3.400 \(\int \frac {\cos ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=221 \[ \frac {(19 B-12 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {(13 B-9 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 B-6 C) \sin (c+d x)}{4 a d \sqrt {a \sec (c+d x)+a}}+\frac {(2 B-C) \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a \sec (c+d x)+a}}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

[Out]

1/4*(19*B-12*C)*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d-1/2*(B-C)*cos(d*x+c)*sin(d*x+c)/d/
(a+a*sec(d*x+c))^(3/2)-1/4*(13*B-9*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)-1/4*(7*B-6*C)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/2*(2*B-C)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x
+c))^(1/2)

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Rubi [A]  time = 0.69, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4072, 4020, 4022, 3920, 3774, 203, 3795} \[ \frac {(19 B-12 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 a^{3/2} d}-\frac {(13 B-9 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(7 B-6 C) \sin (c+d x)}{4 a d \sqrt {a \sec (c+d x)+a}}+\frac {(2 B-C) \sin (c+d x) \cos (c+d x)}{2 a d \sqrt {a \sec (c+d x)+a}}-\frac {(B-C) \sin (c+d x) \cos (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((19*B - 12*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*a^(3/2)*d) - ((13*B - 9*C)*ArcTan[(
Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((B - C)*Cos[c + d*x]*Sin[c
+ d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((7*B - 6*C)*Sin[c + d*x])/(4*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((2*
B - C)*Cos[c + d*x]*Sin[c + d*x])/(2*a*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 4020

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(b*f*(2
*m + 1)), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*(2
*m + n + 1) + (A*b - a*B)*(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] && LtQ[m, -2^(-1)] &&  !GtQ[n, 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 ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=\int \frac {\cos ^2(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\\ &=-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\cos ^2(c+d x) \left (2 a (2 B-C)-\frac {5}{2} a (B-C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(2 B-C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\cos (c+d x) \left (-a^2 (7 B-6 C)+3 a^2 (2 B-C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a^3}\\ &=-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(7 B-6 C) \sin (c+d x)}{4 a d \sqrt {a+a \sec (c+d x)}}+\frac {(2 B-C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\frac {1}{2} a^3 (19 B-12 C)-\frac {1}{2} a^3 (7 B-6 C) \sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a^4}\\ &=-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(7 B-6 C) \sin (c+d x)}{4 a d \sqrt {a+a \sec (c+d x)}}+\frac {(2 B-C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}}+\frac {(19 B-12 C) \int \sqrt {a+a \sec (c+d x)} \, dx}{8 a^2}-\frac {(13 B-9 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(B-C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(7 B-6 C) \sin (c+d x)}{4 a d \sqrt {a+a \sec (c+d x)}}+\frac {(2 B-C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}}-\frac {(19 B-12 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 )}{4 a d}+\frac {(13 B-9 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 )}{2 a d}\\ &=\frac {(19 B-12 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 a^{3/2} d}-\frac {(13 B-9 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 {(B-C) \cos (c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(7 B-6 C) \sin (c+d x)}{4 a d \sqrt {a+a \sec (c+d x)}}+\frac {(2 B-C) \cos (c+d x) \sin (c+d x)}{2 a d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 2.44, size = 395, normalized size = 1.79 \[ \frac {\sec (c+d x) \left ((91 B-48 C) (\sin (c+d x)+\tan (c+d x)) \tanh ^{-1}\left (\sqrt {1-\sec (c+d x)}\right )-40 B \sqrt {1-\sec (c+d x)} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-\sec (c+d x)\right ) (\sin (c+d x)+\tan (c+d x))-13 B \sin (c+d x) \sqrt {1-\sec (c+d x)}+\frac {13}{2} B \sin (2 (c+d x)) \sqrt {1-\sec (c+d x)}+18 B \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec (c+d x)}-52 \sqrt {2} B \sin (c+d x) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )-52 \sqrt {2} B \tan (c+d x) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+24 C \sin (c+d x) \sqrt {1-\sec (c+d x)}+8 C \sin (2 (c+d x)) \sqrt {1-\sec (c+d x)}+36 \sqrt {2} C \sin (c+d x) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )+36 \sqrt {2} C \tan (c+d x) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )\right )}{16 d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(-52*Sqrt[2]*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Sin[c + d*x] + 36*Sqrt[2]*C*ArcTanh[Sqrt[
1 - Sec[c + d*x]]/Sqrt[2]]*Sin[c + d*x] - 13*B*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + 24*C*Sqrt[1 - Sec[c + d*x
]]*Sin[c + d*x] + 18*B*Cos[c + d*x]^2*Sqrt[1 - Sec[c + d*x]]*Sin[c + d*x] + (13*B*Sqrt[1 - Sec[c + d*x]]*Sin[2
*(c + d*x)])/2 + 8*C*Sqrt[1 - Sec[c + d*x]]*Sin[2*(c + d*x)] - 52*Sqrt[2]*B*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqr
t[2]]*Tan[c + d*x] + 36*Sqrt[2]*C*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Tan[c + d*x] + (91*B - 48*C)*ArcTanh
[Sqrt[1 - Sec[c + d*x]]]*(Sin[c + d*x] + Tan[c + d*x]) - 40*B*Hypergeometric2F1[1/2, 3, 3/2, 1 - Sec[c + d*x]]
*Sqrt[1 - Sec[c + d*x]]*(Sin[c + d*x] + Tan[c + d*x])))/(16*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3
/2))

