\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{3/2}} \, dx\) [964]

Optimal result

Integrand size = 43, antiderivative size = 510 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (40 a^3 b B-25 a b^3 B-6 a^2 b^2 (5 A-4 C)-48 a^4 C+3 b^4 (5 A+3 C)\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^5 \sqrt {a+b} d}+\frac {2 \left (a^2 b (40 B-36 C)-48 a^3 C-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^4 \sqrt {a+b} d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \]

[Out]

2/15*(40*B*a^3*b-25*B*a*b^3-6*a^2*b^2*(5*A-4*C)-48*a^4*C+3*b^4*(5*A+3*C))*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)/b^5/d/
(a+b)^(1/2)+2/15*(a^2*b*(40*B-36*C)-48*a^3*C-6*a*b^2*(5*A-5*B+2*C)-b^3*(15*A-5*B+9*C))*cot(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)/b^4/d/(a+b)^(1/2)-2*(A*b^2-a*(B*b-C*a))*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)+2/
15*(20*B*a^2*b-5*B*b^3-3*a*b^2*(5*A-3*C)-24*a^3*C)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d+2/5*(5*A*
b^2-5*B*a*b+6*C*a^2-C*b^2)*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/(a^2-b^2)/d

Rubi [A] (verified)

Time = 1.51 (sec) , antiderivative size = 510, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.140, Rules used = {4183, 4177, 4167, 4090, 3917, 4089} \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 \tan (c+d x) \sec ^2(c+d x) \left (A b^2-a (b B-a C)\right )}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {2 \tan (c+d x) \sec (c+d x) \left (6 a^2 C-5 a b B+5 A b^2-b^2 C\right ) \sqrt {a+b \sec (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}+\frac {2 \cot (c+d x) \left (-48 a^3 C+a^2 b (40 B-36 C)-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{15 b^4 d \sqrt {a+b}}+\frac {2 \tan (c+d x) \left (-24 a^3 C+20 a^2 b B-3 a b^2 (5 A-3 C)-5 b^3 B\right ) \sqrt {a+b \sec (c+d x)}}{15 b^3 d \left (a^2-b^2\right )}+\frac {2 \cot (c+d x) \left (-48 a^4 C+40 a^3 b B-6 a^2 b^2 (5 A-4 C)-25 a b^3 B+3 b^4 (5 A+3 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{15 b^5 d \sqrt {a+b}} \]

[In]

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(2*(40*a^3*b*B - 25*a*b^3*B - 6*a^2*b^2*(5*A - 4*C) - 48*a^4*C + 3*b^4*(5*A + 3*C))*Cot[c + d*x]*EllipticE[Arc
Sin[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^5*Sqrt[a + b]*d) + (2*(a^2*b*(40*B - 36*C) - 48*a^3*C - 6*a*b^2*(5*A - 5*B +
2*C) - b^3*(15*A - 5*B + 9*C))*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^4*Sqrt[a + b]*d) -
(2*(A*b^2 - a*(b*B - a*C))*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(20*a
^2*b*B - 5*b^3*B - 3*a*b^2*(5*A - 3*C) - 24*a^3*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(15*b^3*(a^2 - b^2)*
d) + (2*(5*A*b^2 - 5*a*b*B + 6*a^2*C - b^2*C)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b^2*(a^2
- b^2)*d)

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 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]

Rule 4167

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 4177

Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(
e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^
(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m +
2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B*(m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C,
m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4183

