3.6.59 \(\int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx\) [559]
Optimal. Leaf size=176 \[ \frac {(5 A+3 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}} \]
[Out]
-1/4*(A-B)*sin(d*x+c)/d/cos(d*x+c)^(5/2)/(a+a*sec(d*x+c))^(5/2)+1/16*(5*A+3*B)*sin(d*x+c)/a/d/cos(d*x+c)^(3/2)
/(a+a*sec(d*x+c))^(3/2)+1/32*(5*A+3*B)*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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/a^(5/2)/d*2^(1/2)
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Rubi [A]
time = 0.25, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 5, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3034, 4097,
3895, 3893, 212} \begin {gather*} \frac {(5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \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 )}{16 \sqrt {2} a^{5/2} d}+\frac {(5 A+3 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a \sec (c+d x)+a)^{3/2}}-\frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a \sec (c+d x)+a)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)),x]
[Out]
((5*A + 3*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])]*Sqrt[Cos[c
+ d*x]]*Sqrt[Sec[c + d*x]])/(16*Sqrt[2]*a^(5/2)*d) - ((A - B)*Sin[c + d*x])/(4*d*Cos[c + d*x]^(5/2)*(a + a*Sec
[c + d*x])^(5/2)) + ((5*A + 3*B)*Sin[c + d*x])/(16*a*d*Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(3/2))
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 3034
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[(a + b*Csc[e + f*x])^m*((
c + d*Csc[e + f*x])^n/(g*Csc[e + f*x])^p), x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] && !IntegerQ[p] && !(IntegerQ[m] && IntegerQ[n])
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 3895
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b*d*Co
t[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(a*f*(2*m + 1))), x] + Dist[d*((m + 1)/(b*(2*m + 1
))), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && E
qQ[a^2 - b^2, 0] && EqQ[m + n, 0] && LtQ[m, -2^(-1)] && IntegerQ[2*m]
Rule 4097
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[(a*A*m + b*B*(m + 1))/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])
^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && EqQ[m + n + 1, 0
] && LeQ[m, -1]
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\\ &=-\frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {\left ((5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx}{8 a}\\ &=-\frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}+\frac {\left ((5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=-\frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}-\frac {\left ((5 A+3 B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 )}{16 a^2 d}\\ &=\frac {(5 A+3 B) \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 ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \sin (c+d x)}{4 d \cos ^{\frac {5}{2}}(c+d x) (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sin (c+d x)}{16 a d \cos ^{\frac {3}{2}}(c+d x) (a+a \sec (c+d x))^{3/2}}\\ \end {align*}
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Mathematica [A]
time = 0.89, size = 108, normalized size = 0.61 \begin {gather*} \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left ((5 A+3 B) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right )+\frac {1}{2} (A+7 B+(5 A+3 B) \cos (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{4 d \cos ^{\frac {5}{2}}(c+d x) (a (1+\sec (c+d x)))^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sec[c + d*x])/(Cos[c + d*x]^(3/2)*(a + a*Sec[c + d*x])^(5/2)),x]
[Out]
(Cos[(c + d*x)/2]^2*((5*A + 3*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3 + ((A + 7*B + (5*A + 3*B)*Cos[c
+ d*x])*Tan[(c + d*x)/2])/2))/(4*d*Cos[c + d*x]^(5/2)*(a*(1 + Sec[c + d*x]))^(5/2))
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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(338\) vs.
\(2(147)=294\).
time = 13.45, size = 339, normalized size = 1.93
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| method |
result |
size |
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| default |
\(\frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \left (-1+\cos \left (d x +c \right )\right )^{2} \left (5 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )-5 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+3 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )-3 B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right )+5 A \sin \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )+4 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+3 B \sin \left (d x +c \right ) \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right )-4 B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \cos \left (d x +c \right )+A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+7 B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right ) \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{16 d \sin \left (d x +c \right )^{5} \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, a^{3}}\) |
\(339\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
[Out]
1/16/d*cos(d*x+c)^(1/2)*(-1+cos(d*x+c))^2*(5*A*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)*cos
(d*x+c)-5*A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+3*B*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*cos(d*
x+c)*sin(d*x+c)-3*B*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)^2+5*A*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x
+c)))^(1/2))+4*A*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+3*B*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))
^(1/2))-4*B*(-2/(1+cos(d*x+c)))^(1/2)*cos(d*x+c)+A*(-2/(1+cos(d*x+c)))^(1/2)+7*B*(-2/(1+cos(d*x+c)))^(1/2))*(a
*(1+cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)^5/(-2/(1+cos(d*x+c)))^(1/2)/a^3
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 5356 vs.
