3.5.98 \(\int \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [498]

Optimal. Leaf size=215 \[ \frac {a^{3/2} (75 A+88 B+112 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+88 B+112 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (39 A+56 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a (3 A+8 B) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d} \]

[Out]

1/64*a^(3/2)*(75*A+88*B+112*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/4*A*cos(d*x+c)^3*(a+a*sec
(d*x+c))^(3/2)*sin(d*x+c)/d+1/64*a^2*(75*A+88*B+112*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/96*a^2*(39*A+56*B
+48*C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/24*a*(3*A+8*B)*cos(d*x+c)^2*sin(d*x+c)*(a+a*sec(d*x+c)
)^(1/2)/d

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Rubi [A]
time = 0.41, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4171, 4102, 4100, 3890, 3859, 209} \begin {gather*} \frac {a^{3/2} (75 A+88 B+112 C) \text {ArcTan}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{64 d}+\frac {a^2 (75 A+88 B+112 C) \sin (c+d x)}{64 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (39 A+56 B+48 C) \sin (c+d x) \cos (c+d x)}{96 d \sqrt {a \sec (c+d x)+a}}+\frac {a (3 A+8 B) \sin (c+d x) \cos ^2(c+d x) \sqrt {a \sec (c+d x)+a}}{24 d}+\frac {A \sin (c+d x) \cos ^3(c+d x) (a \sec (c+d x)+a)^{3/2}}{4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a^(3/2)*(75*A + 88*B + 112*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(64*d) + (a^2*(75*A +
88*B + 112*C)*Sin[c + d*x])/(64*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(39*A + 56*B + 48*C)*Cos[c + d*x]*Sin[c + d
*x])/(96*d*Sqrt[a + a*Sec[c + d*x]]) + (a*(3*A + 8*B)*Cos[c + d*x]^2*Sqrt[a + a*Sec[c + d*x]]*Sin[c + d*x])/(2
4*d) + (A*Cos[c + d*x]^3*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(4*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 3890

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

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 4102

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*n)), x]
- Dist[b/(a*d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[a*A*(m - n - 1) - b*B*n - (a*
B*n + A*b*(m + n))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, 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 \cos ^4(c+d x) (a+a \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac {\int \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (3 A+8 B)+\frac {1}{2} a (3 A+8 C) \sec (c+d x)\right ) \, dx}{4 a}\\ &=\frac {a (3 A+8 B) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac {\int \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (39 A+56 B+48 C)+\frac {3}{4} a^2 (9 A+8 B+16 C) \sec (c+d x)\right ) \, dx}{12 a}\\ &=\frac {a^2 (39 A+56 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a (3 A+8 B) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac {1}{64} (a (75 A+88 B+112 C)) \int \cos (c+d x) \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^2 (75 A+88 B+112 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (39 A+56 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a (3 A+8 B) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}+\frac {1}{128} (a (75 A+88 B+112 C)) \int \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^2 (75 A+88 B+112 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (39 A+56 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a (3 A+8 B) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}-\frac {\left (a^2 (75 A+88 B+112 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 )}{64 d}\\ &=\frac {a^{3/2} (75 A+88 B+112 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{64 d}+\frac {a^2 (75 A+88 B+112 C) \sin (c+d x)}{64 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (39 A+56 B+48 C) \cos (c+d x) \sin (c+d x)}{96 d \sqrt {a+a \sec (c+d x)}}+\frac {a (3 A+8 B) \cos ^2(c+d x) \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{24 d}+\frac {A \cos ^3(c+d x) (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]
time = 1.84, size = 157, normalized size = 0.73 \begin {gather*} \frac {a \sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {a (1+\sec (c+d x))} \left (3 \sqrt {2} (75 A+88 B+112 C) \text {ArcSin}\left (\sqrt {2} \sin \left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\cos (c+d x)}+(285 A+296 B+336 C+2 (93 A+88 B+48 C) \cos (c+d x)+4 (15 A+8 B) \cos (2 (c+d x))+12 A \cos (3 (c+d x))) \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )\right )}{384 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + a*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(a*Sec[(c + d*x)/2]*Sqrt[a*(1 + Sec[c + d*x])]*(3*Sqrt[2]*(75*A + 88*B + 112*C)*ArcSin[Sqrt[2]*Sin[(c + d*x)/2
]]*Sqrt[Cos[c + d*x]] + (285*A + 296*B + 336*C + 2*(93*A + 88*B + 48*C)*Cos[c + d*x] + 4*(15*A + 8*B)*Cos[2*(c
+ d*x)] + 12*A*Cos[3*(c + d*x)])*(-Sin[(c + d*x)/2] + Sin[(3*(c + d*x))/2])))/(384*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1105\) vs. \(2(191)=382\).
time = 27.90, size = 1106, normalized size = 5.14

method	result	size
default	\(\text {Expression too large to display}\)	\(1106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBOSE)

[Out]

1/3072/d*(225*A*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*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))+264*B*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(7/2)*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^(1/2)+336*C*2^(1
/2)*sin(d*x+c)*cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*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))+675*A*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*
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))+792*B*cos(d*x+c)^2*sin(d*x+c)*
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*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^(1/2)+1008*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*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))+675*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c
)/(1+cos(d*x+c)))^(7/2)*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))+792*B*
cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*si
n(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+1008*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)
*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))+225*A*(-2*cos(d*x+c)/(1+cos(d
*x+c)))^(7/2)*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^(1/2)*sin(d*x+
c)+264*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*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))*sin(d*x+c)+336*C*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(7/2)*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)-768*A*cos(d*x+c)^8-1152*A*cos(d*x+c)^7-1024
*B*cos(d*x+c)^7-480*A*cos(d*x+c)^6-1792*B*cos(d*x+c)^6-1536*C*cos(d*x+c)^6-1200*A*cos(d*x+c)^5-1408*B*cos(d*x+
c)^5-3840*C*cos(d*x+c)^5+3600*A*cos(d*x+c)^4+4224*B*cos(d*x+c)^4+5376*C*cos(d*x+c)^4)*(a*(1+cos(d*x+c))/cos(d*
x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^3*a

