3.3.88 \(\int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx\) [288]

Optimal. Leaf size=248 \[ -\frac {(15 A+7 C) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {(9 A+5 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+5 C) \sin (c+d x)}{10 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(49 A+25 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}} \]

[Out]

-1/2*(A+C)*sin(d*x+c)/d/sec(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(3/2)-1/4*(15*A+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*C)*sin(d*x+c)/a/d/sec(d*x+c)^(3
/2)/(a+a*sec(d*x+c))^(1/2)-1/10*(13*A+5*C)*sin(d*x+c)/a/d/sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(1/2)+1/10*(49*A+2
5*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 = 0.48, antiderivative size = 248, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {4170, 4107, 4098, 3893, 212} \begin {gather*} -\frac {(15 A+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 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a \sec (c+d x)+a}}-\frac {(A+C) \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}+\frac {(49 A+25 C) \sin (c+d x) \sqrt {\sec (c+d x)}}{10 a d \sqrt {a \sec (c+d x)+a}}-\frac {(13 A+5 C) \sin (c+d x)}{10 a d \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + 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 + 7*C)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqr
t[2]*a^(3/2)*d) - ((A + C)*Sin[c + d*x])/(2*d*Sec[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2)) + ((9*A + 5*C)*Si
n[c + d*x])/(10*a*d*Sec[c + d*x]^(3/2)*Sqrt[a + a*Sec[c + d*x]]) - ((13*A + 5*C)*Sin[c + d*x])/(10*a*d*Sqrt[Se
c[c + d*x]]*Sqrt[a + a*Sec[c + d*x]]) + ((49*A + 25*C)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(10*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 4170

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(-a)*(A + C)*Cot[e + f*x]*(a + b*Csc[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[b*C*n + A*b
*(2*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, C, n}, x
] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \frac {A+C \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \, dx &=-\frac {(A+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 C)+a (3 A+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+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 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 {3}{4} a^2 (13 A+5 C)-a^2 (9 A+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+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 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+5 C) \sin (c+d x)}{10 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}-\frac {2 \int \frac {-\frac {3}{8} a^3 (49 A+25 C)+\frac {3}{4} a^3 (13 A+5 C) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4}\\ &=-\frac {(A+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 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+5 C) \sin (c+d x)}{10 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(49 A+25 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(15 A+7 C) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac {(A+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 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+5 C) \sin (c+d x)}{10 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(49 A+25 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}+\frac {(15 A+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+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+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 C) \sin (c+d x)}{10 a d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+a \sec (c+d x)}}-\frac {(13 A+5 C) \sin (c+d x)}{10 a d \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}}+\frac {(49 A+25 C) \sqrt {\sec (c+d x)} \sin (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 5.51, size = 331, normalized size = 1.33 \begin {gather*} \frac {(1+\sec (c+d x))^{3/2} \left (A+C \sec ^2(c+d x)\right ) \left (\frac {2 (47 A+25 C+(39 A+20 C) \cos (c+d x)-2 A \cos (2 (c+d x))+A \cos (3 (c+d x))) \sec ^3\left (\frac {1}{2} (c+d x)\right ) \sqrt {1+\sec (c+d x)} \left (-\sin \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {3}{2} (c+d x)\right )\right )}{\sec ^{\frac {3}{2}}(c+d x)}-5 \sqrt {2} (15 A+7 C) \cos ^2(c+d x) \cot (c+d x) \left (\log \left (1-2 \sec (c+d x)-3 \sec ^2(c+d x)-2 \sqrt {2} \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt {\tan ^2(c+d x)}\right )-\log \left (1-2 \sec (c+d x)-3 \sec ^2(c+d x)+2 \sqrt {2} \sqrt {\sec (c+d x)} \sqrt {1+\sec (c+d x)} \sqrt {\tan ^2(c+d x)}\right )\right ) \sqrt {\tan ^2(c+d x)}\right )}{40 d (A+2 C+A \cos (2 (c+d x))) (a (1+\sec (c+d x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((1 + Sec[c + d*x])^(3/2)*(A + C*Sec[c + d*x]^2)*((2*(47*A + 25*C + (39*A + 20*C)*Cos[c + d*x] - 2*A*Cos[2*(c
+ d*x)] + A*Cos[3*(c + d*x)])*Sec[(c + d*x)/2]^3*Sqrt[1 + Sec[c + d*x]]*(-Sin[(c + d*x)/2] + Sin[(3*(c + d*x))
/2]))/Sec[c + d*x]^(3/2) - 5*Sqrt[2]*(15*A + 7*C)*Cos[c + d*x]^2*Cot[c + d*x]*(Log[1 - 2*Sec[c + d*x] - 3*Sec[
c + d*x]^2 - 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[Tan[c + d*x]^2]] - Log[1 - 2*Sec[c + d*x
] - 3*Sec[c + d*x]^2 + 2*Sqrt[2]*Sqrt[Sec[c + d*x]]*Sqrt[1 + Sec[c + d*x]]*Sqrt[Tan[c + d*x]^2]])*Sqrt[Tan[c +
d*x]^2]))/(40*d*(A + 2*C + A*Cos[2*(c + d*x)])*(a*(1 + Sec[c + d*x]))^(3/2))

