3.748 \(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=488 \[ -\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (2 a^2 C+A b^2-b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}+\frac {4 a \left (8 a^4 C+a^2 b^2 (A-14 C)-b^4 (3 A-4 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^5 d \sqrt {a+b} \left (a^2-b^2\right )}-\frac {4 a \left (-3 a^4 C+5 a^2 b^2 C+2 A b^4\right ) \tan (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (16 a^4 C+12 a^3 b C+2 a^2 b^2 (A-8 C)+3 a b^3 (A-3 C)-b^4 (3 A+C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^4 d \sqrt {a+b} \left (a^2-b^2\right )} \]

[Out]

4/3*a*(a^2*b^2*(A-14*C)-b^4*(3*A-4*C)+8*a^4*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/(a^2-b^2)/d/(a+b)^(1/2)+2/3*(
2*a^2*b^2*(A-8*C)+3*a*b^3*(A-3*C)+16*a^4*C+12*a^3*b*C-b^4*(3*A+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/(a^2-b^2)
/d/(a+b)^(1/2)-2/3*(A*b^2+C*a^2)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)-4/3*a*(2*A*b^4-3
*C*a^4+5*C*a^2*b^2)*tan(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)+2/3*(A*b^2+2*C*a^2-C*b^2)*(a+b*sec(d*x
+c))^(1/2)*tan(d*x+c)/b^3/(a^2-b^2)/d

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Rubi [A]  time = 1.30, antiderivative size = 488, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4099, 4090, 4082, 4005, 3832, 4004} \[ -\frac {2 \left (a^2 C+A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 \left (2 a^2 C+A b^2-b^2 C\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}-\frac {4 a \left (5 a^2 b^2 C-3 a^4 C+2 A b^4\right ) \tan (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (2 a^2 b^2 (A-8 C)+12 a^3 b C+16 a^4 C+3 a b^3 (A-3 C)-b^4 (3 A+C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^4 d \sqrt {a+b} \left (a^2-b^2\right )}+\frac {4 a \left (a^2 b^2 (A-14 C)+8 a^4 C-b^4 (3 A-4 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^5 d \sqrt {a+b} \left (a^2-b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(4*a*(a^2*b^2*(A - 14*C) - b^4*(3*A - 4*C) + 8*a^4*C)*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))])/(3
*b^5*Sqrt[a + b]*(a^2 - b^2)*d) + (2*(2*a^2*b^2*(A - 8*C) + 3*a*b^3*(A - 3*C) + 16*a^4*C + 12*a^3*b*C - b^4*(3
*A + C))*Cot[c + d*x]*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Se
c[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^4*Sqrt[a + b]*(a^2 - b^2)*d) - (2*(A*b^2 +
a^2*C)*Sec[c + d*x]^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) - (4*a*(2*A*b^4 - 3*a^4*C
+ 5*a^2*b^2*C)*Tan[c + d*x])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(A*b^2 + 2*a^2*C - b^2*C)*S
qrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3*b^3*(a^2 - b^2)*d)

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]

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 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 4090

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[(a*(A*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a + b*Csc[e
+ f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*C
sc[e + f*x])^(m + 1)*Simp[b*(m + 1)*(-(a*(b*B - a*C)) + A*b^2) + (b*B*(a^2 + b^2*(m + 1)) - a*(A*b^2*(m + 2) +
C*(a^2 + b^2*(m + 1))))*Csc[e + f*x] - b*C*(m + 1)*(a^2 - b^2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

