3.467 \(\int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=423 \[ \frac {2 \left (a^2 A+5 a b B-6 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a^2 d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (-5 a^3 B+9 a^2 A b+20 a b^2 B-24 A b^3\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{15 a^3 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}-\frac {2 \left (-5 a^3 B+12 a^2 A b-40 a b^2 B+48 A b^3\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^4 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (9 a^4 A-25 a^3 b B+24 a^2 A b^2+40 a b^3 B-48 A b^4\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^4 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]
[Out]
2*b*(A*b-B*a)*sin(d*x+c)/a/(a^2-b^2)/d/sec(d*x+c)^(3/2)/(a+b*sec(d*x+c))^(1/2)-2/15*(12*A*a^2*b+48*A*b^3-5*B*a
^3-40*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^
(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)*sec(d*x+c)^(1/2)/a^4/d/(a+b*sec(d*x+c))^(1/2)+2/5*(A*a^2-6*A*b^2+5*B*a*b
)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a^2/(a^2-b^2)/d/sec(d*x+c)^(3/2)-2/15*(9*A*a^2*b-24*A*b^3-5*B*a^3+20*B*a*b
^2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a^3/(a^2-b^2)/d/sec(d*x+c)^(1/2)+2/15*(9*A*a^4+24*A*a^2*b^2-48*A*b^4-25*
B*a^3*b+40*B*a*b^3)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a
+b))^(1/2))*(a+b*sec(d*x+c))^(1/2)/a^4/(a^2-b^2)/d/((b+a*cos(d*x+c))/(a+b))^(1/2)/sec(d*x+c)^(1/2)
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Rubi [A] time = 1.22, antiderivative size = 423, normalized size of antiderivative = 1.00,
number of steps used = 10, number of rules used = 9, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used
= {4030, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {2 \left (a^2 A+5 a b B-6 A b^2\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{5 a^2 d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 b (A b-a B) \sin (c+d x)}{a d \left (a^2-b^2\right ) \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (9 a^2 A b-5 a^3 B+20 a b^2 B-24 A b^3\right ) \sin (c+d x) \sqrt {a+b \sec (c+d x)}}{15 a^3 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)}}-\frac {2 \left (12 a^2 A b-5 a^3 B-40 a b^2 B+48 A b^3\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^4 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (24 a^2 A b^2+9 a^4 A-25 a^3 b B+40 a b^3 B-48 A b^4\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^4 d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]
Antiderivative was successfully verified.
[In]
Int[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2)),x]
[Out]
(-2*(12*a^2*A*b + 48*A*b^3 - 5*a^3*B - 40*a*b^2*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^4*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(9*a^4*A + 24*a^2*A*b^2 - 48*A*b^4
- 25*a^3*b*B + 40*a*b^3*B)*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(15*a^4*(a^2 - b^2)
*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*Sqrt[Sec[c + d*x]]) + (2*b*(A*b - a*B)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Se
c[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]) + (2*(a^2*A - 6*A*b^2 + 5*a*b*B)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d
*x])/(5*a^2*(a^2 - b^2)*d*Sec[c + d*x]^(3/2)) - (2*(9*a^2*A*b - 24*A*b^3 - 5*a^3*B + 20*a*b^2*B)*Sqrt[a + b*Se
c[c + d*x]]*Sin[c + d*x])/(15*a^3*(a^2 - b^2)*d*Sqrt[Sec[c + d*x]])
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 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 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]
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 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 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 4030
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(b*(A*b - a*B)*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)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*
x])^n*Simp[A*(a^2*(m + 1) - b^2*(m + n + 1)) + a*b*B*n - a*(A*b - a*B)*(m + 1)*Csc[e + f*x] + b*(A*b - a*B)*(m
+ n + 2)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[A*b - a*B, 0] && 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 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]
Rubi steps
\begin {align*} \int \frac {A+B \sec (c+d x)}{\sec ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}} \, dx &=\frac {2 b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}-\frac {2 \int \frac {\frac {1}{2} \left (-a^2 A+6 A b^2-5 a b B\right )+\frac {1}{2} a (A b-a B) \sec (c+d x)-2 b (A b-a B) \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{a \left (a^2-b^2\right )}\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\frac {1}{4} \left (-9 a^2 A b+24 A b^3+5 a^3 B-20 a b^2 B\right )+\frac {1}{4} a \left (3 a^2 A+2 A b^2-5 a b B\right ) \sec (c+d x)+\frac {1}{2} b \left (a^2 A-6 A b^2+5 a b B\right ) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{5 a^2 \left (a^2-b^2\right )}\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}-\frac {8 \int \frac {\frac {1}{8} \left (-9 a^4 A-24 a^2 A b^2+48 A b^4+25 a^3 b B-40 a b^3 B\right )+\frac {1}{8} a \left (3 a^2 A b+12 A b^3-5 a^3 B-10 a b^2 B\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{15 a^3 \left (a^2-b^2\right )}\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}-\frac {\left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 B\right ) \int \frac {\sqrt {\sec (c+d x)}}{\sqrt {a+b \sec (c+d x)}} \, dx}{15 a^4}+\frac {\left (9 a^4 A+24 a^2 A b^2-48 A b^4-25 a^3 b B+40 a b^3 B\right ) \int \frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \, dx}{15 a^4 \left (a^2-b^2\right )}\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}-\frac {\left (\left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 