3.392 \(\int \frac {\cos ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=206 \[ \frac {(7 B-2 C) \sin (c+d x)}{8 d \sqrt {a \sec (c+d x)+a}}-\frac {(9 B-14 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (B-C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {(B-6 C) \sin (c+d x) \cos (c+d x)}{12 d \sqrt {a \sec (c+d x)+a}}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}} \]

[Out]

-1/8*(9*B-14*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)+(B-C)*arctan(1/2*a^(1/2)*tan(d*x+c
)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2)/d/a^(1/2)+1/8*(7*B-2*C)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-1/12*(B-
6*C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)+1/3*B*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]  time = 0.65, antiderivative size = 206, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 6, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4072, 4022, 3920, 3774, 203, 3795} \[ \frac {(7 B-2 C) \sin (c+d x)}{8 d \sqrt {a \sec (c+d x)+a}}-\frac {(9 B-14 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (B-C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {(B-6 C) \sin (c+d x) \cos (c+d x)}{12 d \sqrt {a \sec (c+d x)+a}}+\frac {B \sin (c+d x) \cos ^2(c+d x)}{3 d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((9*B - 14*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(8*Sqrt[a]*d) + (Sqrt[2]*(B - C)*ArcTa
n[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[a]*d) + ((7*B - 2*C)*Sin[c + d*x])/(8*d*Sq
rt[a + a*Sec[c + d*x]]) - ((B - 6*C)*Cos[c + d*x]*Sin[c + d*x])/(12*d*Sqrt[a + a*Sec[c + d*x]]) + (B*Cos[c + d
*x]^2*Sin[c + d*x])/(3*d*Sqrt[a + a*Sec[c + d*x]])

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])

Rule 3774

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 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 3920

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]

Rule 4022

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), 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 - A*b*(m + n + 1)*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] && LtQ[n, 0]

Rule 4072

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx &=\int \frac {\cos ^3(c+d x) (B+C \sec (c+d x))}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=\frac {B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\cos ^2(c+d x) \left (-\frac {1}{2} a (B-6 C)+\frac {5}{2} a B \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{3 a}\\ &=-\frac {(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (\frac {3}{4} a^2 (7 B-2 C)-\frac {3}{4} a^2 (B-6 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{6 a^2}\\ &=\frac {(7 B-2 C) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}-\frac {(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {-\frac {3}{8} a^3 (9 B-14 C)+\frac {3}{8} a^3 (7 B-2 C) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{6 a^3}\\ &=\frac {(7 B-2 C) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}-\frac {(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}-\frac {(9 B-14 C) \int \sqrt {a+a \sec (c+d x)} \, dx}{16 a}+(B-C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\\ &=\frac {(7 B-2 C) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}-\frac {(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}+\frac {(9 B-14 C) \operatorname {Subst}\left (\int \frac {1}{a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 d}-\frac {(2 (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 {(9 B-14 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{8 \sqrt {a} d}+\frac {\sqrt {2} (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 {(7 B-2 C) \sin (c+d x)}{8 d \sqrt {a+a \sec (c+d x)}}-\frac {(B-6 C) \cos (c+d x) \sin (c+d x)}{12 d \sqrt {a+a \sec (c+d x)}}+\frac {B \cos ^2(c+d x) \sin (c+d x)}{3 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.76, size = 150, normalized size = 0.73 \[ \frac {\tan (c+d x) \left (\cos (c+d x) \sqrt {1-\sec (c+d x)} \left (-2 (B-6 C) \cos (c+d x)+8 B \cos ^2(c+d x)+21 B-6 C\right )+(42 C-27 B) \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )+24 \sqrt {2} (B-C) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )\right )}{24 d \sqrt {1-\sec (c+d x)} \sqrt {a (\sec (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-27*B + 42*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]] + 24*Sqrt[2]*(B - C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]
+ Cos[c + d*x]*(21*B - 6*C - 2*(B - 6*C)*Cos[c + d*x] + 8*B*Cos[c + d*x]^2)*Sqrt[1 - Sec[c + d*x]])*Tan[c + d*
x])/(24*d*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 2.62, size = 539, normalized size = 2.62 \[ \left [-\frac {24 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right ) + {\left (B - C\right )} a\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 ) - 3 \, {\left ({\left (9 \, B - 14 \, C\right )} \cos \left (d x + c\right ) + 9 \, B - 14 \, C\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 (8 \, B \cos \left (d x + c\right )^{3} - 2 \, {\left (B - 6 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B - 2 \, C\right )} \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 )}{48 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}, \frac {3 \, {\left ({\left (9 \, B - 14 \, C\right )} \cos \left (d x + c\right ) + 9 \, B - 14 \, C\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 (8 \, B \cos \left (d x + c\right )^{3} - 2 \, {\left (B - 6 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (7 \, B - 2 \, C\right )} \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 ) - \frac {24 \, \sqrt {2} {\left ({\left (B - C\right )} a \cos \left (d x + c\right ) + {\left (B - C\right )} a\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}}}{24 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(24*sqrt(2)*((B - C)*a*cos(d*x + c) + (B - C)*a)*sqrt(-1/a)*log((2*sqrt(2)*sqrt((a*cos(d*x + c) + a)/co
s(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)) - 3*((9*B - 14*C)*cos(d*x + c) + 9*B - 14*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*(8*B*cos(d*x + c)^3 - 2*(B - 6*C)*cos(d*x + c)^2 + 3*(7*B - 2*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos
(d*x + c))*sin(d*x + c))/(a*d*cos(d*x + c) + a*d), 1/24*(3*((9*B - 14*C)*cos(d*x + c) + 9*B - 14*C)*sqrt(a)*ar
ctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (8*B*cos(d*x + c)^3 - 2*(B
- 6*C)*cos(d*x + c)^2 + 3*(7*B - 2*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c) - 24
*sqrt(2)*((B - C)*a*cos(d*x + c) + (B - C)*a)*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) + a*d)]

