3.358 \(\int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx\)
Optimal. Leaf size=312 \[ \frac{2 (a-b) \sqrt{a+b} (15 a A-3 a B-5 A b+9 b B) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 b d}-\frac{2 (a-b) \sqrt{a+b} \left (3 a^2 B+20 a A b+9 b^2 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^2 d}+\frac{2 (3 a B+5 A b) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 d}+\frac{2 B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d} \]
[Out]
(-2*(a - b)*Sqrt[a + b]*(20*a*A*b + 3*a^2*B + 9*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))])/(
15*b^2*d) + (2*(a - b)*Sqrt[a + b]*(15*a*A - 5*A*b - 3*a*B + 9*b*B)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*S
ec[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))])/(15*b*d) + (2*(5*A*b + 3*a*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*B*(a + b*Sec[c +
d*x])^(3/2)*Tan[c + d*x])/(5*d)
________________________________________________________________________________________
Rubi [A] time = 0.570295, antiderivative size = 312, normalized size of antiderivative = 1.,
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 =
{4002, 4005, 3832, 4004} \[ -\frac{2 (a-b) \sqrt{a+b} \left (3 a^2 B+20 a A b+9 b^2 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^2 d}+\frac{2 (3 a B+5 A b) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 d}+\frac{2 (a-b) \sqrt{a+b} (15 a A-3 a B-5 A b+9 b B) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b d}+\frac{2 B \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]
[Out]
(-2*(a - b)*Sqrt[a + b]*(20*a*A*b + 3*a^2*B + 9*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))])/(
15*b^2*d) + (2*(a - b)*Sqrt[a + b]*(15*a*A - 5*A*b - 3*a*B + 9*b*B)*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*S
ec[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))])/(15*b*d) + (2*(5*A*b + 3*a*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*d) + (2*B*(a + b*Sec[c +
d*x])^(3/2)*Tan[c + d*x])/(5*d)
Rule 4002
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
Rule 4005
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]
Rule 3832
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(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)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4004
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 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rubi steps
\begin{align*} \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \, dx &=\frac{2 B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{2}{5} \int \sec (c+d x) \sqrt{a+b \sec (c+d x)} \left (\frac{1}{2} (5 a A+3 b B)+\frac{1}{2} (5 A b+3 a B) \sec (c+d x)\right ) \, dx\\ &=\frac{2 (5 A b+3 a B) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{4}{15} \int \frac{\sec (c+d x) \left (\frac{1}{4} \left (15 a^2 A+5 A b^2+12 a b B\right )+\frac{1}{4} \left (20 a A b+3 a^2 B+9 b^2 B\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 (5 A b+3 a B) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}+\frac{1}{15} ((a-b) (15 a A-5 A b-3 a B+9 b B)) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx+\frac{1}{15} \left (20 a A b+3 a^2 B+9 b^2 B\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx\\ &=-\frac{2 (a-b) \sqrt{a+b} \left (20 a A b+3 a^2 B+9 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}}}{15 b^2 d}+\frac{2 (a-b) \sqrt{a+b} (15 a A-5 A b-3 a B+9 b 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}}}{15 b d}+\frac{2 (5 A b+3 a B) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac{2 B (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 18.7551, size = 502, normalized size = 1.61 \[ \frac{\cos ^2(c+d x) (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \left (\frac{2 \left (3 a^2 B+20 a A b+9 b^2 B\right ) \sin (c+d x)}{15 b}+\frac{2}{15} \sec (c+d x) (6 a B \sin (c+d x)+5 A b \sin (c+d x))+\frac{2}{5} b B \tan (c+d x) \sec (c+d x)\right )}{d (a \cos (c+d x)+b) (A \cos (c+d x)+B)}-\frac{2 \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)) \left (-2 b (a+b) (3 a (5 A+B)+b (5 A+9 B)) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+\left (3 a^2 B+20 a A b+9 b^2 B\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)+2 (a+b) \left (3 a^2 B+20 a A b+9 b^2 B\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{15 b d \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b)^2 (A \cos (c+d x)+B)} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x]),x]
[Out]
(-2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x])*(2*(a + b)*(20*a*A*b
+ 3*a^2*B + 9*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]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(a + b)*(3*a*(5*A + B) + b*(5*A + 9*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)] + (20*a*A*b + 3*a^2*B + 9*b^2*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*