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-d)*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*
(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 1)/(b*f*(a^2 - b^2)*(m + 1))), x] + Dist[d/(b*(a^2 - b^2)*
(m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 1)*Simp[A*b^2*(n - 1) - a*(b*B - a*C)*(n - 1)
+ b*(a*A - b*B + a*C)*(m + 1)*Csc[e + f*x] - (b*(A*b - a*B)*(m + n + 1) + C*(a^2*n + b^2*(m + 1)))*Csc[e + f*
x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {\sec ^2(c+d x) \left (2 \left (A b^2-a (b B-a C)\right )+\frac {1}{2} b (b B-a (A+C)) \sec (c+d x)-\frac {1}{2} \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{b \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {4 \int \frac {\sec (c+d x) \left (-\frac {1}{2} a \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right )+\frac {1}{4} b \left (5 A b^2-5 a b B+2 a^2 C+3 b^2 C\right ) \sec (c+d x)-\frac {1}{4} \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{5 b^2 \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\sec (c+d x) \left (\frac {1}{8} b \left (10 a^2 b B+5 b^3 B-12 a^3 C-3 a b^2 (5 A+C)\right )+\frac {1}{8} \left (40 a^3 b B-25 a b^3 B-6 a^2 b^2 (5 A-4 C)-48 a^4 C+3 b^4 (5 A+3 C)\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (40 a^3 b B-25 a b^3 B-6 a^2 b^2 (5 A-4 C)-48 a^4 C+3 b^4 (5 A+3 C)\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )}+\frac {\left (a^2 b (40 B-36 C)-48 a^3 C-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 b^3 (a+b)} \\ & = \frac {2 \left (40 a^3 b B-25 a b^3 B-6 a^2 b^2 (5 A-4 C)-48 a^4 C+3 b^4 (5 A+3 C)\right ) \cot (c+d x) E\left (\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}}}{15 b^5 \sqrt {a+b} d}+\frac {2 \left (a^2 b (40 B-36 C)-48 a^3 C-6 a b^2 (5 A-5 B+2 C)-b^3 (15 A-5 B+9 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\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}}}{15 b^4 \sqrt {a+b} d}-\frac {2 \left (A b^2-a (b B-a C)\right ) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-3 a b^2 (5 A-3 C)-24 a^3 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2-5 a b B+6 a^2 C-b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 24.40 (sec) , antiderivative size = 887, normalized size of antiderivative = 1.74 \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {4 (b+a \cos (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left ((a+b) \left (-40 a^3 b B+25 a b^3 B+6 a^2 b^2 (5 A-4 C)+48 a^4 C-3 b^4 (5 A+3 C)\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+b (a+b) \left (-48 a^3 C-6 a b^2 (5 A+5 B+2 C)+b^3 (15 A+5 B+9 C)+4 a^2 b (10 B+9 C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {1-\tan ^2\left (\frac {1}{2} (c+d x)\right )} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+\left (-40 a^3 b B+25 a b^3 B+6 a^2 b^2 (5 A-4 C)+48 a^4 C-3 b^4 (5 A+3 C)\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right ) \left (a \left (-1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )-b \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{15 b^4 \left (-a^2+b^2\right ) d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) \sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2} \left (1+\tan ^2\left (\frac {1}{2} (c+d x)\right )\right )^{3/2} \sqrt {\frac {a+b-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )}{1+\tan ^2\left (\frac {1}{2} (c+d x)\right )}}}+\frac {(b+a \cos (c+d x))^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac {4 \left (-30 a^2 A b^2+15 A b^4+40 a^3 b B-25 a b^3 B-48 a^4 C+24 a^2 b^2 C+9 b^4 C\right ) \sin (c+d x)}{15 b^4 \left (-a^2+b^2\right )}+\frac {4 \sec (c+d x) (5 b B \sin (c+d x)-9 a C \sin (c+d x))}{15 b^3}+\frac {4 \left (a^2 A b^2 \sin (c+d x)-a^3 b B \sin (c+d x)+a^4 C \sin (c+d x)\right )}{b^3 \left (-a^2+b^2\right ) (b+a \cos (c+d x))}+\frac {4 C \sec (c+d x) \tan (c+d x)}{5 b^2}\right )}{d (A+2 C+2 B \cos (c+d x)+A \cos (2 c+2 d x)) (a+b \sec (c+d x))^{3/2}} \]

[In]

Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(4*(b + a*Cos[c + d*x])^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*((a
+ b)*(-40*a^3*b*B + 25*a*b^3*B + 6*a^2*b^2*(5*A - 4*C) + 48*a^4*C - 3*b^4*(5*A + 3*C))*EllipticE[ArcSin[Tan[(c
+ d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d
*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + b*(a + b)*(-48*a^3*C - 6*a*b^2*(5*A + 5*B + 2*C) + b^3*(15*A + 5*B
+ 9*C) + 4*a^2*b*(10*B + 9*C))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[1 - Tan[(c + d*x)/2]
^2]*(1 + Tan[(c + d*x)/2]^2)*Sqrt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + (-40*a^3*b*
B + 25*a*b^3*B + 6*a^2*b^2*(5*A - 4*C) + 48*a^4*C - 3*b^4*(5*A + 3*C))*Tan[(c + d*x)/2]*(-1 + Tan[(c + d*x)/2]
^2)*(a*(-1 + Tan[(c + d*x)/2]^2) - b*(1 + Tan[(c + d*x)/2]^2))))/(15*b^4*(-a^2 + b^2)*d*(A + 2*C + 2*B*Cos[c +
d*x] + A*Cos[2*c + 2*d*x])*Sqrt[Sec[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[
(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(1 + Tan[(c + d*x)/2]^2)]) + ((b + a*Cos[c + d*x])^2*(A
+ B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(-30*a^2*A*b^2 + 15*A*b^4 + 40*a^3*b*B - 25*a*b^3*B - 48*a^4*C + 24*a
^2*b^2*C + 9*b^4*C)*Sin[c + d*x])/(15*b^4*(-a^2 + b^2)) + (4*Sec[c + d*x]*(5*b*B*Sin[c + d*x] - 9*a*C*Sin[c +
d*x]))/(15*b^3) + (4*(a^2*A*b^2*Sin[c + d*x] - a^3*b*B*Sin[c + d*x] + a^4*C*Sin[c + d*x]))/(b^3*(-a^2 + b^2)*(
b + a*Cos[c + d*x])) + (4*C*Sec[c + d*x]*Tan[c + d*x])/(5*b^2)))/(d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c +
2*d*x])*(a + b*Sec[c + d*x])^(3/2))

Maple [B] (warning: unable to verify)

Leaf count of result is larger than twice the leaf count of optimal. \(7392\) vs. \(2(478)=956\).

Time = 39.37 (sec) , antiderivative size = 7393, normalized size of antiderivative = 14.50

method	result	size
parts	\(\text {Expression too large to display}\)	\(7393\)
default	\(\text {Expression too large to display}\)	\(7452\)

[In]

int(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

result too large to display

Fricas [F]

\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^5 + B*sec(d*x + c)^4 + A*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a)/(b^2*sec(d*x + c)^2
+ 2*a*b*sec(d*x + c) + a^2), x)

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[In]

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x))**(3/2), x)

Maxima [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

Giac [F]

\[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^3/(b*sec(d*x + c) + a)^(3/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^3(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(3/2)),x)

[Out]

int((A + B/cos(c + d*x) + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(3/2)), x)