3.3.42 \(\int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx\) [242]
Optimal. Leaf size=180 \[ \frac {a^{5/2} (20 A+19 B) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (4 A-9 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (4 A+7 B) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d} \]
[Out]
1/4*a^(5/2)*(20*A+19*B)*arcsinh(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d+1/2*a*B*(a+a*sec(d*x+c))^(3/2)*si
n(d*x+c)*sec(d*x+c)^(1/2)/d+1/4*a^3*(4*A-9*B)*sin(d*x+c)*sec(d*x+c)^(1/2)/d/(a+a*sec(d*x+c))^(1/2)+1/4*a^2*(4*
A+7*B)*sin(d*x+c)*sec(d*x+c)^(1/2)*(a+a*sec(d*x+c))^(1/2)/d
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Rubi [A]
time = 0.32, antiderivative size = 180, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {4103, 4100,
3886, 221} \begin {gather*} \frac {a^{5/2} (20 A+19 B) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{4 d}+\frac {a^3 (4 A-9 B) \sin (c+d x) \sqrt {\sec (c+d x)}}{4 d \sqrt {a \sec (c+d x)+a}}+\frac {a^2 (4 A+7 B) \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {a \sec (c+d x)+a}}{4 d}+\frac {a B \sin (c+d x) \sqrt {\sec (c+d x)} (a \sec (c+d x)+a)^{3/2}}{2 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]],x]
[Out]
(a^(5/2)*(20*A + 19*B)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(4*d) + (a^3*(4*A - 9*B)*Sqrt
[Sec[c + d*x]]*Sin[c + d*x])/(4*d*Sqrt[a + a*Sec[c + d*x]]) + (a^2*(4*A + 7*B)*Sqrt[Sec[c + d*x]]*Sqrt[a + a*S
ec[c + d*x]]*Sin[c + d*x])/(4*d) + (a*B*Sqrt[Sec[c + d*x]]*(a + a*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(2*d)
Rule 221
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[Rt[b, 2]*(x/Sqrt[a])]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]
Rule 3886
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-2*(a/(b
*f))*Sqrt[a*(d/b)], Subst[Int[1/Sqrt[1 + x^2/a], x], x, b*(Cot[e + f*x]/Sqrt[a + b*Csc[e + f*x]])], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[a*(d/b), 0]
Rule 4100
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]*(csc[(e_.) + (f_.)*(x_)]*(
B_.) + (A_)), x_Symbol] :> Simp[A*b^2*Cot[e + f*x]*((d*Csc[e + f*x])^n/(a*f*n*Sqrt[a + b*Csc[e + f*x]])), x] +
Dist[(A*b*(2*n + 1) + 2*a*B*n)/(2*a*d*n), Int[Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^(n + 1), x], x] /; Fr
eeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[A*b*(2*n + 1) + 2*a*B*n, 0] &&
LtQ[n, 0]
Rule 4103
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-b)*B*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*((d*Csc[e + f*x])^n/(f*(m +
n))), x] + Dist[1/(d*(m + n)), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n*Simp[a*A*d*(m + n) + B*(b*d
*n) + (A*b*d*(m + n) + a*B*d*(2*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] && GtQ[m, 1/2] && !LtQ[n, -1]
Rubi steps
\begin {align*} \int \frac {(a+a \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sqrt {\sec (c+d x)}} \, dx &=\frac {a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac {1}{2} \int \frac {(a+a \sec (c+d x))^{3/2} \left (\frac {1}{2} a (4 A-B)+\frac {1}{2} a (4 A+7 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {a^2 (4 A+7 B) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac {1}{2} \int \frac {\sqrt {a+a \sec (c+d x)} \left (\frac {1}{4} a^2 (4 A-9 B)+\frac {1}{4} a^2 (20 A+19 B) \sec (c+d x)\right )}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {a^3 (4 A-9 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (4 A+7 B) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}+\frac {1}{8} \left (a^2 (20 A+19 B)\right ) \int \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \, dx\\ &=\frac {a^3 (4 A-9 