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fricas [A]  time = 5.91, size = 644, normalized size = 2.91 \[ \left [\frac {\sqrt {2} {\left ({\left (13 \, B - 9 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (13 \, B - 9 \, C\right )} \cos \left (d x + c\right ) + 13 \, B - 9 \, C\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 ) + {\left ({\left (19 \, B - 12 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (19 \, B - 12 \, C\right )} \cos \left (d x + c\right ) + 19 \, B - 12 \, 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 (2 \, B \cos \left (d x + c\right )^{3} - {\left (3 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (7 \, B - 6 \, 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 )}{8 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}, \frac {\sqrt {2} {\left ({\left (13 \, B - 9 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (13 \, B - 9 \, C\right )} \cos \left (d x + c\right ) + 13 \, B - 9 \, C\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 ) - {\left ({\left (19 \, B - 12 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (19 \, B - 12 \, C\right )} \cos \left (d x + c\right ) + 19 \, B - 12 \, 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 (2 \, B \cos \left (d x + c\right )^{3} - {\left (3 \, B - 4 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (7 \, B - 6 \, 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 )}{4 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/8*(sqrt(2)*((13*B - 9*C)*cos(d*x + c)^2 + 2*(13*B - 9*C)*cos(d*x + c) + 13*B - 9*C)*sqrt(-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)) + ((19*B - 12*C)*cos(d*x + c)^2 + 2*(19*B - 12*C)*cos(d*x +
c) + 19*B - 12*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*(2*B*cos(d*x + c)^3 - (3*B - 4*C)*cos(d*x +
c)^2 - (7*B - 6*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 +
2*a^2*d*cos(d*x + c) + a^2*d), 1/4*(sqrt(2)*((13*B - 9*C)*cos(d*x + c)^2 + 2*(13*B - 9*C)*cos(d*x + c) + 13*B
- 9*C)*sqrt(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))) -
((19*B - 12*C)*cos(d*x + c)^2 + 2*(19*B - 12*C)*cos(d*x + c) + 19*B - 12*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))) + (2*B*cos(d*x + c)^3 - (3*B - 4*C)*cos(d*x + c)^2 -
(7*B - 6*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2
*d*cos(d*x + c) + a^2*d)]

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giac [B]  time = 3.55, size = 673, normalized size = 3.05 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*(sqrt(2)*(13*B - 9*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)*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + (19*B - 12*C)*log(abs(147573952589676412928*(sqrt(-a)*tan(1/2*d*x + 1
/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 - 295147905179352825856*sqrt(2)*abs(a) - 442721857769029238784*
a)/abs(147573952589676412928*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 2951479
05179352825856*sqrt(2)*abs(a) - 442721857769029238784*a))/(sqrt(-a)*abs(a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) -
2*(sqrt(2)*B*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - sqrt(2)*C*a*sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*sqrt(-a*tan(1/2*
d*x + 1/2*c)^2 + a)*tan(1/2*d*x + 1/2*c)/a^3 - 4*sqrt(2)*(29*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*
d*x + 1/2*c)^2 + a))^6*B - 12*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^6*C - 133*
(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*B*a + 76*(sqrt(-a)*tan(1/2*d*x + 1/2*c
) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^4*C*a + 55*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2
*c)^2 + a))^2*B*a^2 - 36*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))^2*C*a^2 - 7*B*a
^3 + 4*C*a^3)/(((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*sqrt(-a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))/d

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maple [B]  time = 2.09, size = 1075, normalized size = 4.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16/d*(-1+cos(d*x+c))*(19*B*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)*cos(d*x+c)^2-12*C*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)*cos(d*x+c)
^2+38*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)*(-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))+26*B*sin(d*x+c)*cos(d*x+c)^2*(-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))-24*C*2^(1/2)*cos(d*x+c)*sin(d*x
+c)*(-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))-18*C*sin(d*x+c)*cos(d*x+c)^2*(-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))+19*B*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)+52*B*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))-12
*C*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)/co
s(d*x+c)*2^(1/2))*sin(d*x+c)-36*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+cos(d*x+c)))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+26*B*(-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)-8*B*cos(d*x+c)^5-18*C*(-2*co
s(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))*s
in(d*x+c)+20*B*cos(d*x+c)^4-16*C*cos(d*x+c)^4+16*B*cos(d*x+c)^3-8*C*cos(d*x+c)^3-28*B*cos(d*x+c)^2+24*C*cos(d*
x+c)^2)*(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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \cos \left (d x + c\right )^{3}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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