3.453 \(\int \frac{(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx\)
Optimal. Leaf size=340 \[ \frac{2 \left (a^2-b^2\right ) \left (25 a^2 A+56 a b B+15 A b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{105 a d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (25 a^2 A+77 a b B+45 A b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{105 d \sqrt{\sec (c+d x)}}+\frac{2 \left (145 a^2 A b+63 a^3 B+161 a b^2 B+15 A b^3\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{105 a d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a (7 a B+10 A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]
[Out]
(2*(a^2 - b^2)*(25*a^2*A + 15*A*b^2 + 56*a*b*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a
)/(a + b)]*Sqrt[Sec[c + d*x]])/(105*a*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(145*a^2*A*b + 15*A*b^3 + 63*a^3*B + 16
1*a*b^2*B)*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(105*a*d*Sqrt[(b + a*Cos[c + d*x])/
(a + b)]*Sqrt[Sec[c + d*x]]) + (2*a*(10*A*b + 7*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(35*d*Sec[c + d*x]
^(3/2)) + (2*(25*a^2*A + 45*A*b^2 + 77*a*b*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]
) + (2*a*A*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))
________________________________________________________________________________________
Rubi [A] time = 1.15505, antiderivative size = 340, 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 = {4025, 4094, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ \frac{2 \left (25 a^2 A+77 a b B+45 A b^2\right ) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{105 d \sqrt{\sec (c+d x)}}+\frac{2 \left (a^2-b^2\right ) \left (25 a^2 A+56 a b B+15 A b^2\right ) \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{105 a d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (145 a^2 A b+63 a^3 B+161 a b^2 B+15 A b^3\right ) \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{105 a d \sqrt{\sec (c+d x)} \sqrt{\frac{a \cos (c+d x)+b}{a+b}}}+\frac{2 a (7 a B+10 A b) \sin (c+d x) \sqrt{a+b \sec (c+d x)}}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a A \sin (c+d x) (a+b \sec (c+d x))^{3/2}}{7 d \sec ^{\frac{5}{2}}(c+d x)} \]
Antiderivative was successfully verified.
[In]
Int[((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(7/2),x]
[Out]
(2*(a^2 - b^2)*(25*a^2*A + 15*A*b^2 + 56*a*b*B)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a
)/(a + b)]*Sqrt[Sec[c + d*x]])/(105*a*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(145*a^2*A*b + 15*A*b^3 + 63*a^3*B + 16
1*a*b^2*B)*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(105*a*d*Sqrt[(b + a*Cos[c + d*x])/
(a + b)]*Sqrt[Sec[c + d*x]]) + (2*a*(10*A*b + 7*a*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(35*d*Sec[c + d*x]
^(3/2)) + (2*(25*a^2*A + 45*A*b^2 + 77*a*b*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(105*d*Sqrt[Sec[c + d*x]]
) + (2*a*A*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(7*d*Sec[c + d*x]^(5/2))
Rule 4025
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]
Rule 4094