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (4 A+7 B) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}-\frac {\left (a^2 (20 A+19 B)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{a}}} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}\\ &=\frac {a^{5/2} (20 A+19 B) \sinh ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{4 d}+\frac {a^3 (4 A-9 B) \sqrt {\sec (c+d x)} \sin (c+d x)}{4 d \sqrt {a+a \sec (c+d x)}}+\frac {a^2 (4 A+7 B) \sqrt {\sec (c+d x)} \sqrt {a+a \sec (c+d x)} \sin (c+d x)}{4 d}+\frac {a B \sqrt {\sec (c+d x)} (a+a \sec (c+d x))^{3/2} \sin (c+d x)}{2 d}\\ \end {align*}
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Mathematica [A]
time = 1.94, size = 137, normalized size = 0.76 \begin {gather*} \frac {a^3 \left (20 A \text {ArcSin}\left (\sqrt {1-\sec (c+d x)}\right ) \tan (c+d x)-19 B \text {ArcSin}\left (\sqrt {\sec (c+d x)}\right ) \tan (c+d x)+\sqrt {-((-1+\sec (c+d x)) \sec (c+d x))} (8 A \sin (c+d x)+(4 A+11 B+2 B \sec (c+d x)) \tan (c+d x))\right )}{4 d \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[((a + a*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sqrt[Sec[c + d*x]],x]
[Out]
(a^3*(20*A*ArcSin[Sqrt[1 - Sec[c + d*x]]]*Tan[c + d*x] - 19*B*ArcSin[Sqrt[Sec[c + d*x]]]*Tan[c + d*x] + Sqrt[-
((-1 + Sec[c + d*x])*Sec[c + d*x])]*(8*A*Sin[c + d*x] + (4*A + 11*B + 2*B*Sec[c + d*x])*Tan[c + d*x])))/(4*d*S
qrt[1 - Sec[c + d*x]]*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. \(385\) vs.
\(2(154)=308\).
time = 7.79, size = 386, normalized size = 2.14
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| method |
result |
size |
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\(\frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (20 A \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-20 A \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )+19 B \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )-\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-19 B \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}\, \sin \left (d x +c \right ) \sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2}{1+\cos \left (d x +c \right )}}\, \left (1+\cos \left (d x +c \right )+\sin \left (d x +c \right )\right ) \sqrt {2}}{4}\right )-32 A \left (\cos ^{3}\left (d x +c \right )\right )+16 A \left (\cos ^{2}\left (d x +c \right )\right )-44 B \left (\cos ^{2}\left (d x +c \right )\right )+16 A \cos \left (d x +c \right )+36 B \cos \left (d x +c \right )+8 B \right ) \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, a^{2}}{16 d \sin \left (d x +c \right ) \cos \left (d x +c \right )}\) |
\(386\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
[Out]
1/16/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(20*A*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)*arc
tan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(1+cos(d*x+c)-sin(d*x+c))*2^(1/2))-20*A*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)*(-2/
(1+cos(d*x+c)))^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(1+cos(d*x+c)+sin(d*x+c))*2^(1/2))+19*B*cos(d*x+c)^
2*2^(1/2)*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(1+cos(d*x+c)-sin(d*x+c))*
2^(1/2))-19*B*cos(d*x+c)^2*2^(1/2)*sin(d*x+c)*(-2/(1+cos(d*x+c)))^(1/2)*arctan(1/4*(-2/(1+cos(d*x+c)))^(1/2)*(
1+cos(d*x+c)+sin(d*x+c))*2^(1/2))-32*A*cos(d*x+c)^3+16*A*cos(d*x+c)^2-44*B*cos(d*x+c)^2+16*A*cos(d*x+c)+36*B*c
os(d*x+c)+8*B)*(1/cos(d*x+c))^(1/2)/sin(d*x+c)/cos(d*x+c)*a^2
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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 14322 vs.
\(2 (154) = 308\).
