3.439 \(\int \frac{\sqrt{a+b \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=267 \[ -\frac{2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{15 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (9 a^2 A+5 a b B-2 A b^2\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^2 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 (5 a B+A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{15 a d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]
[Out]
(-2*(a^2 - b^2)*(2*A*b - 5*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[
Sec[c + d*x]])/(15*a^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^2*A - 2*A*b^2 + 5*a*b*B)*EllipticE[(c + d*x)/2, (
2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(15*a^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*
A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*(A*b + 5*a*B)*Sqrt[a + b*Sec[c + d*x]]*
Sin[c + d*x])/(15*a*d*Sqrt[Sec[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 0.747935, antiderivative size = 267, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 9, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.257, Rules used =
{4032, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ -\frac{2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (9 a^2 A+5 a b B-2 A b^2\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{15 a^2 d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 (5 a B+A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{15 a d \sqrt{\sec (c+d x)}}+\frac{2 A \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{5 d \sec ^{\frac{3}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]
[Out]
(-2*(a^2 - b^2)*(2*A*b - 5*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[
Sec[c + d*x]])/(15*a^2*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^2*A - 2*A*b^2 + 5*a*b*B)*EllipticE[(c + d*x)/2, (
2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(15*a^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*
A*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*(A*b + 5*a*B)*Sqrt[a + b*Sec[c + d*x]]*
Sin[c + d*x])/(15*a*d*Sqrt[Sec[c + d*x]])
Rule 4032
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1
/(d*n), Int[(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^(n + 1)*Simp[A*b*m - a*B*n - (b*B*n + a*A*(n + 1))*C
sc[e + f*x] - 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[0, m, 1] && LeQ[n, -1]
Rule 4104
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*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m +
1)*(d*Csc[e + f*x])^n)/(a*f*n), x] + Dist[1/(a*d*n), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n + 1)*Simp[
a*B*n - A*b*(m + n + 1) + a*(A + A*n + C*n)*Csc[e + f*x] + A*b*(m + n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ
[{a, b, d, e, f, A, B, C, m}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1]
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{\sqrt{a+b \sec (c+d x)} (A+B \sec (c+d x))}{\sec ^{\frac{5}{2}}(c+d x)} \, dx &=\frac{2 A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2}{5} \int \frac{\frac{1}{2} (A b+5 a B)+\frac{1}{2} (3 a A+5 b B) \sec (c+d x)+A b \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b+5 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d \sqrt{\sec (c+d x)}}-\frac{4 \int \frac{\frac{1}{4} \left (-9 a^2 A+2 A b^2-5 a b B\right )-\frac{1}{4} a (7 A b+5 a B) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{15 a}\\ &=\frac{2 A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b+5 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d \sqrt{\sec (c+d x)}}-\frac{\left (\left (a^2-b^2\right ) (2 A b-5 a B)\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 a^2}+\frac{\left (9 a^2 A-2 A b^2+5 a b B\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{15 a^2}\\ &=\frac{2 A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b+5 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d \sqrt{\sec (c+d x)}}-\frac{\left (\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{15 a^2 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (9 a^2 A-2 A b^2+5 a b B\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{15 a^2 \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{2 A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b+5 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d \sqrt{\sec (c+d x)}}-\frac{\left (\left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{15 a^2 \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (9 a^2 A-2 A b^2+5 a b B\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}} \, dx}{15 a^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=-\frac{2 \left (a^2-b^2\right ) (2 A b-5 a B) \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{\sec (c+d x)}}{15 a^2 d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (9 a^2 A-2 A b^2+5 a b B\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{15 a^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{2 A \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{5 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 (A b+5 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{15 a d \sqrt{\sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 1.18059, size = 200, normalized size = 0.75 \[ \frac{2 \sqrt{a+b \sec (c+d x)} \left (\left (a^2-b^2\right ) (5 a B-2 A b) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )+(a+b) \left (9 a^2 A+5 a b B-2 A b^2\right ) \sqrt{\frac{a \cos (c+d x)+b}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )+a \sin (c+d x) (a \cos (c+d x)+b) (3 a A \cos (c+d x)+5 a B+A b)\right )}{15 a^2 d \sqrt{\sec (c+d x)} (a \cos (c+d x)+b)} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[a + b*Sec[c + d*x]]*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(5/2),x]
