3.371 \(\int \frac{\sec ^3(c+d x) (A+B \sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx\)
Optimal. Leaf size=329 \[ \frac{2 \sqrt{a+b} \left (-8 a^2 B+2 a b (5 A+B)+b^2 (5 A-9 B)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 b^3 d}+\frac{2 (a-b) \sqrt{a+b} \left (-8 a^2 B+10 a A b-9 b^2 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^4 d}+\frac{2 (5 A b-4 a B) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 b^2 d}+\frac{2 B \tan (c+d x) \sec (c+d x) \sqrt{a+b \sec (c+d x)}}{5 b d} \]
[Out]
(2*(a - b)*Sqrt[a + b]*(10*a*A*b - 8*a^2*B - 9*b^2*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/S
qrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(1
5*b^4*d) + (2*Sqrt[a + b]*(b^2*(5*A - 9*B) - 8*a^2*B + 2*a*b*(5*A + 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))])/(15*b^3*d) + (2*(5*A*b - 4*a*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*b^2*d) + (2*B*Sec[c
+ d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d)
________________________________________________________________________________________
Rubi [A] time = 0.617519, antiderivative size = 329, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.152, Rules used =
{4033, 4082, 4005, 3832, 4004} \[ \frac{2 \sqrt{a+b} \left (-8 a^2 B+2 a b (5 A+B)+b^2 (5 A-9 B)\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^3 d}+\frac{2 (a-b) \sqrt{a+b} \left (-8 a^2 B+10 a A b-9 b^2 B\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^4 d}+\frac{2 (5 A b-4 a B) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 b^2 d}+\frac{2 B \tan (c+d x) \sec (c+d x) \sqrt{a+b \sec (c+d x)}}{5 b d} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x]
[Out]
(2*(a - b)*Sqrt[a + b]*(10*a*A*b - 8*a^2*B - 9*b^2*B)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/S
qrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(1
5*b^4*d) + (2*Sqrt[a + b]*(b^2*(5*A - 9*B) - 8*a^2*B + 2*a*b*(5*A + 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))])/(15*b^3*d) + (2*(5*A*b - 4*a*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*b^2*d) + (2*B*Sec[c
+ d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d)
Rule 4033
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(B*d^2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2))/(
b*f*(m + n)), x] + Dist[d^2/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 2)*Simp[a*B*(n - 2)
+ B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e,
f, A, B, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + n, 0] && !IGtQ[m, 1]
Rule 4082
Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m
+ 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && !LtQ[m, -1]
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 \frac{\sec ^3(c+d x) (A+B \sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx &=\frac{2 B \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b d}+\frac{2 \int \frac{\sec (c+d x) \left (a B+\frac{3}{2} b B \sec (c+d x)+\frac{1}{2} (5 A b-4 a B) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{5 b}\\ &=\frac{2 (5 A b-4 a B) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac{2 B \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b d}+\frac{4 \int \frac{\sec (c+d x) \left (\frac{1}{4} b (5 A b+2 a B)-\frac{1}{4} \left (10 a A b-8 a^2 B-9 b^2 B\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b^2}\\ &=\frac{2 (5 A b-4 a B) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac{2 B \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b d}-\frac{\left (10 a A b-8 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}{15 b^2}+\frac{\left (b^2 (5 A-9 B)-8 a^2 B+2 a b (5 A+B)\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b^2}\\ &=\frac{2 (a-b) \sqrt{a+b} \left (10 a A b-8 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^4 d}+\frac{2 \sqrt{a+b} \left (b^2 (5 A-9 B)-8 a^2 B+2 a b (5 A+B)\right ) \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^3 d}+\frac{2 (5 A b-4 a B) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac{2 B \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b d}\\ \end{align*}
Mathematica [B] time = 23.224, size = 3000, normalized size = 9.12 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x]
[Out]
((b + a*Cos[c + d*x])*Sec[c + d*x]*((2*(-10*a*A*b + 8*a^2*B + 9*b^2*B)*Sin[c + d*x])/(15*b^3) + (2*Sec[c + d*x
]*(5*A*b*Sin[c + d*x] - 4*a*B*Sin[c + d*x]))/(15*b^2) + (2*B*Sec[c + d*x]*Tan[c + d*x])/(5*b)))/(d*Sqrt[a + b*
Sec[c + d*x]]) - (2*((2*a*A)/(3*b*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (3*B)/(5*Sqrt[b + a*Cos[c + d
*x]]*Sqrt[Sec[c + d*x]]) - (8*a^2*B)/(15*b^2*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) + (A*Sqrt[Sec[c + d*
x]])/(3*Sqrt[b + a*Cos[c + d*x]]) + (2*a^2*A*Sqrt[Sec[c + d*x]])/(3*b^2*Sqrt[b + a*Cos[c + d*x]]) - (8*a^3*B*S
qrt[Sec[c + d*x]])/(15*b^3*Sqrt[b + a*Cos[c + d*x]]) - (7*a*B*Sqrt[Sec[c + d*x]])/(15*b*Sqrt[b + a*Cos[c + d*x
]]) + (2*a^2*A*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b^2*Sqrt[b + a*Cos[c + d*x]]) - (8*a^3*B*Cos[2*(c + d*x
)]*Sqrt[Sec[c + d*x]])/(15*b^3*Sqrt[b + a*Cos[c + d*x]]) - (3*a*B*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*b*Sq
rt[b + a*Cos[c + d*x]]))*Sqrt[Sec[c + d*x]]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b)*(-10*a*A*b + 8*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]))]*E
llipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(8*a^2*B + 2*a*b*(-5*A + B) + b^2*(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)] + (-10*a*A*b + 8*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^3*d*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[a + b*Sec[c + d*x]]*(-(a*Sqrt[Cos[(c + d*x
)/2]^2*Sec[c + d*x]]*Sin[c + d*x]*(2*(a + b)*(-10*a*A*b + 8*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*(8*a^2*B + 2*a*b*(-5*A + B) + b^2*(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)] + (-10*a*A*b + 8
