3.7.18 \(\int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\) [618]

Optimal. Leaf size=263 \[ -\frac {(15 A-11 B+7 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B+5 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-35 B+15 C) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(147 A-95 B+75 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \sec (c+d x)}} \]

[Out]

-1/2*(A-B+C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2)-1/4*(15*A-11*B+7*C)*arctanh(1/2*sin(d*x+c)*a
^(1/2)*sec(d*x+c)^(1/2)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)+1/10*(9*A-5*B+5*C)*sin(d*x+c)/a/d/se
c(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(1/2)-1/30*(39*A-35*B+15*C)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(
1/2)+1/30*(147*A-95*B+75*C)*sin(d*x+c)*sec(d*x+c)^(1/2)/a/d/(a+a*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.51, antiderivative size = 263, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {4169, 4107, 4098, 3893, 212} \begin {gather*} -\frac {(15 A-11 B+7 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sin (c+d x) \sqrt {\sec (c+d x)}}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {(9 A-5 B+5 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {(A-B+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}+\frac {(147 A-95 B+75 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{30 a d \sqrt {a \sec (c+d x)+a}}-\frac {(39 A-35 B+15 C) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

-1/2*((15*A - 11*B + 7*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])
])/(Sqrt[2]*a^(3/2)*d) - ((A - B + C)*Sin[c + d*x])/(2*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)) + ((9*
A - 5*B + 5*C)*Sin[c + d*x])/(10*a*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - ((39*A - 35*B + 15*C)*Sin[
c + d*x])/(30*a*d*Sqrt[Sec[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + ((147*A - 95*B + 75*C)*Sqrt[Sec[c + d*x]]*Sin
[c + d*x])/(30*a*d*Sqrt[a + a*Sec[c + d*x]])

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3893

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

Rule 4098

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[(
a*A*m - b*B*n)/(b*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, A
, B, m, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0] &&  !LeQ[m, -1]

Rule 4107

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 4169

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*A - b*B + a*C))*Cot[e + f*x]*(a + b*C
sc[e + f*x])^m*((d*Csc[e + f*x])^n/(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*Simp[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(m + n + 1) - a*(A*(m + n + 1) - C*(m -
n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A-B+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {\frac {1}{2} a (9 A-5 B+5 C)-a (3 A-3 B+C) \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac {(A-B+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B+5 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}+\frac {\int \frac {-\frac {1}{4} a^2 (39 A-35 B+15 C)+a^2 (9 A-5 B+5 C) \sec (c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}} \, dx}{5 a^3}\\ &=-\frac {(A-B+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B+5 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-35 B+15 C) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {2 \int \frac {\frac {1}{8} a^3 (147 A-95 B+75 C)-\frac {1}{4} a^3 (39 A-35 B+15 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {(A-B+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B+5 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-35 B+15 C) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(147 A-95 B+75 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \sec (c+d x)}}-\frac {(15 A-11 B+7 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A-B+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B+5 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-35 B+15 C) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(147 A-95 B+75 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \sec (c+d x)}}+\frac {(15 A-11 B+7 C) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,-\frac {a \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{2 a d}\\ &=-\frac {(15 A-11 B+7 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A-B+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A-5 B+5 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(39 A-35 B+15 C) \sin (c+d x)}{30 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(147 A-95 B+75 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{30 a d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 2.12, size = 148, normalized size = 0.56 \begin {gather*} \frac {\sec \left (\frac {1}{2} (c+d x)\right ) \sqrt {\sec (c+d x)} \left (-30 (15 A-11 B+7 C) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )+2 (141 A-85 B+75 C+3 (39 A+20 (-B+C)) \cos (c+d x)+(-6 A+10 B) \cos (2 (c+d x))+3 A \cos (3 (c+d x))) \sin \left (\frac {1}{2} (c+d x)\right )\right )}{60 a d \sqrt {a (1+\sec (c+d x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(Sec[c + d*x]^(5/2)*(a + a*Sec[c + d*x])^(3/2)),x]

[Out]

