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

Optimal. Leaf size=216 \[ \frac {(11 B-15 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(35 B-39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{30 a^2 d}+\frac {(B-C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {(5 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt {a \sec (c+d x)+a}}-\frac {(65 B-93 C) \tan (c+d x)}{15 a d \sqrt {a \sec (c+d x)+a}} \]

[Out]

1/4*(11*B-15*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*(B-C)*sec(
d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)-1/15*(65*B-93*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/10*(5*B-
9*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/30*(35*B-39*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2
/d

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Rubi [A]  time = 0.73, antiderivative size = 216, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {4072, 4019, 4021, 4010, 4001, 3795, 203} \[ \frac {(11 B-15 C) \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(35 B-39 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{30 a^2 d}+\frac {(B-C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}-\frac {(5 B-9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt {a \sec (c+d x)+a}}-\frac {(65 B-93 C) \tan (c+d x)}{15 a d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[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]

((11*B - 15*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)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) - ((65*B - 93*C)*Tan[c + d*x])/(15*a*d*Sqrt
[a + a*Sec[c + d*x]]) - ((5*B - 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) + ((35*B -
39*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(30*a^2*d)

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 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 4001

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 4010

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)), I
nt[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] /; Free
Q[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4019

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/
(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 - 1)*Simp[A
*(a*d*(n - 1)) - B*(b*d*(n - 1)) - d*(a*B*(m - n + 1) + 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] && LtQ[m, -2^(-1)] && GtQ[n, 0]

Rule 4021

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(B*d*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1))/(f*(m + n
)), x] + Dist[d/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m +
n) + a*B*m)*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] && GtQ[n, 1]

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 {\sec ^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 {\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\sec ^3(c+d x) \left (3 a (B-C)-\frac {1}{2} a (5 B-9 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(5 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {\sec ^2(c+d x) \left (-a^2 (5 B-9 C)+\frac {1}{4} a^2 (35 B-39 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(5 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(35 B-39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac {2 \int \frac {\sec (c+d x) \left (\frac {1}{8} a^3 (35 B-39 C)-\frac {1}{4} a^3 (65 B-93 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(65 B-93 C) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(35 B-39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}+\frac {(11 B-15 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=\frac {(B-C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(65 B-93 C) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(35 B-39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}-\frac {(11 B-15 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 {(11 B-15 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) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {(65 B-93 C) \tan (c+d x)}{15 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 B-9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(35 B-39 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{30 a^2 d}\\ \end {align*}

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Mathematica [A]  time = 2.35, size = 160, normalized size = 0.74 \[ \frac {\tan (c+d x) \left (\sqrt {1-\sec (c+d x)} \left (4 (5 B-3 C) \sec ^2(c+d x)-12 (5 B-9 C) \sec (c+d x)-95 B+12 C \sec ^3(c+d x)+147 C\right )+15 \sqrt {2} (11 B-15 C) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right )\right )}{30 d \sqrt {1-\sec (c+d x)} (a (\sec (c+d x)+1))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sec[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]

((15*Sqrt[2]*(11*B - 15*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sec[c + d*x] + Sqrt[1 -
Sec[c + d*x]]*(-95*B + 147*C - 12*(5*B - 9*C)*Sec[c + d*x] + 4*(5*B - 3*C)*Sec[c + d*x]^2 + 12*C*Sec[c + d*x]^
3))*Tan[c + d*x])/(30*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

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fricas [A]  time = 0.56, size = 504, normalized size = 2.33 \[ \left [\frac {15 \, \sqrt {2} {\left ({\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{2}\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 (95 \, B - 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (5 \, B - 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 12 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{120 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, -\frac {15 \, \sqrt {2} {\left ({\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (11 \, B - 15 \, C\right )} \cos \left (d x + c\right )^{2}\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 (95 \, B - 147 \, C\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (5 \, B - 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, {\left (5 \, B - 3 \, C\right )} \cos \left (d x + c\right ) - 12 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{60 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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]

[1/120*(15*sqrt(2)*((11*B - 15*C)*cos(d*x + c)^4 + 2*(11*B - 15*C)*cos(d*x + c)^3 + (11*B - 15*C)*cos(d*x + c)
^2)*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)) - 4*((95*B - 147*C)*cos(d*x + c)
^3 + 12*(5*B - 9*C)*cos(d*x + c)^2 - 4*(5*B - 3*C)*cos(d*x + c) - 12*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c)
)*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2), -1/60*(15*sqrt(2)*((11
*B - 15*C)*cos(d*x + c)^4 + 2*(11*B - 15*C)*cos(d*x + c)^3 + (11*B - 15*C)*cos(d*x + c)^2)*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))) + 2*((95*B - 147*C)*cos(d*x +
c)^3 + 12*(5*B - 9*C)*cos(d*x + c)^2 - 4*(5*B - 3*C)*cos(d*x + c) - 12*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x +
c))*sin(d*x + c))/(a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)]

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giac [A]  time = 2.60, size = 312, normalized size = 1.44 \[ \frac {\frac {15 \, \sqrt {2} {\left (11 \, B - 15 \, C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {{\left ({\left ({\left (\frac {15 \, \sqrt {2} {\left (B a^{3} - C a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {\sqrt {2} {\left (245 \, B a^{3} - 381 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, \sqrt {2} {\left (73 \, B a^{3} - 105 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {15 \, \sqrt {2} {\left (9 \, B a^{3} - 17 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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]

1/60*(15*sqrt(2)*(11*B - 15*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(tan(1/2*d*x + 1/2*c)^2 - 1)) - (((15*sqrt(2)*(B*a^3 - C*a^3)*tan(1/2*d*x + 1/2*c)^2/(a^2*sgn(t
an(1/2*d*x + 1/2*c)^2 - 1)) - sqrt(2)*(245*B*a^3 - 381*C*a^3)/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d
*x + 1/2*c)^2 + 5*sqrt(2)*(73*B*a^3 - 105*C*a^3)/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)^2
- 15*sqrt(2)*(9*B*a^3 - 17*C*a^3)/(a^2*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)))*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x
+ 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

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maple [B]  time = 1.84, size = 793, normalized size = 3.67 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(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]

-1/240/d*(-1+cos(d*x+c))*(165*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)^3*sin(d*x+c)-225*C*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(1+c
os(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))+495*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)-675*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*c
os(d*x+c)/(1+cos(d*x+c)))^(5/2)*sin(d*x+c)*cos(d*x+c)^2+495*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)-675*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)+165*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))*sin(d*x+c)-225*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)+760*B*cos(d*x+c)^4-1176*C*cos(d*x+c)^4-280*B*cos(d*x+c
)^3+312*C*cos(d*x+c)^3-640*B*cos(d*x+c)^2+960*C*cos(d*x+c)^2+160*B*cos(d*x+c)-192*C*cos(d*x+c)+96*C)*(a*(1+cos
(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)^3/a^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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))*sec(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 \[ \int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(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))*sec(c + d*x)**4/(a*(sec(c + d*x) + 1))**(3/2), x)

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