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giac [B]  time = 3.00, size = 846, normalized size = 4.11 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/48*(24*sqrt(2)*(B - C)*log((sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2)/(sqrt(-a
)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 3*(9*B - 14*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)))/(sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) - 3*(9*B - 14*C)*log(ab
s((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)))/(sqrt(-a)*sgn(
tan(1/2*d*x + 1/2*c)^2 - 1)) + 4*sqrt(2)*(165*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^10*B*sqrt(-a) - 102*(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)
- 1323*(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 + 954*(sqrt(-a)*ta
n(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^8*C*sqrt(-a)*a + 3906*(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^2 - 2268*(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^2 - 2118*(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^3 + 1044*(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^3 + 393*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*B*sqrt(-a)*a^4 - 222*(s
qrt(-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^4 - 31*B*sqrt(-a)*a^5 + 18*
C*sqrt(-a)*a^5)/(((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)^3*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 1.89, size = 1067, normalized size = 5.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x)

[Out]

1/192/d*(27*B*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))-42*C*2^(1/2)*sin(d*x+c)*cos(d*x+c)^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+cos(d*x+c)))^(5/2)+54*B*cos(d*x+
c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))+48*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))*cos(d*x+c)^2*sin(d*x+c)-84*C*2^(1/2)*sin(d*x+c)*cos(d*x+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)))^(5/2)-48*C*
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))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(
5/2)*sin(d*x+c)*cos(d*x+c)^2+27*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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)+96*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))*cos(d*x+c)*sin(d*x+c)-42*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^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2
)*sin(d*x+c)-96*C*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))*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(5/2)*sin(d*x+c)*cos(d*x+c)+48*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*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))*sin(d*x+c)-48*C*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))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(5/2)*sin(d*x+c)-64*B*cos(d*x+c)^6+80*B*
cos(d*x+c)^5-96*C*cos(d*x+c)^5-184*B*cos(d*x+c)^4+144*C*cos(d*x+c)^4+168*B*cos(d*x+c)^3-48*C*cos(d*x+c)^3)*(a*
(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)^2/a

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{4}}{\sqrt {a \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))*cos(d*x + c)^4/sqrt(a*sec(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^4\,\left (\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^4*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2),x)

[Out]

int((cos(c + d*x)^4*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x))^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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