3.631 \(\int \frac{A+B \sec (c+d x)}{\sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=346 \[ \frac{2 \left (3 a^2 A-a b B-2 A b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \left (5 a^2 A b-2 a^3 B-2 a b^2 B-A b^3\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 (A b-a B) \sin (c+d x)}{3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (6 a^2 A b-3 a^3 B-a b^2 B-2 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]
[Out]
(2*(3*a^2*A - 2*A*b^2 - a*b*B)*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*(6*a^2*A*b - 2*A*b^3 - 3*a^3*B - a*b^2*B)*Sq
rt[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 - a*B)*Sin[c + d*x])/(3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Sec[c
+ d*x])^(3/2)) - (2*(5*a^2*A*b - A*b^3 - 2*a^3*B - 2*a*b^2*B)*Sin[c + d*x])/(3*a*(a^2 - b^2)^2*d*Sqrt[Cos[c +
d*x]]*Sqrt[a + b*Sec[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 1.00769, antiderivative size = 346, normalized size of antiderivative
= 1., number of steps used = 10, number of rules used = 10, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\)
= 0.286, Rules used = {2955, 4027, 4100, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ -\frac{2 \left (5 a^2 A b-2 a^3 B-2 a b^2 B-A b^3\right ) \sin (c+d x)}{3 a d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 (A b-a B) \sin (c+d x)}{3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 \left (3 a^2 A-a b B-2 A b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (6 a^2 A b-3 a^3 B-a b^2 B-2 A b^3\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)),x]
[Out]
(2*(3*a^2*A - 2*A*b^2 - a*b*B)*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*(6*a^2*A*b - 2*A*b^3 - 3*a^3*B - a*b^2*B)*Sq
rt[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 - a*B)*Sin[c + d*x])/(3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Sec[c
+ d*x])^(3/2)) - (2*(5*a^2*A*b - A*b^3 - 2*a^3*B - 2*a*b^2*B)*Sin[c + d*x])/(3*a*(a^2 - b^2)^2*d*Sqrt[Cos[c +
d*x]]*Sqrt[a + b*Sec[c + d*x]])
Rule 2955
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.)*((g_.)*sin[(e_.
) + (f_.)*(x_)])^(p_.), x_Symbol] :> Dist[(g*Csc[e + f*x])^p*(g*Sin[e + f*x])^p, Int[((a + b*Csc[e + f*x])^m*(
c + d*Csc[e + f*x])^n)/(g*Csc[e + f*x])^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[b*c - a*d
, 0] && !IntegerQ[p] && !(IntegerQ[m] && IntegerQ[n])
Rule 4027
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> -Simp[(d*(A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n
- 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e +
f*x])^(n - 1)*Simp[d*(n - 1)*(A*b - a*B) + d*(a*A - b*B)*(m + 1)*Csc[e + f*x] - d*(A*b - a*B)*(m + n + 1)*Csc
[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m,
-1] && LtQ[0, n, 1]
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)}{\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)} (A+B \sec (c+d x))}{(a+b \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{2 (A b-a B) \sin (c+d x)}{3 \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} (-A b+a B)-\frac{3}{2} (a A-b B) \sec (c+d x)+(A b-a B) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 \left (a^2-b^2\right )}\\ &=-\frac{2 (A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sin (c+d x)}{3 a \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} \left (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\right )+\frac{1}{4} a \left (3 a^2 A+A b^2-4 a b B\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{3 a \left (a^2-b^2\right )^2}\\ &=-\frac{2 (A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (3 a^2 A-2 A b^2-a b B\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 (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\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 (A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (3 a^2 A-2 A b^2-a b B\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 (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\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 (A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sin (c+d x)}{3 a \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (3 a^2 A-2 A b^2-a b B\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 (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\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 (3 a^2 A-2 A b^2-a b B\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 (6 a^2 A b-2 A b^3-3 a^3 B-a b^2 B\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 (A b-a B) \sin (c+d x)}{3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac{2 \left (5 a^2 A b-A b^3-2 a^3 B-2 a b^2 B\right ) \sin (c+d x)}{3 a \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 = 16.3687, size = 463, normalized size = 1.34 \[ \frac{(a \cos (c+d x)+b)^2 \left (\frac{2 \sin (c+d x) \left (a \left (-6 a^2 A b+3 a^3 B+a b^2 B+2 A b^3\right ) \cos (c+d x)+b \left (-5 a^2 A b+2 a^3 B+2 a b^2 B+A b^3\right )\right )}{a \left (a^2-b^2\right )^2 (a \cos (c+d x)+b)}+\frac{2 \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-i a (a+b) \left (3 a^2 (A-B)+a b (3 A-B)-2 A b^2\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{b-a}{a+b}\right )-\left (-6 a^2 A b+3 a^3 B+a b^2 B+2 A b^3\right ) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i (a+b) \left (-6 a^2 A b+3 a^3 B+a b^2 B+2 A b^3\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{\left (a^3-a b^2\right )^2 \sec ^{\frac{3}{2}}(c+d x)}\right )}{3 d \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(A + B*Sec[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)),x]