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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)^4*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]
time = 3.26, size = 420, normalized size = 1.95 \begin {gather*} \left [\frac {3 \, {\left ({\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 88 \, B + 112 \, C\right )} a\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 (48 \, A a \cos \left (d x + c\right )^{4} + 8 \, {\left (15 \, A + 8 \, B\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (75 \, A + 88 \, B + 48 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{384 \, {\left (d \cos \left (d x + c\right ) + d\right )}}, -\frac {3 \, {\left ({\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right ) + {\left (75 \, A + 88 \, B + 112 \, C\right )} a\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 (48 \, A a \cos \left (d x + c\right )^{4} + 8 \, {\left (15 \, A + 8 \, B\right )} a \cos \left (d x + c\right )^{3} + 2 \, {\left (75 \, A + 88 \, B + 48 \, C\right )} a \cos \left (d x + c\right )^{2} + 3 \, {\left (75 \, A + 88 \, B + 112 \, C\right )} a \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{192 \, {\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(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

[1/384*(3*((75*A + 88*B + 112*C)*a*cos(d*x + c) + (75*A + 88*B + 112*C)*a)*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*(48*A*a*cos(d*x + c)^4 + 8*(15*A + 8*B)*a*cos(d*x + c)^3 + 2*(75*A + 88*B + 48*C)*a*cos(d*x + c)^
2 + 3*(75*A + 88*B + 112*C)*a*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x +
c) + d), -1/192*(3*((75*A + 88*B + 112*C)*a*cos(d*x + c) + (75*A + 88*B + 112*C)*a)*sqrt(a)*arctan(sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - (48*A*a*cos(d*x + c)^4 + 8*(15*A + 8*B)*a
*cos(d*x + c)^3 + 2*(75*A + 88*B + 48*C)*a*cos(d*x + c)^2 + 3*(75*A + 88*B + 112*C)*a*cos(d*x + c))*sqrt((a*co
s(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c) + d)]

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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)**4*(a+a*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)+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. 1530 vs. \(2 (191) = 382\).
time = 2.22, size = 1530, normalized size = 7.12 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+a*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

-1/384*(3*(75*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 88*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 112*C*sqrt(-a)*a*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*(75*A*sqrt(-a)*a*sgn(cos(d*x + c)) + 88*B*sqrt(-a)*a*sgn(cos(d*x + c)) + 112*C*sqrt(-a)*a*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))) + 4*
sqrt(2)*(225*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*A*sqrt(-a)*a^2*sgn(cos(d
*x + c)) + 264*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*B*sqrt(-a)*a^2*sgn(cos
(d*x + c)) + 336*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^14*C*sqrt(-a)*a^2*sgn(c
os(d*x + c)) - 6261*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*A*sqrt(-a)*a^3*sg
n(cos(d*x + c)) - 4008*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*B*sqrt(-a)*a^3
*sgn(cos(d*x + c)) - 8592*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^12*C*sqrt(-a)*
a^3*sgn(cos(d*x + c)) + 35925*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*A*sqrt(
-a)*a^4*sgn(cos(d*x + c)) + 33960*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10*B*s
qrt(-a)*a^4*sgn(cos(d*x + c)) + 70032*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^10
*C*sqrt(-a)*a^4*sgn(cos(d*x + c)) - 127449*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a
))^8*A*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 131784*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^8*B*sqrt(-a)*a^5*sgn(cos(d*x + c)) - 208080*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*
c)^2 + a))^8*C*sqrt(-a)*a^5*sgn(cos(d*x + c)) + 101667*(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^6*sgn(cos(d*x + c)) + 108312*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*
x + 1/2*c)^2 + a))^6*B*sqrt(-a)*a^6*sgn(cos(d*x + c)) + 154608*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/
2*d*x + 1/2*c)^2 + a))^6*C*sqrt(-a)*a^6*sgn(cos(d*x + c)) - 26079*(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^7*sgn(cos(d*x + c)) - 29432*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*
tan(1/2*d*x + 1/2*c)^2 + a))^4*B*sqrt(-a)*a^7*sgn(cos(d*x + c)) - 44208*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(
-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*sqrt(-a)*a^7*sgn(cos(d*x + c)) + 3303*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqr
t(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*A*sqrt(-a)*a^8*sgn(cos(d*x + c)) + 3384*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - s
qrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*sqrt(-a)*a^8*sgn(cos(d*x + c)) + 5424*(sqrt(-a)*tan(1/2*d*x + 1/2*c) -
sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*sqrt(-a)*a^8*sgn(cos(d*x + c)) - 147*A*sqrt(-a)*a^9*sgn(cos(d*x + c)
) - 152*B*sqrt(-a)*a^9*sgn(cos(d*x + c)) - 240*C*sqrt(-a)*a^9*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)^4)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\cos \left (c+d\,x\right )}^4\,{\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 ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^4*(a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

int(cos(c + d*x)^4*(a + a/cos(c + d*x))^(3/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2), x)

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