3.4.88 \(\int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\) [388]
Optimal. Leaf size=353 \[ -\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 (3 a A-A b+a B-3 b B) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b (a+b)^{3/2} d}-\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]
[Out]
-2/3*(4*A*a*b-B*a^2-3*B*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(
1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b^2/(a+b)^(3/2)/d+2/3*(3*A*a-A*b+B*a-3*B*b)*c
ot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b
*(1+sec(d*x+c))/(a-b))^(1/2)/(a-b)/b/(a+b)^(3/2)/d-2/3*(A*b-B*a)*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)
-2/3*(4*A*a*b-B*a^2-3*B*b^2)*tan(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)
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Rubi [A]
time = 0.39, antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps
used = 5, number of rules used = 4, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.129, Rules used = {4088, 4090,
3917, 4089} \begin {gather*} -\frac {2 \left (a^2 (-B)+4 a A b-3 b^2 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^2 d (a-b) (a+b)^{3/2}}-\frac {2 \left (a^2 (-B)+4 a A b-3 b^2 B\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {2 (A b-a B) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 (3 a A+a B-A b-3 b B) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b d (a-b) (a+b)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(-2*(4*a*A*b - a^2*B - 3*b^2*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(
a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*(a - b)*b^2*(a + b)^(
3/2)*d) + (2*(3*a*A - A*b + a*B - 3*b*B)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]],
(a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*(a - b)*b*(a
+ b)^(3/2)*d) - (2*(A*b - a*B)*Tan[c + d*x])/(3*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (2*(4*a*A*b - a^2
*B - 3*b^2*B)*Tan[c + d*x])/(3*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])
Rule 3917
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(Rt[a + b, 2]/(b*
f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin
[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4088
Int[csc[(e_.) + (f_.)*(x_)]*(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 + 1)/(f*(m + 1)*(a^2 - b^2))), x] + D
ist[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a
*B)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1]
Rule 4089
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rule 4090
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[Csc[e + f*x]*((1 +
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]
Rubi steps
\begin {align*} \int \frac {\sec (c+d x) (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx &=-\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {\sec (c+d x) \left (-\frac {3}{2} (a A-b B)+\frac {1}{2} (A b-a B) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\sec (c+d x) \left (\frac {1}{4} \left (3 a^2 A+A b^2-4 a b B\right )+\frac {1}{4} \left (4 a A b-a^2 B-3 b^2 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}\\ &=-\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 a A b-a^2 B-3 b^2 B\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2}+\frac {(a (3 A+B)-b (A+3 B)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) (a+b)^2}\\ &=-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 (a (3 A+B)-b (A+3 B)) \cot (c+d x) F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b (a+b)^{3/2} d}-\frac {2 (A b-a B) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a A b-a^2 B-3 b^2 B\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 18.15, size = 603, normalized size = 1.71 \begin {gather*} \frac {(b+a \cos (c+d x))^3 \sec ^2(c+d x) (A+B \sec (c+d x)) \left (-\frac {2 \left (-4 a A b+a^2 B+3 b^2 B\right ) \sin (c+d x)}{3 b \left (-a^2+b^2\right )^2}+\frac {2 \left (A b^2 \sin (c+d x)-a b B \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {2 \left (-5 a^2 A b \sin (c+d x)+A b^3 \sin (c+d x)+2 a^3 B \sin (c+d x)+2 a b^2 B \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (B+A \cos (c+d x)) (a+b \sec (c+d x))^{5/2}}+\frac {2 (b+a \cos (c+d x))^2 \sec ^{\frac {3}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} (A+B \sec (c+d x)) \left (2 (a+b) \left (-4 a A b+a^2 B+3 b^2 B\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) (3 a A+A b-a B-3 b B) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+\left (-4 a A b+a^2 B+3 b^2 B\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b \left (a^2-b^2\right )^2 d (B+A \cos (c+d x)) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[(Sec[c + d*x]*(A + B*Sec[c + d*x]))/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]^2*(A + B*Sec[c + d*x])*((-2*(-4*a*A*b + a^2*B + 3*b^2*B)*Sin[c + d*x])/(3
*b*(-a^2 + b^2)^2) + (2*(A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) +
(2*(-5*a^2*A*b*Sin[c + d*x] + A*b^3*Sin[c + d*x] + 2*a^3*B*Sin[c + d*x] + 2*a*b^2*B*Sin[c + d*x]))/(3*a*(a^2 -
b^2)^2*(b + a*Cos[c + d*x]))))/(d*(B + A*Cos[c + d*x])*(a + b*Sec[c + d*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^
2*Sec[c + d*x]^(3/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(A + B*Sec[c + d*x])*(2*(a + b)*(-4*a*A*b + a^2*B +
3*b^2*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*Ellipt
icE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(3*a*A + A*b - a*B - 3*b*B)*Sqrt[Cos[c + d*x]/(1
+ Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (
a - b)/(a + b)] + (-4*a*A*b + a^2*B + 3*b^2*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d
*x)/2]))/(3*b*(a^2 - b^2)^2*d*(B + A*Cos[c + d*x])*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*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. \(4212\) vs.
