3.1371 \(\int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=378 \[ -\frac{2 \left (a^2 (-(3 A+C))+a b B+2 A b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \sin (c+d x) \left (-5 a^2 b^2 (A+C)+2 a^3 b B+a^4 C+2 a b^3 B+A b^4\right )}{3 a b d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \sqrt{\cos (c+d x)} \left (-2 a^2 b (3 A+2 C)+3 a^3 B+a b^2 B+2 A b^3\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

(-2*(2*A*b^2 + a*b*B - a^2*(3*A + C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]
)/(3*a^2*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - (2*(2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*
b*(3*A + 2*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^2*(a^2
- b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) - (2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*S
qrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)) + (2*(A*b^4 + 2*a^3*b*B + 2*a*b^3*B + a^4*C - 5*a^2*b^2*(A + C))
*Sin[c + d*x])/(3*a*b*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.22934, antiderivative size = 378, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {4265, 4098, 4100, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac{2 \sin (c+d x) \left (-5 a^2 b^2 (A+C)+2 a^3 b B+a^4 C+2 a b^3 B+A b^4\right )}{3 a b d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \sin (c+d x) \left (A b^2-a (b B-a C)\right )}{3 b d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (a^2 (-(3 A+C))+a b B+2 A b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \sqrt{\cos (c+d x)} \left (-2 a^2 b (3 A+2 C)+3 a^3 B+a b^2 B+2 A b^3\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*A*b^2 + a*b*B - a^2*(3*A + C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]
)/(3*a^2*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - (2*(2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*
b*(3*A + 2*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^2*(a^2
- b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) - (2*(A*b^2 - a*(b*B - a*C))*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*S
qrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)) + (2*(A*b^4 + 2*a^3*b*B + 2*a*b^3*B + a^4*C - 5*a^2*b^2*(A + C))
*Sin[c + d*x])/(3*a*b*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]])

Rule 4265

Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rule 4098

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]

Rule 4100

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[((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)/(a*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] &&  !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])

Rule 4035

Int[(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(
b_.) + (a_)]), x_Symbol] :> Dist[A/a, Int[Sqrt[a + b*Csc[e + f*x]]/Sqrt[d*Csc[e + f*x]], x], x] - Dist[(A*b -
a*B)/(a*d), Int[Sqrt[d*Csc[e + f*x]]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && Ne
Q[A*b - a*B, 0] && NeQ[a^2 - b^2, 0]

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 3858

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(Sqrt[d*
Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/Sqrt[b + a*Sin[e + f*x]], x], x] /; Fr
eeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rubi steps

\begin{align*} \int \frac{A+B \sec (c+d x)+C \sec ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{\left (2 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\frac{1}{2} \left (-A b^2+a (b B-a C)\right )+\frac{3}{2} b (b B-a (A+C)) \sec (c+d x)+\frac{1}{2} \left (2 A b^2-2 a b B-a^2 C+3 b^2 C\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (4 \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{-\frac{1}{4} b \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right )-\frac{1}{4} a b \left (4 a b B-a^2 (3 A+C)-b^2 (A+3 C)\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{3 a b \left (a^2-b^2\right )^2}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^2+a b B-a^2 (3 A+C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )}-\frac{\left (\left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^2+a b B-a^2 (3 A+C)\right ) \sqrt{b+a \cos (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3 a^2 \left (a^2-b^2\right )^2 \sqrt{b+a \cos (c+d x)}}\\ &=-\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^2+a b B-a^2 (3 A+C)\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{3 a^2 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{\left (\left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{3 a^2 \left (a^2-b^2\right )^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=-\frac{2 \left (2 A b^2+a b B-a^2 (3 A+C)\right ) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (2 A b^3+3 a^3 B+a b^2 B-2 a^2 b (3 A+2 C)\right ) \sqrt{\cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}-\frac{2 \left (A b^2-a (b B-a C)\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (A b^4+2 a^3 b B+2 a b^3 B+a^4 C-5 a^2 b^2 (A+C)\right ) \sin (c+d x)}{3 a b \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 19.3603, size = 673, normalized size = 1.78 \[ \frac{(a \cos (c+d x)+b)^3 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\frac{4 \left (a^2 C \sin (c+d x)-a b B \sin (c+d x)+A b^2 \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (a \cos (c+d x)+b)^2}+\frac{4 \left (-6 a^2 A b \sin (c+d x)-4 a^2 b C \sin (c+d x)+3 a^3 B \sin (c+d x)+a b^2 B \sin (c+d x)+2 A b^3 \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}\right )}{d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{5/2} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)}+\frac{4 \cos ^{\frac{3}{2}}(c+d x) \sqrt{\sec (c+d x)} \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a \cos (c+d x)+b)^2 \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (-i a (a+b) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (a^2 (3 A-3 B+C)+a b (3 A-B+3 C)-2 A b^2\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{b-a}{a+b}\right )-\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} \left (-2 a^2 b (3 A+2 C)+3 a^3 B+a b^2 B+2 A b^3\right ) (a \cos (c+d x)+b)-i (a+b) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (-2 a^2 b (3 A+2 C)+3 a^3 B+a b^2 B+2 A b^3\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{3 d \left (a^3-a b^2\right )^2 (a+b \sec (c+d x))^{5/2} (A \cos (2 c+2 d x)+A+2 B \cos (c+d x)+2 C)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((4*(A*b^2*Sin[c + d*x] - a*b*B*Sin[c + d*x] +
a^2*C*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (4*(-6*a^2*A*b*Sin[c + d*x] + 2*A*b^3*Sin[c +
d*x] + 3*a^3*B*Sin[c + d*x] + a*b^2*B*Sin[c + d*x] - 4*a^2*b*C*Sin[c + d*x]))/(3*a*(a^2 - b^2)^2*(b + a*Cos[c
+ d*x]))))/(d*Sqrt[Cos[c + d*x]]*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2)
) + (4*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])^2*Sqrt[Sec[c + d*x]]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(A
+ B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((-I)*(a + b)*(2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*b*(3*A + 2*C))*Ellipt
icE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)
/2]^2)/(a + b)] - I*a*(a + b)*(-2*A*b^2 + a^2*(3*A - 3*B + C) + a*b*(3*A - B + 3*C))*EllipticF[I*ArcSinh[Tan[(
c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - (
2*A*b^3 + 3*a^3*B + a*b^2*B - 2*a^2*b*(3*A + 2*C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*
x)/2]))/(3*(a^3 - a*b^2)^2*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*c + 2*d*x])*(a + b*Sec[c + d*x])^(5/2))

