3.2.78 \(\int \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} (A+C \sec ^2(c+d x)) \, dx\) [178]

Optimal. Leaf size=188 \[ \frac {a^{5/2} (19 A+8 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d} \]

[Out]

1/4*a^(5/2)*(19*A+8*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d-1/6*a*(3*A-4*C)*(a+a*sec(d*x+c))^(3
/2)*sin(d*x+c)/d+1/2*A*cos(d*x+c)*(a+a*sec(d*x+c))^(5/2)*sin(d*x+c)/d+1/12*a^3*(27*A-56*C)*sin(d*x+c)/d/(a+a*s
ec(d*x+c))^(1/2)-1/2*a^2*(A-8*C)*sin(d*x+c)*(a+a*sec(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.39, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4172, 4103, 4100, 3859, 209} \begin {gather*} \frac {a^{5/2} (19 A+8 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 d}+\frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a \sec (c+d x)+a}}-\frac {a^2 (A-8 C) \sin (c+d x) \sqrt {a \sec (c+d x)+a}}{2 d}-\frac {a (3 A-4 C) \sin (c+d x) (a \sec (c+d x)+a)^{3/2}}{6 d}+\frac {A \sin (c+d x) \cos (c+d x) (a \sec (c+d x)+a)^{5/2}}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^(5/2)*(19*A + 8*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(27*A - 56*C)*Sin[
c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) - (a^2*(A - 8*C)*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2*d) - (a*(
3*A - 4*C)*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(6*d) + (A*Cos[c + d*x]*(a + a*Sec[c + d*x])^(5/2)*Sin[c +
d*x])/(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 3859

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 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 4172

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(f*n)), x] - Dis
t[1/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*m - b*(A*(m + n + 1) + C*n)*Csc[e +
f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, 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 \cos ^2(c+d x) (a+a \sec (c+d x))^{5/2} \left (A+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^{5/2} \left (\frac {5 a A}{2}-\frac {1}{2} a (3 A-4 C) \sec (c+d x)\right ) \, dx}{2 a}\\ &=-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac {\int \cos (c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{4} a^2 (21 A-8 C)-\frac {3}{4} a^2 (A-8 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=-\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac {2 \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{8} a^3 (27 A-56 C)+\frac {5}{8} a^3 (3 A+8 C) \sec (c+d x)\right ) \, dx}{3 a}\\ &=\frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}+\frac {1}{8} \left (a^2 (19 A+8 C)\right ) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}-\frac {\left (a^3 (19 A+8 C)\right ) \text {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}\\ &=\frac {a^{5/2} (19 A+8 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (27 A-56 C) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}-\frac {a^2 (A-8 C) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{2 d}-\frac {a (3 A-4 C) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{6 d}+\frac {A \cos (c+d x) (a+a \sec (c+d x))^{5/2} \sin (c+d x)}{2 d}\\ \end {align*}

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Mathematica [A]
time = 0.90, size = 137, normalized size = 0.73 \begin {gather*} \frac {a^2 \sec \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {a (1+\sec (c+d x))} \left (6 \sqrt {2} (19 A+8 C) \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^{\frac {3}{2}}(c+d x)+2 (33 A+16 C+(9 A+128 C) \cos (c+d x)+33 A \cos (2 (c+d x))+3 A \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{48 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[(c + d*x)/2]*Sec[c + d*x]*Sqrt[a*(1 + Sec[c + d*x])]*(6*Sqrt[2]*(19*A + 8*C)*ArcSin[Sqrt[2]*Sin[(c +
d*x)/2]]*Cos[c + d*x]^(3/2) + 2*(33*A + 16*C + (9*A + 128*C)*Cos[c + d*x] + 33*A*Cos[2*(c + d*x)] + 3*A*Cos[3*
(c + d*x)])*Sin[(c + d*x)/2]))/(48*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(401\) vs. \(2(164)=328\).
time = 24.16, size = 402, normalized size = 2.14

method	result	size
default	\(\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (57 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right ) \sqrt {2}+24 C \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right )+57 A \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )+24 C \sqrt {2}\, \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )-24 A \left (\cos ^{4}\left (d x +c \right )\right )-108 A \left (\cos ^{3}\left (d x +c \right )\right )+132 A \left (\cos ^{2}\left (d x +c \right )\right )-256 C \left (\cos ^{2}\left (d x +c \right )\right )+224 C \cos \left (d x +c \right )+32 C \right ) a^{2}}{48 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}\)	\(402\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*sec(d*x+c))^(5/2)*(A+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/48/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(57*A*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*cos(d*x+c)*arcta
nh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+24*C*cos(d*x+c)*sin(d*x+
c)*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(3/2)+57*A*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)+24*C*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x
+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(3/2)*2^(1/2)*sin(d*x+c)-24*A*cos(d*x+c)^4-108*A*cos(d*
x+c)^3+132*A*cos(d*x+c)^2-256*C*cos(d*x+c)^2+224*C*cos(d*x+c)+32*C)/sin(d*x+c)/cos(d*x+c)*a^2