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Maple [A]
time = 20.08, size = 328, normalized size = 1.32

method	result	size
default	\(\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (-1+\cos \left (d x +c \right )\right ) \left (8 A \left (\cos ^{4}\left (d x +c \right )\right )-75 A \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )-35 C \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )-16 A \left (\cos ^{3}\left (d x +c \right )\right )-75 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-35 C \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right )+80 A \left (\cos ^{2}\left (d x +c \right )\right )+40 C \left (\cos ^{2}\left (d x +c \right )\right )+26 A \cos \left (d x +c \right )+10 C \cos \left (d x +c \right )-98 A -50 C \right ) \left (\cos ^{3}\left (d x +c \right )\right ) \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{20 d \sin \left (d x +c \right )^{3} a^{2}}\)	\(328\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/20/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))*(8*A*cos(d*x+c)^4-75*A*cos(d*x+c)*sin(d*x+c)*(-2/(1
+cos(d*x+c)))^(1/2)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))-35*C*cos(d*x+c)*sin(d*x+c)*(-2/(1+cos(d*x
+c)))^(1/2)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))-16*A*cos(d*x+c)^3-75*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)-35*C*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/2*sin(d*x+
c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)+80*A*cos(d*x+c)^2+40*C*cos(d*x+c)^2+26*A*cos(d*x+c)+10*C*cos(d*x+c)-9
8*A-50*C)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/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+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 = 1.97, size = 488, normalized size = 1.97 \begin {gather*} \left [\frac {5 \, \sqrt {2} {\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A + 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 (4 \, A \cos \left (d x + c\right )^{4} - 4 \, A \cos \left (d x + c\right )^{3} + 4 \, {\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (49 \, A + 25 \, 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 )}}}{40 \, {\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 {5 \, \sqrt {2} {\left ({\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, A + 7 \, C\right )} \cos \left (d x + c\right ) + 15 \, A + 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 (4 \, A \cos \left (d x + c\right )^{4} - 4 \, A \cos \left (d x + c\right )^{3} + 4 \, {\left (9 \, A + 5 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (49 \, A + 25 \, 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 )}}}{20 \, {\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+C*sec(d*x+c)^2)/sec(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

[1/40*(5*sqrt(2)*((15*A + 7*C)*cos(d*x + c)^2 + 2*(15*A + 7*C)*cos(d*x + c) + 15*A + 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*c
os(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) + 4*(4*A*cos(d*x + c)^4 - 4*A*cos(d*x + c)^3 + 4*(9*
A + 5*C)*cos(d*x + c)^2 + (49*A + 25*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqr
t(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), 1/20*(5*sqrt(2)*((15*A + 7*C)*cos(d*x
+ c)^2 + 2*(15*A + 7*C)*cos(d*x + c) + 15*A + 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*(4*A*cos(d*x + c)^4 - 4*A*cos(d*x + c)^3 + 4*(9*A + 5*C
)*cos(d*x + c)^2 + (49*A + 25*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+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+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 + 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 {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 + 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 + 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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