(Sec[(c + d*x)/2]*Sqrt[Sec[c + d*x]]*(-30*(15*A - 11*B + 7*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^2 + 2
*(141*A - 85*B + 75*C + 3*(39*A + 20*(-B + C))*Cos[c + d*x] + (-6*A + 10*B)*Cos[2*(c + d*x)] + 3*A*Cos[3*(c +
d*x)])*Sin[(c + d*x)/2]))/(60*a*d*Sqrt[a*(1 + Sec[c + d*x])])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(459\) vs. \(2(226)=452\).
time = 0.25, size = 460, normalized size = 1.75

method	result	size
default	\(-\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (225 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )-24 A \left (\cos ^{4}\left (d x +c \right )\right )-165 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+105 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \cos \left (d x +c \right )+225 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, A \sin \left (d x +c \right )+48 A \left (\cos ^{3}\left (d x +c \right )\right )-165 \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, B \sin \left (d x +c \right )-40 B \left (\cos ^{3}\left (d x +c \right )\right )+105 C \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-240 A \left (\cos ^{2}\left (d x +c \right )\right )+160 B \left (\cos ^{2}\left (d x +c \right )\right )-120 C \left (\cos ^{2}\left (d x +c \right )\right )-78 A \cos \left (d x +c \right )+70 B \cos \left (d x +c \right )-30 C \cos \left (d x +c \right )+294 A -190 B +150 C \right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \left (\cos ^{3}\left (d x +c \right )\right )}{60 d \sin \left (d x +c \right )^{3} a^{2}}\)	\(460\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/60/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(225*A*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(
1/2))*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)-24*A*cos(d*x+c)^4-165*B*arctan(1/2*sin(d*x+c)*(-2/(1+cos
(d*x+c)))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)+105*C*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)
))^(1/2))*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*cos(d*x+c)+225*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))
*(-2/(1+cos(d*x+c)))^(1/2)*A*sin(d*x+c)+48*A*cos(d*x+c)^3-165*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))
*(-2/(1+cos(d*x+c)))^(1/2)*B*sin(d*x+c)-40*B*cos(d*x+c)^3+105*C*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2
))*(-2/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-240*A*cos(d*x+c)^2+160*B*cos(d*x+c)^2-120*C*cos(d*x+c)^2-78*A*cos(d*x+
c)+70*B*cos(d*x+c)-30*C*cos(d*x+c)+294*A-190*B+150*C)*(1/cos(d*x+c))^(5/2)*cos(d*x+c)^3/sin(d*x+c)^3/a^2

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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Fricas [A]
time = 4.09, size = 530, normalized size = 2.02 \begin {gather*} \left [\frac {15 \, \sqrt {2} {\left ({\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A - 11 \, B + 7 \, C\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} + 2 \, \sqrt {2} \sqrt {a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right ) - 3 \, a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) + \frac {4 \, {\left (12 \, A \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (9 \, A - 5 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (147 \, A - 95 \, B + 75 \, 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 )}{\sqrt {\cos \left (d x + c\right )}}}{120 \, {\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 {15 \, \sqrt {2} {\left ({\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A - 11 \, B + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A - 11 \, B + 7 \, C\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )}}{a \sin \left (d x + c\right )}\right ) + \frac {2 \, {\left (12 \, A \cos \left (d x + c\right )^{4} - 4 \, {\left (3 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{3} + 12 \, {\left (9 \, A - 5 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (147 \, A - 95 \, B + 75 \, 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 )}{\sqrt {\cos \left (d x + c\right )}}}{60 \, {\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 ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/120*(15*sqrt(2)*((15*A - 11*B + 7*C)*cos(d*x + c)^2 + 2*(15*A - 11*B + 7*C)*cos(d*x + c) + 15*A - 11*B + 7*
C)*sqrt(a)*log(-(a*cos(d*x + c)^2 + 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c
))*sin(d*x + c) - 2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(12*A*cos(d*x + c)^4 - 4*
(3*A - 5*B)*cos(d*x + c)^3 + 12*(9*A - 5*B + 5*C)*cos(d*x + c)^2 + (147*A - 95*B + 75*C)*cos(d*x + c))*sqrt((a
*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c)
+ a^2*d), 1/60*(15*sqrt(2)*((15*A - 11*B + 7*C)*cos(d*x + c)^2 + 2*(15*A - 11*B + 7*C)*cos(d*x + c) + 15*A -
11*B + 7*C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin
(d*x + c))) + 2*(12*A*cos(d*x + c)^4 - 4*(3*A - 5*B)*cos(d*x + c)^3 + 12*(9*A - 5*B + 5*C)*cos(d*x + c)^2 + (1
47*A - 95*B + 75*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(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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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/sec(d*x+c)**(5/2)/(a+a*sec(d*x+c))**(3/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3007 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^(3/2)*sec(d*x + c)^(5/2)), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(5/2)),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/((a + a/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(5/2)), x)

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