3.4.94 \(\int \frac {\sec ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [394]

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

[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 = 0.48, 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 = {4157, 4104, 4106, 4095, 4086, 3880, 209} \begin {gather*} \frac {(11 B-15 C) \text {ArcTan}\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}} \end {gather*}

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 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 3880

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 4086

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 4095

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)),
Int[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] /; Fr
eeQ[{a, b, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] &&  !LtQ[m, -1]

Rule 4104

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 4106

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 4157

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) \text {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 = 2.36, size = 160, normalized size = 0.74 \begin {gather*} \frac {\left (15 \sqrt {2} (11 B-15 C) \tanh ^{-1}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+\sqrt {1-\sec (c+d x)} \left (-95 B+147 C-12 (5 B-9 C) \sec (c+d x)+4 (5 B-3 C) \sec ^2(c+d x)+12 C \sec ^3(c+d x)\right )\right ) \tan (c+d x)}{30 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \end {gather*}

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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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(792\) vs. \(2(189)=378\).
time = 14.07, size = 793, normalized size = 3.67

method	result	size
default	\(-\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (165 B \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right )-225 C \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right )+495 B \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-675 C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right )+495 B \sin \left (d x +c \right ) \cos \left (d x +c \right ) \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}}-675 C \sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right )+165 B \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \sin \left (d x +c \right )-225 C \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \ln \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sin \left (d x +c \right )}\right ) \sin \left (d x +c \right )+760 B \left (\cos ^{4}\left (d x +c \right )\right )-1176 C \left (\cos ^{4}\left (d x +c \right )\right )-280 B \left (\cos ^{3}\left (d x +c \right )\right )+312 C \left (\cos ^{3}\left (d x +c \right )\right )-640 B \left (\cos ^{2}\left (d x +c \right )\right )+960 C \left (\cos ^{2}\left (d x +c \right )\right )+160 B \cos \left (d x +c \right )-192 C \cos \left (d x +c \right )+96 C \right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{240 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{2} a^{2}}\)	\(793\)

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,method=_RETURNVERBOSE)

[Out]

-1/240/d*(-1+cos(d*x+c))*(165*B*sin(d*x+c)*cos(d*x+c)^3*(-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))-225*C*sin(d*x+c)*cos(d*x+c)^3*(-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*sin(d*x+
c)*cos(d*x+c)^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))*(-2*cos(d*x+c)/(
1+cos(d*x+c)))^(5/2)-675*C*sin(d*x+c)*cos(d*x+c)^2*(-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))+495*B*sin(d*x+c)*cos(d*x+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)-675*C*sin(d*x+c)*cos(
d*x+c)*(-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))+165*B*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)-225*C*(-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)+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)/sin(d*x+c)^3/cos(d*x+c)^2/a^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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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Fricas [A]
time = 2.74, size = 504, normalized size = 2.33 \begin {gather*} \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 ] \end {gather*}

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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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \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 \end {gather*}

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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Giac [A]
time = 1.75, size = 277, normalized size = 1.28 \begin {gather*} -\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 (\cos \left (d x + c\right )\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 (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (245 \, B a^{3} - 381 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\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 (\cos \left (d x + c\right )\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 (\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 )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{60 \, d} \end {gather*}

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(cos(d*x + c))) - (((15*sqrt(2)*(B*a^3 - C*a^3)*tan(1/2*d*x + 1/2*c)^2/(a^2*sgn(cos(d*x + c)))
- sqrt(2)*(245*B*a^3 - 381*C*a^3)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 5*sqrt(2)*(73*B*a^3 - 105
*C*a^3)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 15*sqrt(2)*(9*B*a^3 - 17*C*a^3)/(a^2*sgn(cos(d*x + c
))))*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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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