3.3.65 \(\int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx\) [265]
Optimal. Leaf size=156 \[ \frac {(19 A+5 B) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {\sec (c+d x)} \sin (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(9 A-B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]
[Out]
-1/4*(A-B)*sec(d*x+c)^(3/2)*sin(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(9*A-B)*sec(d*x+c)^(3/2)*sin(d*x+c)/a/d/(
a+a*sec(d*x+c))^(3/2)+1/32*(19*A+5*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))/a^(5/2)/d*2^(1/2)
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Rubi [A]
time = 0.37, antiderivative size = 203, normalized size of antiderivative = 1.30, 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 = {4104, 4105,
4098, 3893, 212} \begin {gather*} \frac {(19 A+5 B) \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 {(9 A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{16 a^2 d \sqrt {a \sec (c+d x)+a}}+\frac {(5 A+3 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{16 a d (a \sec (c+d x)+a)^{3/2}}+\frac {(A-B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d (a \sec (c+d x)+a)^{5/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]
[Out]
((19*A + 5*B)*ArcTanh[(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[
2]*a^(5/2)*d) + ((A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((5*A + 3*B)*Sqrt
[Sec[c + d*x]]*Sin[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(3/2)) - ((9*A - B)*Sqrt[Sec[c + d*x]]*Sin[c + d*x])
/(16*a^2*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 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 4105
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[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*
(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[
A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0]
Rubi steps
\begin {align*} \int \frac {\sqrt {\sec (c+d x)} (A+B \sec (c+d x))}{(a+a \sec (c+d x))^{5/2}} \, dx &=\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {\int \frac {-\frac {1}{2} a (A-B)+2 a (A+B) \sec (c+d x)}{\sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2}\\ &=\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {\int \frac {-\frac {1}{4} a^2 (9 A-B)+\frac {1}{2} a^2 (5 A+3 B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)}} \, dx}{8 a^4}\\ &=\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(9 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}+\frac {(19 A+5 B) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2}\\ &=\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(9 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}-\frac {(19 A+5 B) \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 {(19 A+5 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 )}{16 \sqrt {2} a^{5/2} d}+\frac {(A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(5 A+3 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(9 A-B) \sqrt {\sec (c+d x)} \sin (c+d x)}{16 a^2 d \sqrt {a+a \sec (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 1.03, size = 103, normalized size = 0.66 \begin {gather*} \frac {\sqrt {\sec (c+d x)} \left (2 (19 A+5 B) \tanh ^{-1}\left (\sin \left (\frac {1}{2} (c+d x)\right )\right ) \cos ^3\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+(-13 A+5 B+(-9 A+B) \sec (c+d x)) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{16 a d (a (1+\sec (c+d x)))^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[Sec[c + d*x]]*(A + B*Sec[c + d*x]))/(a + a*Sec[c + d*x])^(5/2),x]
[Out]
(Sqrt[Sec[c + d*x]]*(2*(19*A + 5*B)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2]^3*Sec[c + d*x] + (-13*A + 5*B +
(-9*A + B)*Sec[c + d*x])*Tan[(c + d*x)/2]))/(16*a*d*(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. \(346\) vs.
\(2(131)=262\).
time = 7.90, size = 347, normalized size = 2.22
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method |
result |
size |
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default |
\(\frac {\sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \cos \left (d x +c \right ) \left (-1+\cos \left (d x +c \right )\right )^{2} \left (19 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )+13 A \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+5 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right ) \cos \left (d x +c \right )-5 B \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+19 A \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right )-4 A \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+5 B \arctan \left (\frac {\sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}}{2}\right ) \sin \left (d x +c \right )+4 B \cos \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}-9 A \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}+B \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\right )}{16 d \sin \left (d x +c \right )^{5} \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, a^{3}}\) |
\(347\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c))*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
[Out]
1/16/d*(1/cos(d*x+c))^(1/2)*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)*(-1+cos(d*x+c))^2*(19*A*arctan(1/2*
sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)*cos(d*x+c)+13*A*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2)+5*B*ar
ctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)*cos(d*x+c)-5*B*cos(d*x+c)^2*(-2/(1+cos(d*x+c)))^(1/2
)+19*A*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)-4*A*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)+5*
B*arctan(1/2*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2))*sin(d*x+c)+4*B*cos(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)-9*A*(-2
/(1+cos(d*x+c)))^(1/2)+B*(-2/(1+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. 5924 vs.