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*(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*(C*n + A*(n + 1)))*Csc[e + f*x] - b*(C*n + A*(m + n + 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] && GtQ[m, 0] && 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{(a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x))}{\sec ^{\frac{7}{2}}(c+d x)} \, dx &=\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{2}{7} \int \frac{\sqrt{a+b \sec (c+d x)} \left (-\frac{1}{2} a (10 A b+7 a B)-\frac{1}{2} \left (5 a^2 A+7 A b^2+14 a b B\right ) \sec (c+d x)-\frac{1}{2} b (2 a A+7 b B) \sec ^2(c+d x)\right )}{\sec ^{\frac{5}{2}}(c+d x)} \, dx\\ &=\frac{2 a (10 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}-\frac{4}{35} \int \frac{-\frac{1}{4} a \left (25 a^2 A+45 A b^2+77 a b B\right )-\frac{1}{4} \left (65 a^2 A b+35 A b^3+21 a^3 B+105 a b^2 B\right ) \sec (c+d x)-\frac{1}{4} b \left (30 a A b+14 a^2 B+35 b^2 B\right ) \sec ^2(c+d x)}{\sec ^{\frac{3}{2}}(c+d x) \sqrt{a+b \sec (c+d x)}} \, dx\\ &=\frac{2 a (10 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{8 \int \frac{\frac{1}{8} a \left (145 a^2 A b+15 A b^3+63 a^3 B+161 a b^2 B\right )+\frac{1}{8} a \left (25 a^3 A+135 a A b^2+119 a^2 b B+105 b^3 B\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{105 a}\\ &=\frac{2 a (10 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{\left (\left (a^2-b^2\right ) \left (25 a^2 A+15 A b^2+56 a b B\right )\right ) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+b \sec (c+d x)}} \, dx}{105 a}+\frac{\left (145 a^2 A b+15 A b^3+63 a^3 B+161 a b^2 B\right ) \int \frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{\sec (c+d x)}} \, dx}{105 a}\\ &=\frac{2 a (10 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{\left (\left (a^2-b^2\right ) \left (25 a^2 A+15 A b^2+56 a b B\right ) \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{105 a \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (145 a^2 A b+15 A b^3+63 a^3 B+161 a b^2 B\right ) \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{105 a \sqrt{b+a \cos (c+d x)} \sqrt{\sec (c+d x)}}\\ &=\frac{2 a (10 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}+\frac{\left (\left (a^2-b^2\right ) \left (25 a^2 A+15 A b^2+56 a b B\right ) \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}{105 a \sqrt{a+b \sec (c+d x)}}+\frac{\left (\left (145 a^2 A b+15 A b^3+63 a^3 B+161 a b^2 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}{105 a \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}\\ &=\frac{2 \left (a^2-b^2\right ) \left (25 a^2 A+15 A b^2+56 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 ) \sqrt{\sec (c+d x)}}{105 a d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (145 a^2 A b+15 A b^3+63 a^3 B+161 a b^2 B\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right ) \sqrt{a+b \sec (c+d x)}}{105 a d \sqrt{\frac{b+a \cos (c+d x)}{a+b}} \sqrt{\sec (c+d x)}}+\frac{2 a (10 A b+7 a B) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{35 d \sec ^{\frac{3}{2}}(c+d x)}+\frac{2 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt{a+b \sec (c+d x)} \sin (c+d x)}{105 d \sqrt{\sec (c+d x)}}+\frac{2 a A (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d \sec ^{\frac{5}{2}}(c+d x)}\\ \end{align*}
Mathematica [A] time = 1.79112, size = 257, normalized size = 0.76 \[ \frac{(a+b \sec (c+d x))^{5/2} \left (2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \left (a \left (25 a^3 A+119 a^2 b B+135 a A b^2+105 b^3 B\right ) \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )+\left (145 a^2 A b+63 a^3 B+161 a b^2 B+15 A b^3\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )-b \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )\right )\right )+a \sin (c+d x) (a \cos (c+d x)+b) \left (15 a^2 A \cos (2 (c+d x))+65 a^2 A+6 a (7 a B+15 A b) \cos (c+d x)+154 a b B+90 A b^2\right )\right )}{105 a d \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b)^3} \]
Antiderivative was successfully verified.