[Out]
(2*Sqrt[a + b*Sec[c + d*x]]*((a + b)*(9*a^2*A - 2*A*b^2 + 5*a*b*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Elliptic
E[(c + d*x)/2, (2*a)/(a + b)] + (a^2 - b^2)*(-2*A*b + 5*a*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c +
d*x)/2, (2*a)/(a + b)] + a*(b + a*Cos[c + d*x])*(A*b + 5*a*B + 3*a*A*Cos[c + d*x])*Sin[c + d*x]))/(15*a^2*d*(
b + a*Cos[c + d*x])*Sqrt[Sec[c + d*x]])
________________________________________________________________________________________
Maple [B] time = 0.494, size = 2739, normalized size = 10.3 \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))*(a+b*sec(d*x+c))^(1/2)/sec(d*x+c)^(5/2),x)
[Out]
-2/15/d/((a-b)/(a+b))^(1/2)/a^2*(9*A*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(
1/2))*a^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+2*A*EllipticE((-
1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+5*B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a
+b)/(a-b))^(1/2))*a^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+7*A*
cos(d*x+c)*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)*(1/(cos(d*x+c)+1))^(1/2)*a^2*b+2*A*cos(d*x+c)*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)*(1/(cos(d*x+c)+1))^(1/2)*a*b^2-9*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)*(1
/(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*
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)*(1/(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^2-5*B*cos(d*x+c)*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)*(1/(cos(d*x+c)+1))^(1/2)*a^2*b+5*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)*
(1/(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-
5*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)*(1/(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^2-9*A*a^2*b*((a-b)/(a+b))^(1/2)-A*a*b^
2*((a-b)/(a+b))^(1/2)-5*B*a^2*b*((a-b)/(a+b))^(1/2)-5*B*a*b^2*((a-b)/(a+b))^(1/2)-9*A*EllipticF((-1+cos(d*x+c)
)*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/
(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^2+10*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2
)*a^2*b+5*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b+5*B*cos(d*x+c)*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)*(1/(cos(d*x+c)+1))
^(1/2)*a^3+7*A*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*(1/(a+b)*(
b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+2*A*EllipticF((-1+cos(d*x+c))*((a-b)
/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d
*x+c)+1))^(1/2)*sin(d*x+c)-9*A*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*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-2*A*EllipticE((-1+co
s(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))
^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-5*B*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*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+5*B*
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*(1/(a+b)*(b+a*cos(d*x+c))
/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-5*B*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/s
in(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)
*sin(d*x+c)+2*A*b^3*((a-b)/(a+b))^(1/2)+5*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^3-5*B*a^3*((a-b)/(a+b))^(1/2)*c
os(d*x+c)-9*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3-2*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^3+3*A*cos(d*x+c)^4*((a-b
)/(a+b))^(1/2)*a^3+6*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3-9*A*cos(d*x+c)*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)*(1/(cos
(d*x+c)+1))^(1/2)*a^3+9*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)*(1/(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+2*A*cos(d*x+c)*s
in(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(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^3+2*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2-5*B*cos(d*x+c
)*((a-b)/(a+b))^(1/2)*a^2*b+5*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2+4*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*
b)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^3*(1/cos(d*x+c))^(5/2)/sin(d*x+c)/(b+a*cos(d*x+c))
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac{5}{2}}}\,{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))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)/sec(d*x + c)^(5/2), 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}}{\sec \left (d x + c\right )^{\frac{5}{2}}}, 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))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="fricas")
[Out]
integral((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)/sec(d*x + c)^(5/2), 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))**(1/2)/sec(d*x+c)**(5/2),x)
[Out]
Timed out
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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 )} \sqrt{b \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac{5}{2}}}\,{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))^(1/2)/sec(d*x+c)^(5/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)/sec(d*x + c)^(5/2), x)