Rule 4099

Int[((A_.) + 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^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^2*C*(n - 1) + a*b*(A + C)*(m + 1)*Csc[e + f*x] - (A*b^2*(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, C}, x] && NeQ[a^2 - b
^2, 0] && LtQ[m, -1] && GtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx &=-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {\sec ^2(c+d x) \left (2 \left (A b^2+a^2 C\right )-\frac {3}{2} a b (A+C) \sec (c+d x)-\frac {3}{2} \left (A b^2+2 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^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}{2} b \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right )-\frac {1}{2} a \left (2 A b^4-\left (6 a^4-11 a^2 b^2+3 b^4\right ) C\right ) \sec (c+d x)-\frac {3}{4} b \left (a^2-b^2\right ) \left (A b^2+2 a^2 C-b^2 C\right ) \sec ^2(c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\sec (c+d x) \left (\frac {3}{8} b^2 \left (4 a^4 C-b^4 (3 A+C)-a^2 b^2 (A+7 C)\right )+\frac {3}{4} a b \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {\left (2 a \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\right )\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}+\frac {\left (2 a^2 b^2 (A-8 C)+3 a b^3 (A-3 C)+16 a^4 C+12 a^3 b C-b^4 (3 A+C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) b^3 (a+b)^2}\\ &=\frac {4 a \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\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^5 (a+b)^{3/2} d}+\frac {2 \left (2 a^2 b^2 (A-8 C)+3 a b^3 (A-3 C)+16 a^4 C+12 a^3 b C-b^4 (3 A+C)\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}}}{3 (a-b) b^4 (a+b)^{3/2} d}-\frac {2 \left (A b^2+a^2 C\right ) \sec ^2(c+d x) \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \tan (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 22.63, size = 830, normalized size = 1.70 \[ \frac {\sec (c+d x) \left (C \sec ^2(c+d x)+A\right ) \left (-\frac {8 a \left (8 C a^4+A b^2 a^2-14 b^2 C a^2-3 A b^4+4 b^4 C\right ) \sin (c+d x)}{3 b^4 \left (a^2-b^2\right )^2}-\frac {4 \left (C \sin (c+d x) a^3+A b^2 \sin (c+d x) a\right )}{3 b^2 \left (b^2-a^2\right ) (b+a \cos (c+d x))^2}-\frac {4 \left (-7 C \sin (c+d x) a^5-A b^2 \sin (c+d x) a^3+11 b^2 C \sin (c+d x) a^3+5 A b^4 \sin (c+d x) a\right )}{3 b^3 \left (b^2-a^2\right )^2 (b+a \cos (c+d x))}+\frac {4 C \tan (c+d x)}{3 b^3}\right ) (b+a \cos (c+d x))^3}{d (\cos (2 c+2 d x) A+A+2 C) (a+b \sec (c+d x))^{5/2}}+\frac {4 \sqrt {\sec (c+d x)} \left (C \sec ^2(c+d x)+A\right ) \sqrt {\frac {1}{1-\tan ^2\left (\frac {1}{2} (c+d x)\right )}} \left (2 a (a+b) \left (8 C a^4+b^2 (A-14 C) a^2+b^4 (4 C-3 A)\right ) E\left (\sin ^{-1}\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 )} \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )+b (a+b) \left (-16 C a^4+12 b C a^3-2 b^2 (A-8 C) a^2+3 b^3 (A-3 C) a+b^4 (3 A+C)\right ) F\left (\sin ^{-1}\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 )} \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{a+b}} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )+2 a \left (8 C a^4+b^2 (A-14 C) a^2+b^4 (4 C-3 A)\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \left (-b \tan ^4\left (\frac {1}{2} (c+d x)\right )+a \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )-1\right )^2+b\right )\right ) (b+a \cos (c+d x))^{5/2}}{3 b^4 \left (a^2-b^2\right )^2 d (\cos (2 c+2 d x) A+A+2 C) (a+b \sec (c+d x))^{5/2} \left (\tan ^2\left (\frac {1}{2} (c+d x)\right )+1\right )^{3/2} \sqrt {\frac {-a \tan ^2\left (\frac {1}{2} (c+d x)\right )+b \tan ^2\left (\frac {1}{2} (c+d x)\right )+a+b}{\tan ^2\left (\frac {1}{2} (c+d x)\right )+1}}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Sec[c + d*x]^3*(A + C*Sec[c + d*x]^2))/(a + b*Sec[c + d*x])^(5/2),x]

[Out]

(4*(b + a*Cos[c + d*x])^(5/2)*Sqrt[Sec[c + d*x]]*(A + C*Sec[c + d*x]^2)*Sqrt[(1 - Tan[(c + d*x)/2]^2)^(-1)]*(2
*a*(a + b)*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-3*A + 4*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)*(-2*a^2*b^2*(A - 8*C) + 3*a*b^3*(A - 3*C) - 16*a^4*C + 12*a^3*b*C + b^4*(3*A + 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)*Sq
rt[(a + b - a*Tan[(c + d*x)/2]^2 + b*Tan[(c + d*x)/2]^2)/(a + b)] + 2*a*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-
3*A + 4*C))*Tan[(c + d*x)/2]*(b - b*Tan[(c + d*x)/2]^4 + a*(-1 + Tan[(c + d*x)/2]^2)^2)))/(3*b^4*(a^2 - b^2)^2
*d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)*(1 + Tan[(c + d*x)/2]^2)^(3/2)*Sqrt[(a + b - a*Ta
n[(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])^3*Sec[c + d*x]*(A
+ C*Sec[c + d*x]^2)*((-8*a*(a^2*A*b^2 - 3*A*b^4 + 8*a^4*C - 14*a^2*b^2*C + 4*b^4*C)*Sin[c + d*x])/(3*b^4*(a^2
- b^2)^2) - (4*(a*A*b^2*Sin[c + d*x] + a^3*C*Sin[c + d*x]))/(3*b^2*(-a^2 + b^2)*(b + a*Cos[c + d*x])^2) - (4*(
-(a^3*A*b^2*Sin[c + d*x]) + 5*a*A*b^4*Sin[c + d*x] - 7*a^5*C*Sin[c + d*x] + 11*a^3*b^2*C*Sin[c + d*x]))/(3*b^3
*(-a^2 + b^2)^2*(b + a*Cos[c + d*x])) + (4*C*Tan[c + d*x])/(3*b^3)))/(d*(A + 2*C + A*Cos[2*c + 2*d*x])*(a + b*
Sec[c + d*x])^(5/2))

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \sec \left (d x + c\right )^{5} + A \sec \left (d x + c\right )^{3}\right )} \sqrt {b \sec \left (d x + c\right ) + a}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((C*sec(d*x + c)^5 + A*sec(d*x + c)^3)*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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + A\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + A)*sec(d*x + c)^3/(b*sec(d*x + c) + a)^(5/2), x)

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maple [B]  time = 3.30, size = 7051, normalized size = 14.45 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x)

[Out]

result too large to display

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maxima [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/2)),x)

[Out]

int((A + C/cos(c + d*x)^2)/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(A+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(5/2),x)

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a + b*sec(c + d*x))**(5/2), x)

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