 \, \sqrt {2} 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} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(((3*(sqrt(2)*A*a*sgn(cos(d*x + c)) + sqrt(2)*C*a*sgn(cos(d*x + c)))*tan(1/2*d*x + 1/2*c)^2/a - 2*(3*sqrt
(2)*A*a*sgn(cos(d*x + c)) + 23*sqrt(2)*C*a*sgn(cos(d*x + c)))/a)*tan(1/2*d*x + 1/2*c)^2 + 3*(sqrt(2)*A*a*sgn(c
os(d*x + c)) + 9*sqrt(2)*C*a*sgn(cos(d*x + c)))/a)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)*sqrt(-
a*tan(1/2*d*x + 1/2*c)^2 + a)) - 3*(3*sqrt(2)*A + 11*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)*a*sgn(cos(d*x + c))))/d

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

[Out]

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

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