x)/2]^2*Tan[(c + d*x)/2]))/(15*b*d*(b + a*Cos[c + d*x])^2*(B + A*Cos[c + d*x])*Sqrt[Sec[(c + d*x)/2]^2]*Sec[c
+ d*x]^(5/2)) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(3/2)*(A + B*Sec[c + d*x])*((2*(20*a*A*b + 3*a^2*B + 9*b^
2*B)*Sin[c + d*x])/(15*b) + (2*Sec[c + d*x]*(5*A*b*Sin[c + d*x] + 6*a*B*Sin[c + d*x]))/15 + (2*b*B*Sec[c + d*x
]*Tan[c + d*x])/5))/(d*(b + a*Cos[c + d*x])*(B + A*Cos[c + d*x]))
________________________________________________________________________________________
Maple [B] time = 0.71, size = 2683, normalized size = 8.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x)
[Out]
2/15/d/b*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(-15*A*sin(d*x+c)*cos(d*x+c)^3
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-5*A*cos(d*x+c)^3*b^3-20*A*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^
(1/2))*a*b^2-15*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x
+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+20*A*sin(d*x+c)*cos(d*x+c)^3*(co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d
*x+c),((a-b)/(a+b))^(1/2))*a^2*b+20*A*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*
cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-3*B*sin(d*x+
c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-
1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-12*B*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))
*a*b^2+3*B*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))
^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+9*B*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((
a-b)/(a+b))^(1/2))*a*b^2-20*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+
c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+20*A*sin(d*x+c)*cos(
d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d
*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+20*A*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-
3*B*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*
EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b-12*B*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(
a+b))^(1/2))*a*b^2+3*B*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(co
s(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+9*B*sin(d*x+c)*cos(d*x+c)^2
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+3*B*b^3+5*A*cos(d*x+c)*b^3-3*B*cos(d*x+c)^4*a^3+3*B*cos(d*x+c)^3*a^3-9*B*
cos(d*x+c)^3*b^3+6*B*cos(d*x+c)^2*b^3-5*A*cos(d*x+c)^4*a*b^2-6*B*cos(d*x+c)^4*a^2*b-9*B*cos(d*x+c)^4*a*b^2+20*
A*cos(d*x+c)^3*a^2*b-20*A*cos(d*x+c)^3*a*b^2-3*B*cos(d*x+c)^3*a^2*b+25*A*cos(d*x+c)^2*a*b^2+9*B*cos(d*x+c)^2*a
^2*b+9*B*cos(d*x+c)*a*b^2-20*A*cos(d*x+c)^4*a^2*b-5*A*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^
3-9*B*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2
)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+3*B*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a
+b))^(1/2))*a^3+9*B*sin(d*x+c)*cos(d*x+c)^3*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d
*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3-5*A*sin(d*x+c)*cos(d*x+c)^2*(cos
(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*
x+c),((a-b)/(a+b))^(1/2))*b^3-9*B*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(
d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+3*B*sin(d*x+c)*cos
(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(
d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3+9*B*sin(d*x+c)*cos(d*x+c)^2*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(
a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3)/(b+
a*cos(d*x+c))/cos(d*x+c)^2/sin(d*x+c)^5
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sec \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/2)*sec(d*x + c), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b \sec \left (d x + c\right )^{3} + A a \sec \left (d x + c\right ) +{\left (B a + A b\right )} \sec \left (d x + c\right )^{2}\right )} \sqrt{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")
[Out]
integral((B*b*sec(d*x + c)^3 + A*a*sec(d*x + c) + (B*a + A*b)*sec(d*x + c)^2)*sqrt(b*sec(d*x + c) + a), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (A + B \sec{\left (c + d x \right )}\right ) \left (a + b \sec{\left (c + d x \right )}\right )^{\frac{3}{2}} \sec{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(a+b*sec(d*x+c))**(3/2)*(A+B*sec(d*x+c)),x)
[Out]
Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))**(3/2)*sec(c + d*x), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \sec \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)*(a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(3/2)*sec(d*x + c), x)