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Maple [B]  time = 0.785, size = 2767, normalized size = 7.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x)

[Out]

2/3/d*(cos(d*x+c)+1)^5*(-1+cos(d*x+c))^3*(3*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)
/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*b-3*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b
))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^2-2*A*Elliptic
F((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*a*b^3+6*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/
2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2-2*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*b^4*(1/(cos(d*x+c)+1))^(3/2)-C*
cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^4*(1/(cos(d*x+c)+1))^(3/2)-3*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(
d*x+c)*a^4*(1/(cos(d*x+c)+1))^(3/2)+5*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(cos(d*x+c)+1))^(3/2)-A*((a-
b)/(a+b))^(1/2)*sin(d*x+c)*a*b^3*(1/(cos(d*x+c)+1))^(3/2)-2*B*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*b*(1/(cos(d*x
+c)+1))^(3/2)+B*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(cos(d*x+c)+1))^(3/2)-B*((a-b)/(a+b))^(1/2)*sin(d*x+
c)*a*b^3*(1/(cos(d*x+c)+1))^(3/2)-C*((a-b)/(a+b))^(1/2)*a^3*b*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)+4*C*((a-b)/(
a+b))^(1/2)*a^2*b^2*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)-3*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*El
lipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*b-B*(1/(a+b)*(b+a*cos(d*x+c))
/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3+3*
B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(co
s(d*x+c)+1))^(1/2)*a^3*b-B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(
a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^2+4*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipt
icE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2+C*EllipticF((-1+cos(d*x+c))*(
(a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*b-B*(1
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)
/(a-b))^(1/2))*cos(d*x+c)*a^3*b-B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((
a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^2*b^2-3*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c
)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^3*b+4*
C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(
a+b)/(a-b))^(1/2))*cos(d*x+c)*a^3*b-3*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b)
)^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*a^3*b-2*A*EllipticF((-1+cos(d*x+c))*((a-b)
/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*a^2*
b^2+6*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+
c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^3*b-2*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+co
s(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a*b^3-C*((a-b)/(a+b))^(1/2)*(1/(cos(
d*x+c)+1))^(3/2)*sin(d*x+c)*a^4+3*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*
((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^4-3*B*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+
1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*cos(d*x+c)*a^4+C*(1/(
a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(
a-b))^(1/2))*cos(d*x+c)*a^4+3*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))
*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*a^4+6*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*
a^3*b*(1/(cos(d*x+c)+1))^(3/2)-A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(cos(d*x+c)+1))^(3/2)-3*
A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a*b^3*(1/(cos(d*x+c)+1))^(3/2)+B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*si
n(d*x+c)*a^3*b*(1/(cos(d*x+c)+1))^(3/2)+3*C*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b*sin(d*x+c)*(1/(cos(d*x+c)+1))
^(3/2)-2*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d
*x+c),(-(a+b)/(a-b))^(1/2))*b^4-3*C*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1
/2))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^2)*cos(d*x+c)^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(
1/2)*((a-b)/(a+b))^(1/2)*(1/(cos(d*x+c)+1))^(3/2)/(a+b)/(a-b)^2/a^2/(b+a*cos(d*x+c))^2/sin(d*x+c)^6

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{b^{3} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right ) \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) \sec \left (d x + c\right ) + a^{3} \cos \left (d x + c\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

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

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Sympy [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((A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(5/2)/cos(d*x+c)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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