[In]
Integrate[((a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]))/Sec[c + d*x]^(7/2),x]
[Out]
((a + b*Sec[c + d*x])^(5/2)*(2*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*(a*(25*a^3*A + 135*a*A*b^2 + 119*a^2*b*B + 1
05*b^3*B)*EllipticF[(c + d*x)/2, (2*a)/(a + b)] + (145*a^2*A*b + 15*A*b^3 + 63*a^3*B + 161*a*b^2*B)*((a + b)*E
llipticE[(c + d*x)/2, (2*a)/(a + b)] - b*EllipticF[(c + d*x)/2, (2*a)/(a + b)])) + a*(b + a*Cos[c + d*x])*(65*
a^2*A + 90*A*b^2 + 154*a*b*B + 6*a*(15*A*b + 7*a*B)*Cos[c + d*x] + 15*a^2*A*Cos[2*(c + d*x)])*Sin[c + d*x]))/(
105*a*d*(b + a*Cos[c + d*x])^3*Sec[c + d*x]^(5/2))
________________________________________________________________________________________
Maple [B] time = 0.557, size = 3980, normalized size = 11.7 \begin{align*} \text{output too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x)
[Out]
-2/105/d/a/((a-b)/(a+b))^(1/2)*(25*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(
1/2))*a^4*(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)+105*B*sin(d*x+c)
*cos(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^3-15*A*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)
/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^4*(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)-145*A*sin(d*x+c)*cos(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*b-25*A*a^3*b*((a-b)/(a+b
))^(1/2)-145*A*a^2*b^2*((a-b)/(a+b))^(1/2)-45*A*a*b^3*((a-b)/(a+b))^(1/2)-63*B*a^3*b*((a-b)/(a+b))^(1/2)-77*B*
a^2*b^2*((a-b)/(a+b))^(1/2)-161*B*a*b^3*((a-b)/(a+b))^(1/2)-161*B*sin(d*x+c)*cos(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/(c
os(d*x+c)+1))^(1/2)*a^2*b^2-63*B*sin(d*x+c)*cos(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*b+161*B*s
in(d*x+c)*cos(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+co
s(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b^2-161*B*sin(d*x+c)*cos(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)/si
n(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3+25*A*sin(d*x+c)*cos(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^4-
15*A*sin(d*x+c)*cos(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^4-63*B*sin(d*x+c)*cos(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^4+63*B*sin(d*x+c)*cos(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^4-1
45*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*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)+135*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^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)-15*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^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)+145*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*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)+21*B*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^4+42*B*cos(d*x+c)^2*((a-b)/(a+
b))^(1/2)*a^4+15*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^4-63*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^4+15*A*cos(d*x+c)^
5*((a-b)/(a+b))^(1/2)*a^4+10*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^4-25*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^4-14
5*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^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)+15*A*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*(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)+119*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*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)-161*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^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)-63*B*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*(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)+161*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^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)-161*B*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*(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)+60*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*b+90*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b^2+98*
B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^3*b+110*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b+60*A*cos(d*x+c)^2*((a-b)
/(a+b))^(1/2)*a*b^3+238*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b^2-145*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b+
55*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b^2-15*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^3-35*B*cos(d*x+c)*((a-b)/(
a+b))^(1/2)*a^3*b-161*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b^2+161*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^3-15*A
*b^4*((a-b)/(a+b))^(1/2)+135*A*sin(d*x+c)*cos(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^2-15*A*si
n(d*x+c)*cos(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^3+145*A*sin(d*x+c)*cos(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*b-145*A*sin(d*x+c)*cos(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+15*A*sin(d*x+c)*cos(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)*Elliptic
E((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3+119*B*sin(d*x+c)*cos(d*x+c)*Ellip
ticF((-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*b-63*B*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+
b)/(a-b))^(1/2))*a^4*(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)+63*B*
EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^4*(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)+105*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)*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
*sin(d*x+c))*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^4*(1/cos(d*x+c))^(7/2)/sin(d*x+c)/(b+a*cos(d*x+c))
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\sec \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(7/2), x)
________________________________________________________________________________________
Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B b^{2} \sec \left (d x + c\right )^{3} + A a^{2} +{\left (2 \, B a b + A b^{2}\right )} \sec \left (d x + c\right )^{2} +{\left (B a^{2} + 2 \, A a b\right )} \sec \left (d x + c\right )\right )} \sqrt{b \sec \left (d x + c\right ) + a}}{\sec \left (d x + c\right )^{\frac{7}{2}}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="fricas")
[Out]
integral((B*b^2*sec(d*x + c)^3 + A*a^2 + (2*B*a*b + A*b^2)*sec(d*x + c)^2 + (B*a^2 + 2*A*a*b)*sec(d*x + c))*sq
rt(b*sec(d*x + c) + a)/sec(d*x + c)^(7/2), x)
________________________________________________________________________________________
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))**(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)**(7/2),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \sec \left (d x + c\right ) + A\right )}{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}{\sec \left (d x + c\right )^{\frac{7}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c))/sec(d*x+c)^(7/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)/sec(d*x + c)^(7/2), x)