time = 3.60, size = 14322, normalized size = 79.57 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
1/16*(4*(8*a^2*cos(1/2*d*x + 1/2*c)^4*sin(1/2*d*x + 1/2*c) + 16*a^2*cos(1/2*d*x + 1/2*c)^2*sin(1/2*d*x + 1/2*c
)^3 + 8*a^2*sin(1/2*d*x + 1/2*c)^5 + 5*(sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 +
2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^
2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*
a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1
/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2
*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(1/2*d*x + 1/2*c)^4 + 10*(sqrt(2)*a^2*log(2*cos(1/2*d*
x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)
- sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sq
rt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt
(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*
sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(1/2*d*x + 1
/2*c)^2*sin(1/2*d*x + 1/2*c)^2 + 5*(sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sq
rt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 +
2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a^2*
log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d
*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x
+ 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(1/2*d*x + 1/2*c)^4 + (8*a^2*sin(1/2*d*x + 1/2*c)^3 + (5*s
qrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2
)*sin(1/2*d*x + 1/2*c) + 2) - 5*sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2
)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 5*sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*
sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 5*sqrt(2)*a^2*
log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d
*x + 1/2*c) + 2) + 8*a^2*sin(1/2*d*x + 1/2*c))*cos(3/2*d*x + 3/2*c)^2 + 5*(sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2
*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt
(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*s
in(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos
(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2
*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*cos(1/2*d*x + 1/2*c)^2
+ (5*sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2
*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 5*sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2
*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) + 5*sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)
^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - 5*sqrt(
2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*si
n(1/2*d*x + 1/2*c) + 2) + 8*a^2*sin(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2*c)^2 + 5*(sqrt(2)*a^2*log(2*cos(1/2*d*
x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2)
- sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sq
rt(2)*sin(1/2*d*x + 1/2*c) + 2) + sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt
(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*
sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2))*sin(1/2*d*x + 1
/2*c)^2 + 2*(8*a^2*cos(1/2*d*x + 1/2*c)*sin(1/2*d*x + 1/2*c) + 5*(sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2
*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*l
og(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 + 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqrt(2)*sin(1/2*d*
x + 1/2*c) + 2) + sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/2*c)^2 - 2*sqrt(2)*cos(1/2*d*x
+ 1/2*c) + 2*sqrt(2)*sin(1/2*d*x + 1/2*c) + 2) - sqrt(2)*a^2*log(2*cos(1/2*d*x + 1/2*c)^2 + 2*sin(1/2*d*x + 1/
2*c)^2 - 2*sqrt(2)*cos(1/2*d*x + 1/2*c) - 2*sqr...
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Fricas [A]
time = 3.28, size = 454, normalized size = 2.52 \begin {gather*} \left [\frac {{\left ({\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {a} \log \left (\frac {a \cos \left (d x + c\right )^{3} - 7 \, a \cos \left (d x + c\right )^{2} - \frac {4 \, {\left (\cos \left (d x + c\right )^{2} - 2 \, \cos \left (d x + c\right )\right )} \sqrt {a} \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 )}} + 8 \, a}{\cos \left (d x + c\right )^{3} + \cos \left (d x + c\right )^{2}}\right ) + \frac {4 \, {\left (8 \, A a^{2} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\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 )}}}{16 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}, \frac {{\left ({\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + {\left (20 \, A + 19 \, B\right )} a^{2} \cos \left (d x + c\right )\right )} \sqrt {-a} \arctan \left (\frac {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 )}{a \cos \left (d x + c\right )^{2} - a \cos \left (d x + c\right ) - 2 \, a}\right ) + \frac {2 \, {\left (8 \, A a^{2} \cos \left (d x + c\right )^{2} + {\left (4 \, A + 11 \, B\right )} a^{2} \cos \left (d x + c\right ) + 2 \, B a^{2}\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 )}}}{8 \, {\left (d \cos \left (d x + c\right )^{2} + d \cos \left (d x + c\right )\right )}}\right ] \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
[1/16*(((20*A + 19*B)*a^2*cos(d*x + c)^2 + (20*A + 19*B)*a^2*cos(d*x + c))*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a
*cos(d*x + c)^2 - 4*(cos(d*x + c)^2 - 2*cos(d*x + c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x
+ c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3 + cos(d*x + c)^2)) + 4*(8*A*a^2*cos(d*x + c)^2 + (4*A + 11*B)*a
^2*cos(d*x + c) + 2*B*a^2)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x
+ c)^2 + d*cos(d*x + c)), 1/8*(((20*A + 19*B)*a^2*cos(d*x + c)^2 + (20*A + 19*B)*a^2*cos(d*x + c))*sqrt(-a)*a
rctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c)/(a*cos(d*x + c)^2 - a
*cos(d*x + c) - 2*a)) + 2*(8*A*a^2*cos(d*x + c)^2 + (4*A + 11*B)*a^2*cos(d*x + c) + 2*B*a^2)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(d*cos(d*x + c)^2 + d*cos(d*x + c))]
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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+a*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(1/2),x)
[Out]
Exception raised: SystemError >> excessive stack use: stack is 4371 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+a*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(a*sec(d*x + c) + a)^(5/2)/sqrt(sec(d*x + c)), 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 )\,{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(1/2),x)
[Out]
int(((A + B/cos(c + d*x))*(a + a/cos(c + d*x))^(5/2))/(1/cos(c + d*x))^(1/2), x)
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