\(2 (147) = 294\).
time = 1.29, size = 5356, normalized size = 30.43 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
1/32*((4*(3*sin(3/2*d*x + 3/2*c) + 5*sin(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*sin(5/3*
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 5*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3
/2*c))))*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 40*(2*sin(3*d*x + 3*c) + 3*sin(4/3*arc
tan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*
c))))*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 24*(2*sin(3*d*x + 3*c) + 3*sin(4/3*arctan
2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))
))*cos(5/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 24*(3*sin(3/2*d*x + 3/2*c) - 5*sin(1/3*arcta
n2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))
+ 16*(3*sin(3/2*d*x + 3/2*c) - 5*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*cos(2/3*arctan2
(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 5*(16*cos(3*d*x + 3*c)^2 + 2*(4*cos(3*d*x + 3*c) + 6*cos(4/3*a
rctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/
2*c))) + 1)*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + cos(8/3*arctan2(sin(3/2*d*x + 3/2*c
), cos(3/2*d*x + 3/2*c)))^2 + 12*(4*cos(3*d*x + 3*c) + 4*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3
/2*c))) + 1)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 36*cos(4/3*arctan2(sin(3/2*d*x + 3
/2*c), cos(3/2*d*x + 3/2*c)))^2 + 8*(4*cos(3*d*x + 3*c) + 1)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x
+ 3/2*c))) + 16*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 16*sin(3*d*x + 3*c)^2 + 4*(2
*sin(3*d*x + 3*c) + 3*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 2*sin(2/3*arctan2(sin(3/2
*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sin(8/3*
arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 48*(sin(3*d*x + 3*c) + sin(2/3*arctan2(sin(3/2*d*x +
3/2*c), cos(3/2*d*x + 3/2*c))))*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 36*sin(4/3*arct
an2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 32*sin(3*d*x + 3*c)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c),
cos(3/2*d*x + 3/2*c))) + 16*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 8*cos(3*d*x + 3*
c) + 1)*log(cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 3/2
*c), cos(3/2*d*x + 3/2*c)))^2 + 2*sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 5*(16*co
s(3*d*x + 3*c)^2 + 2*(4*cos(3*d*x + 3*c) + 6*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*
cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/
2*d*x + 3/2*c))) + cos(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 12*(4*cos(3*d*x + 3*c) + 4
*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3
/2*d*x + 3/2*c))) + 36*cos(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 8*(4*cos(3*d*x + 3*c)
+ 1)*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 16*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), c
os(3/2*d*x + 3/2*c)))^2 + 16*sin(3*d*x + 3*c)^2 + 4*(2*sin(3*d*x + 3*c) + 3*sin(4/3*arctan2(sin(3/2*d*x + 3/2*
c), cos(3/2*d*x + 3/2*c))) + 2*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(8/3*arctan2(s
in(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + sin(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 +
48*(sin(3*d*x + 3*c) + sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin(4/3*arctan2(sin(3/2*
d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 36*sin(4/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 32*
sin(3*d*x + 3*c)*sin(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 16*sin(2/3*arctan2(sin(3/2*d*x
+ 3/2*c), cos(3/2*d*x + 3/2*c)))^2 + 8*cos(3*d*x + 3*c) + 1)*log(cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/
2*d*x + 3/2*c)))^2 + sin(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))^2 - 2*sin(1/3*arctan2(sin(3/
2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1) - 48*cos(3/2*d*x + 3/2*c)*sin(3*d*x + 3*c) + 80*cos(1/3*arctan2(si
n(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c)))*sin(3*d*x + 3*c) + 48*cos(3*d*x + 3*c)*sin(3/2*d*x + 3/2*c) - 4*(3*
cos(3/2*d*x + 3/2*c) + 5*cos(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 3*cos(5/3*arctan2(sin(
3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 5*cos(1/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))))*sin
(8/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 20*(4*cos(3*d*x + 3*c) + 6*cos(4/3*arctan2(sin(3/2
*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*cos(2/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 1)*si
n(7/3*arctan2(sin(3/2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) - 12*(4*cos(3*d*x + 3*c) + 6*cos(4/3*arctan2(sin(3/
2*d*x + 3/2*c), cos(3/2*d*x + 3/2*c))) + 4*cos(...
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Fricas [A]
time = 2.35, size = 478, normalized size = 2.72 \begin {gather*} \left [\frac {\sqrt {2} {\left ({\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 5 \, A + 3 \, B\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 ) + 4 \, {\left ({\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + A + 7 \, B\right )} \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 )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {\sqrt {2} {\left ({\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + 5 \, A + 3 \, B\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 ) - 2 \, {\left ({\left (5 \, A + 3 \, B\right )} \cos \left (d x + c\right ) + A + 7 \, B\right )} \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 )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
[1/64*(sqrt(2)*((5*A + 3*B)*cos(d*x + c)^3 + 3*(5*A + 3*B)*cos(d*x + c)^2 + 3*(5*A + 3*B)*cos(d*x + c) + 5*A +
3*B)*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*((5*A + 3*B)*cos(d*x +
c) + A + 7*B)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3
+ 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), -1/32*(sqrt(2)*((5*A + 3*B)*cos(d*x + c)^3 + 3*(5*A
+ 3*B)*cos(d*x + c)^2 + 3*(5*A + 3*B)*cos(d*x + c) + 5*A + 3*B)*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*((5*A + 3*B)*cos(d*x + c) + A + 7*B)*sqrt(
(a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x +
c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/cos(d*x+c)**(3/2)/(a+a*sec(d*x+c))**(5/2),x)
[Out]
Timed out
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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))/cos(d*x+c)^(3/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)/((a*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(3/2)), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^{3/2}\,{\left (a+\frac {a}{\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))/(cos(c + d*x)^(3/2)*(a + a/cos(c + d*x))^(5/2)),x)
[Out]
int((A + B/cos(c + d*x))/(cos(c + d*x)^(3/2)*(a + a/cos(c + d*x))^(5/2)), x)
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