\(2 (131) = 262\).
time = 1.43, size = 5924, normalized size = 37.97 \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))*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
1/32*((19*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x
+ 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(4*d*x + 4*c)^2 + 304*(log(cos(1/2*d*x
+ 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x +
1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(3*d*x + 3*c)^2 + 684*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x +
1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x +
1/2*c) + 1))*cos(2*d*x + 2*c)^2 + 304*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x +
1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c)^
2 + 19*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x +
1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*sin(4*d*x + 4*c)^2 + 304*(log(cos(1/2*d*x + 1
/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/
2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*sin(3*d*x + 3*c)^2 + 684*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2
*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/
2*c) + 1))*sin(2*d*x + 2*c)^2 + 304*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2
*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*sin(d*x + c)^2 +
2*(76*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x +
1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(3*d*x + 3*c) + 114*(log(cos(1/2*d*x + 1/2
*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*
c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(2*d*x + 2*c) + 76*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2
+ 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c)
+ 1))*cos(d*x + c) + 19*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 19
*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 26*sin(7/2*d*x + 7/2*c) -
10*sin(5/2*d*x + 5/2*c) + 10*sin(3/2*d*x + 3/2*c) + 26*sin(1/2*d*x + 1/2*c))*cos(4*d*x + 4*c) + 104*(2*sin(3*
d*x + 3*c) + 3*sin(2*d*x + 2*c) + 2*sin(d*x + c))*cos(7/2*d*x + 7/2*c) + 8*(114*(log(cos(1/2*d*x + 1/2*c)^2 +
sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2
*sin(1/2*d*x + 1/2*c) + 1))*cos(2*d*x + 2*c) + 76*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin
(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*co
s(d*x + c) + 19*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 19*log(cos
(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) - 10*sin(5/2*d*x + 5/2*c) + 10*sin(
3/2*d*x + 3/2*c) + 26*sin(1/2*d*x + 1/2*c))*cos(3*d*x + 3*c) + 40*(3*sin(2*d*x + 2*c) + 2*sin(d*x + c))*cos(5/
2*d*x + 5/2*c) + 12*(76*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - l
og(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*cos(d*x + c) + 19*log(cos(1/
2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 19*log(cos(1/2*d*x + 1/2*c)^2 + sin(
1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) + 10*sin(3/2*d*x + 3/2*c) + 26*sin(1/2*d*x + 1/2*c))*cos(2*d*
x + 2*c) + 8*(19*log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - 19*log(co
s(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1) + 26*sin(1/2*d*x + 1/2*c))*cos(d*x
+ c) + 4*(38*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2
*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*sin(3*d*x + 3*c) + 57*(log(cos(1/2*d*x
+ 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x
+ 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*sin(2*d*x + 2*c) + 38*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/
2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1
/2*c) + 1))*sin(d*x + c) + 13*cos(7/2*d*x + 7/2*c) + 5*cos(5/2*d*x + 5/2*c) - 5*cos(3/2*d*x + 3/2*c) - 13*cos(
1/2*d*x + 1/2*c))*sin(4*d*x + 4*c) - 52*(4*cos(3*d*x + 3*c) + 6*cos(2*d*x + 2*c) + 4*cos(d*x + c) + 1)*sin(7/2
*d*x + 7/2*c) + 16*(57*(log(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - lo
g(cos(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + 1))*sin(2*d*x + 2*c) + 38*(log(co
s(1/2*d*x + 1/2*c)^2 + sin(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c) + 1) - log(cos(1/2*d*x + 1/2*c)^2 + sin
(1/2*d*x + 1/2*c)^2 - 2*sin(1/2*d*x + 1/2*c) + ...
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Fricas [A]
time = 3.78, size = 502, normalized size = 3.22 \begin {gather*} \left [\frac {\sqrt {2} {\left ({\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 19 \, A + 5 \, 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 ) - \frac {4 \, {\left ({\left (13 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (9 \, A - B\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 )}}}{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 (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (19 \, A + 5 \, B\right )} \cos \left (d x + c\right ) + 19 \, A + 5 \, 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 ) + \frac {2 \, {\left ({\left (13 \, A - 5 \, B\right )} \cos \left (d x + c\right )^{2} + {\left (9 \, A - B\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 )}}}{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))*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
[1/64*(sqrt(2)*((19*A + 5*B)*cos(d*x + c)^3 + 3*(19*A + 5*B)*cos(d*x + c)^2 + 3*(19*A + 5*B)*cos(d*x + c) + 19
*A + 5*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*((13*A - 5*B)*cos(
d*x + c)^2 + (9*A - B)*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^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)*((19*A + 5*B)*c
os(d*x + c)^3 + 3*(19*A + 5*B)*cos(d*x + c)^2 + 3*(19*A + 5*B)*cos(d*x + c) + 19*A + 5*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*((13*A - 5*B)*co
s(d*x + c)^2 + (9*A - B)*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^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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (A + B \sec {\left (c + d x \right )}\right ) \sqrt {\sec {\left (c + d x \right )}}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))*sec(d*x+c)**(1/2)/(a+a*sec(d*x+c))**(5/2),x)
[Out]
Integral((A + B*sec(c + d*x))*sqrt(sec(c + d*x))/(a*(sec(c + d*x) + 1))**(5/2), x)
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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))*sec(d*x+c)^(1/2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*sqrt(sec(d*x + c))/(a*sec(d*x + c) + a)^(5/2), x)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}}{{\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))*(1/cos(c + d*x))^(1/2))/(a + a/cos(c + d*x))^(5/2),x)
[Out]
int(((A + B/cos(c + d*x))*(1/cos(c + d*x))^(1/2))/(a + a/cos(c + d*x))^(5/2), x)
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