*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^3*(b + a*Cos[c
+ d*x])^(3/2)*Sqrt[Sec[(c + d*x)/2]^2]) + (Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Tan[(c + d*x)/2]*(2*(a + b)*
(-10*a*A*b + 8*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*(8*a^2*B + 2*a*b*(-5*A + B) + b^2*(
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]))]*Elli
pticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-10*a*A*b + 8*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^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) - (2*S
qrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(((-10*a*A*b + 8*a^2*B + 9*b^2*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(
c + d*x)/2]^4)/2 + ((a + b)*(-10*a*A*b + 8*a^2*B + 9*b^2*B)*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)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 -
Sin[c + d*x]/(1 + Cos[c + d*x])))/Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])] - (b*(8*a^2*B + 2*a*b*(-5*A + B) + b^
2*(5*A + 9*B))*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)]*((Cos[c + d*x]*Sin[c + d*x])/(1 + Cos[c + d*x])^2 - Sin[c + d*x]/(1 + Cos[c + d*x])))/Sqrt[Cos[c
+ d*x]/(1 + Cos[c + d*x])] + ((a + b)*(-10*a*A*b + 8*a^2*B + 9*b^2*B)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*E
llipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b +
a*Cos[c + d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c +
d*x]))] - (b*(8*a^2*B + 2*a*b*(-5*A + B) + b^2*(5*A + 9*B))*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*EllipticF[A
rcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*(-((a*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x]))) + ((b + a*Cos[c +
d*x])*Sin[c + d*x])/((a + b)*(1 + Cos[c + d*x])^2)))/Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*x]))] -
a*(-10*a*A*b + 8*a^2*B + 9*b^2*B)*Cos[c + d*x]*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] - (-10*a*A*b
+ 8*a^2*B + 9*b^2*B)*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Sin[c + d*x]*Tan[(c + d*x)/2] + (-10*a*A*b + 8*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]^2 - (b*(8*a^2*B + 2*a*b*(
-5*A + B) + b^2*(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]))]*Sec[(c + d*x)/2]^2)/(Sqrt[1 - Tan[(c + d*x)/2]^2]*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])
+ ((a + b)*(-10*a*A*b + 8*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]))]*Sec[(c + d*x)/2]^2*Sqrt[1 - ((a - b)*Tan[(c + d*x)/2]^2)/(a + b)])/Sqrt[1 - Tan[(c
+ d*x)/2]^2]))/(15*b^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]) - ((2*(a + b)*(-10*a*A*b + 8*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]))]*Ellip
ticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(8*a^2*B + 2*a*b*(-5*A + B) + b^2*(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)] + (-10*a*A*b + 8*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])*(-(Cos[(c + d*x)/2]*Sec[c + d*x]*Sin[(c + d*x)/2]) + Cos[(c + d*x)/2]^2*Sec[c + d*x]*Ta
n[c + d*x]))/(15*b^3*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]))
)
________________________________________________________________________________________
Maple [B] time = 0.756, size = 2499, normalized size = 7.6 \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)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x)
[Out]
-2/15/d/b^3*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(5*A*cos(d*x+c)^3*b^3-10*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)*Elli
pticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2+10*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^2*b+10*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+8*B*sin(d*x+c)*cos(d*x+c)^3*(c
os(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+2*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^2-8*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^2-10*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+10*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+10*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+8*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+2*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^2-8
*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)*E
llipticE((-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))/sin(d*x+c),((a-b)/(a+
b))^(1/2))*a*b^2-3*B*b^3-5*A*cos(d*x+c)*b^3+8*B*cos(d*x+c)^4*a^3-8*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^2-4*B*cos(d*x+c)^4*a^2*b+9*B*cos(d*x+c)^4*a*b^2+10*A*cos(d*x+c)^3*a^2*b
-10*A*cos(d*x+c)^3*a*b^2+8*B*cos(d*x+c)^3*a^2*b-10*B*cos(d*x+c)^3*a*b^2+5*A*cos(d*x+c)^2*a*b^2-4*B*cos(d*x+c)^
2*a^2*b+B*cos(d*x+c)*a*b^2-10*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^3-8*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*(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)*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^3-8*B*sin(d*x+c)*co
s(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
________________________________________________________________________________________
Maxima [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{B \sec \left (d x + c\right )^{4} + A \sec \left (d x + c\right )^{3}}{\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)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral((B*sec(d*x + c)^4 + A*sec(d*x + c)^3)/sqrt(b*sec(d*x + c) + a), x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (A + B \sec{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\sqrt{a + b \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)**3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(1/2),x)
[Out]
Integral((A + B*sec(c + d*x))*sec(c + d*x)**3/sqrt(a + b*sec(c + d*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{\sqrt{b \sec \left (d x + c\right ) + a}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*sec(d*x + c)^3/sqrt(b*sec(d*x + c) + a), x)