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Maxima [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Fricas [A]
time = 3.01, size = 402, normalized size = 2.14 \begin {gather*} \left [\frac {3 \, {\left ({\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 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 ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) + 2 \, {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 33 \, A a^{2} \cos \left (d x + c\right )^{2} + 64 \, C a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\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 (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, -\frac {3 \, {\left ({\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (19 \, A + 8 \, C\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \arctan \left (\frac {\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 ) - {\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 33 \, A a^{2} \cos \left (d x + c\right )^{2} + 64 \, C a^{2} \cos \left (d x + c\right ) + 8 \, C a^{2}\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 (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*(3*((19*A + 8*C)*a^2*cos(d*x + c)^2 + (19*A + 8*C)*a^2*cos(d*x + c))*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)) + 2*(6*A*a^2*cos(d*x + c)^3 + 33*A*a^2*cos(d*x + c)^2 + 64*C*a^2*cos(d*x + c) + 8*C*a^2)*sqrt((a*cos(
d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c)), -1/12*(3*((19*A + 8*C)*a^2*cos(
d*x + c)^2 + (19*A + 8*C)*a^2*cos(d*x + c))*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c
)/(sqrt(a)*sin(d*x + c))) - (6*A*a^2*cos(d*x + c)^3 + 33*A*a^2*cos(d*x + c)^2 + 64*C*a^2*cos(d*x + c) + 8*C*a^
2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*sec(d*x+c))**(5/2)*(A+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 554 vs. \(2 (164) = 328\).
time = 2.02, size = 554, normalized size = 2.95 \begin {gather*} -\frac {3 \, {\left (19 \, A \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 8 \, C \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\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 )}^{2} - a {\left (2 \, \sqrt {2} + 3\right )} \right |}\right ) - 3 \, {\left (19 \, A \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 8 \, C \sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )} \log \left ({\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 )}^{2} + a {\left (2 \, \sqrt {2} - 3\right )} \right |}\right ) - \frac {16 \, {\left (7 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, \sqrt {2} C a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\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 {12 \, \sqrt {2} {\left (19 \, {\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 )}^{6} A \sqrt {-a} a^{3} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 171 \, {\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 )}^{4} A \sqrt {-a} a^{4} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) + 89 \, {\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 )}^{2} A \sqrt {-a} a^{5} \mathrm {sgn}\left (\cos \left (d x + c\right )\right ) - 9 \, A \sqrt {-a} a^{6} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )\right )}}{{\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 )}^{4} - 6 \, {\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 )}^{2} a + a^{2}\right )}^{2}}}{24 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(3*(19*A*sqrt(-a)*a^2*sgn(cos(d*x + c)) + 8*C*sqrt(-a)*a^2*sgn(cos(d*x + 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))^2 - a*(2*sqrt(2) + 3))) - 3*(19*A*sqrt(-a)*a^2*sgn(cos(d*x
+ c)) + 8*C*sqrt(-a)*a^2*sgn(cos(d*x + 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))^2 + a*(2*sqrt(2) - 3))) - 16*(7*sqrt(2)*C*a^4*sgn(cos(d*x + c))*tan(1/2*d*x + 1/2*c)^2 - 9*sqrt(2
)*C*a^4*sgn(cos(d*x + c)))*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)) + 12*sqrt(2)*(19*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*A*sqrt(-a)*a^3
*sgn(cos(d*x + c)) - 171*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*A*sqrt(-a)*a^
4*sgn(cos(d*x + c)) + 89*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^
5*sgn(cos(d*x + c)) - 9*A*sqrt(-a)*a^6*sgn(cos(d*x + c)))/((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*
x + 1/2*c)^2 + a))^4 - 6*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a + a^2)^2)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^2\,\left (A+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2),x)

[Out]

int(cos(c + d*x)^2*(A + C/cos(c + d*x)^2)*(a + a/cos(c + d*x))^(5/2), x)

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