[Out]
((b + a*Cos[c + d*x])^2*((2*(b*(-5*a^2*A*b + A*b^3 + 2*a^3*B + 2*a*b^2*B) + a*(-6*a^2*A*b + 2*A*b^3 + 3*a^3*B
+ a*b^2*B)*Cos[c + d*x])*Sin[c + d*x])/(a*(a^2 - b^2)^2*(b + a*Cos[c + d*x])) + (2*(Cos[(c + d*x)/2]^2*Sec[c +
d*x])^(3/2)*((-I)*(a + b)*(-6*a^2*A*b + 2*A*b^3 + 3*a^3*B + a*b^2*B)*EllipticE[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 + 3*a^2*(A - B) + a*b*(3*A - B))*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)] - (-6*a^2*A*b + 2*A*b^3 + 3*a^3*B + a*b^2*B)*(b +
a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/((a^3 - a*b^2)^2*Sec[c + d*x]^(3/2))))/(3*d*Cos
[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2))
________________________________________________________________________________________
Maple [B] time = 0.495, size = 2416, normalized size = 7. \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))/(a+b*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x)
[Out]
-2/3/d*(-1+cos(d*x+c))*(cos(d*x+c)+1)^2*(3*A*cos(d*x+c)*sin(d*x+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))*a^3*b+2*A*cos(d*x+c)*sin(d
*x+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))*a^2*b^2+B*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellipti
cF((-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*cos(d*x+c)*sin(d*x+c)*Elliptic
E((-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^2-6*A*cos(d*x+c)*sin(d*x+c)*EllipticE((-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+2*A*cos(d*x+c)*sin(d*x+c)*EllipticE((-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^3-3*A*cos(d*x+c)*sin(d*x+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))*a^4-6*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1)
)^(1/2)+6*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(cos(d*x+c)+1))^(1/2)+2*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)
*a*b^3*(1/(cos(d*x+c)+1))^(1/2)-3*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)-3*B*cos(d*x+
c)^2*((a-b)/(a+b))^(1/2)*a^4*(1/(cos(d*x+c)+1))^(1/2)-2*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^4*(1/(cos(d*x+c)+1)
)^(1/2)-A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(cos(d*x+c)+1))^(1/2)+B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)
*a^3*b*(1/(cos(d*x+c)+1))^(1/2)+2*A*((a-b)/(a+b))^(1/2)*b^4*(1/(cos(d*x+c)+1))^(1/2)+B*cos(d*x+c)*((a-b)/(a+b)
)^(1/2)*a^2*b^2*(1/(cos(d*x+c)+1))^(1/2)-B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^3*(1/(cos(d*x+c)+1))^(1/2)-3*B*s
in(d*x+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*sin(d*x+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+3*B*sin(d*x+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))*a^3*b+B
*sin(d*x+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))*a*b^3+6*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)-3*A*cos
(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^3*(1/(cos(d*x+c)+1))^(1/2)-6*A*sin(d*x+c)*EllipticE((-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^2-3*A*sin(d
*x+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))*a^3*b+3*A*sin(d*x+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))*a^2*b^2+2*A*sin(d*x+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))*a*b^3-3*B
*cos(d*x+c)*sin(d*x+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))*a^4+3*B*cos(d*x+c)*sin(d*x+c)*EllipticE((-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^4+3*B*cos(d*x+c)*((a-
b)/(a+b))^(1/2)*a^4*(1/(cos(d*x+c)+1))^(1/2)-5*A*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(cos(d*x+c)+1))^(1/2)+A*((a-b)
/(a+b))^(1/2)*a*b^3*(1/(cos(d*x+c)+1))^(1/2)+2*B*((a-b)/(a+b))^(1/2)*a^3*b*(1/(cos(d*x+c)+1))^(1/2)-B*((a-b)/(
a+b))^(1/2)*a^2*b^2*(1/(cos(d*x+c)+1))^(1/2)+B*((a-b)/(a+b))^(1/2)*a*b^3*(1/(cos(d*x+c)+1))^(1/2)+2*A*sin(d*x+
c)*EllipticE((-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))/(c
os(d*x+c)+1))^(1/2)*b^4)*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))^(1/2)/a^2/(a+b)/(a-b)^2/(b+a*cos(d*x+c))^2/sin(d*x+c)^3
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{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))/(a+b*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="maxima")
[Out]
integrate((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 (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))/(a+b*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="fricas")
[Out]
integral((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))/(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{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))/(a+b*sec(d*x+c))^(5/2)/cos(d*x+c)^(1/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^(5/2)*sqrt(cos(d*x + c))), x)