\(2(323)=646\).
time = 9.35, size = 4213, normalized size = 11.93
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\(\text {Expression too large to display}\) |
\(4213\) |
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
[Out]
-1/3/d*4^(1/2)*(A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*E
llipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4+3*B*cos(d*x+c)^2*b^4-A*cos(d*x+c)*b^4-3*B*(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),
((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a*b^3+4*B*cos(d*x+c)*a*b^3-4*A*cos(d*x+c)*a^3*(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),(
(a-b)/(a+b))^(1/2))*b-8*A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x
+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2-4*A*sin(d*x+c)*cos(d*x+c)*
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(
d*x+c),((a-b)/(a+b))^(1/2))*a*b^3+7*A*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c)
)/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2+5*A*sin(d*x+c)
*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(
d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3+2*B*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+
a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b+4*B*
sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Elliptic
E((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2+6*B*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c))
)^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*a*b^3-B*cos(d*x+c)*a^3*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d
*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b-5*B*cos(d*x+c)*a^2*b^2*(cos(d*x+c)/(1+cos(d*
x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a
-b)/(a+b))^(1/2))-7*B*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))
/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3-5*A*cos(d*x+c)^3*a^2*b^2+2*B*cos
(d*x+c)^3*a^3*b+2*B*cos(d*x+c)^3*a*b^3-4*A*cos(d*x+c)^2*a*b^3+4*B*cos(d*x+c)^2*a^2*b^2+3*B*(cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b)
)^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a*b^3-B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a
+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a^3*b-4*B*(cos(d*
x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),
((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a^2*b^2+3*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*co
s(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4
+B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a^4+4*A*cos(d*x+c)^3*a^3*b-3*B*cos(d*x+c)^3*a^2*b^2-4*A
*cos(d*x+c)^2*a^3*b+8*A*cos(d*x+c)^2*a^2*b^2-2*B*cos(d*x+c)^2*a^3*b-6*B*cos(d*x+c)^2*a*b^3-3*A*cos(d*x+c)*a^2*
b^2+4*A*cos(d*x+c)*a*b^3-B*cos(d*x+c)*a^2*b^2+A*cos(d*x+c)^3*b^4-3*B*cos(d*x+c)*b^4-B*cos(d*x+c)^3*a^4+B*cos(d
*x+c)^2*a^4-4*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+
cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a^3*b-4*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(
d*x+c)^2*sin(d*x+c)*a^2*b^2+4*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2
)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a^2*b^2+A*(cos(d*x+c)/(1+c
os(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a
+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a*b^3+B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))
/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a^3*b+3*B*(cos
(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+
c),((a-b)/(a+b))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*a^2*b^2+3*A*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a
*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*a^3*b-3*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipti
cF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^4+3*B*sin(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*
cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipti...
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Maxima [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(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [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(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
integral((B*sec(d*x + c)^2 + A*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x +
c)^2 + 3*a^2*b*sec(d*x + c) + a^3), x)
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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 ) \sec {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(5/2),x)
[Out]
Integral((A + B*sec(c + d*x))*sec(c + d*x)/(a + b*sec(c + d*x))**(5/2), x)
________________________________________________________________________________________
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(sec(d*x+c)*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*sec(d*x + c)/(b*sec(d*x + c) + a)^(5/2), x)
________________________________________________________________________________________
Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{\cos \left (c+d\,x\right )\,{\left (a+\frac {b}{\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)*(a + b/cos(c + d*x))^(5/2)),x)
[Out]
int((A + B/cos(c + d*x))/(cos(c + d*x)*(a + b/cos(c + d*x))^(5/2)), x)
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