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}{15 a^4 \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (9 a^4 A+24 a^2 A b^2-48 A b^4-25 a^3 b B+40 a b^3 B\right ) \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{15 a^4 \left (a^2-b^2\right ) \sqrt {b+a \cos (c+d x)} \sqrt {\sec (c+d x)}}\\ &=\frac {2 b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}-\frac {\left (\left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 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}{15 a^4 \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (9 a^4 A+24 a^2 A b^2-48 A b^4-25 a^3 b B+40 a b^3 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^4 \left (a^2-b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}\\ &=-\frac {2 \left (12 a^2 A b+48 A b^3-5 a^3 B-40 a b^2 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)}}{15 a^4 d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (9 a^4 A+24 a^2 A b^2-48 A b^4-25 a^3 b B+40 a b^3 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^4 \left (a^2-b^2\right ) d \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \sqrt {\sec (c+d x)}}+\frac {2 b (A b-a B) \sin (c+d x)}{a \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (a^2 A-6 A b^2+5 a b B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \sec ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (9 a^2 A b-24 A b^3-5 a^3 B+20 a b^2 B\right ) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^3 \left (a^2-b^2\right ) d \sqrt {\sec (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 2.40, size = 316, normalized size = 0.75 \[ -\frac {\sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b) \left (a (a-b) (a+b) \left (2 \left (a^2-b^2\right ) (9 A b-5 a B) \sin (c+d x) (a \cos (c+d x)+b)-3 a A \left (a^2-b^2\right ) \sin (2 (c+d x)) (a \cos (c+d x)+b)+30 b^3 (a B-A b) \sin (c+d x)\right )+2 \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \left (a^2 \left (-5 a^3 B+3 a^2 A b-10 a b^2 B+12 A b^3\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )-\left (9 a^4 A-25 a^3 b B+24 a^2 A b^2+40 a b^3 B-48 A b^4\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )-b F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )\right )\right )\right )}{15 a^4 d \left (a^2-b^2\right )^2 (a+b \sec (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(A + B*Sec[c + d*x])/(Sec[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2)),x]
[Out]
-1/15*((b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*(2*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*(a^2*(3*a^2*A
*b + 12*A*b^3 - 5*a^3*B - 10*a*b^2*B)*EllipticF[(c + d*x)/2, (2*a)/(a + b)] - (9*a^4*A + 24*a^2*A*b^2 - 48*A*b
^4 - 25*a^3*b*B + 40*a*b^3*B)*((a + b)*EllipticE[(c + d*x)/2, (2*a)/(a + b)] - b*EllipticF[(c + d*x)/2, (2*a)/
(a + b)])) + a*(a - b)*(a + b)*(30*b^3*(-(A*b) + a*B)*Sin[c + d*x] + 2*(a^2 - b^2)*(9*A*b - 5*a*B)*(b + a*Cos[
c + d*x])*Sin[c + d*x] - 3*a*A*(a^2 - b^2)*(b + a*Cos[c + d*x])*Sin[2*(c + d*x)])))/(a^4*(a^2 - b^2)^2*d*(a +
b*Sec[c + d*x])^(3/2))
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fricas [F] time = 0.95, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \sec \left (d x + c\right ) + A\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\sec \left (d x + c\right )}}{b^{2} \sec \left (d x + c\right )^{5} + 2 \, a b \sec \left (d x + c\right )^{4} + a^{2} \sec \left (d x + c\right )^{3}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
integral((B*sec(d*x + c) + A)*sqrt(b*sec(d*x + c) + a)*sqrt(sec(d*x + c))/(b^2*sec(d*x + c)^5 + 2*a*b*sec(d*x
+ c)^4 + a^2*sec(d*x + c)^3), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sec \left (d x + c\right ) + A}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)/((b*sec(d*x + c) + a)^(3/2)*sec(d*x + c)^(5/2)), x)
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maple [B] time = 2.50, size = 3156, normalized size = 7.46 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x)
[Out]
2/15/d*(-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^4*((b+a*cos(d*x+
c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^
(1/2)*sin(d*x+c)+48*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*((b+a
*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+9*A*a^3*b*((a-b)/(a+b))^(1/2)+24*
A*a*b^3*((a-b)/(a+b))^(1/2)+5*B*a^3*b*((a-b)/(a+b))^(1/2)-20*B*a^2*b^2*((a-b)/(a+b))^(1/2)-40*B*a*b^3*((a-b)/(
a+b))^(1/2)-5*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^4-3*A*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^4-6*A*cos(d*x+c)^2
*((a-b)/(a+b))^(1/2)*a^4-9*A*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)
*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a^4-48*A*cos(d*x+c)
*((a-b)/(a+b))^(1/2)*b^4+5*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^4+9*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^4+6*A*cos
(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^3*b+48*A*b^4*((a-b)/(a+b))^(1/2)-24*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b^2
+20*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b+48*A*cos(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))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c
)*b^4+9*A*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(
d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a^4-5*B*cos(d*x+c)*((b+a*cos(d*x+c))/(
1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(
-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a^4-24*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*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+12*A*Ellip
ticF((-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+a*cos(d*x+c))/(1+cos(d*x+c
))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+36*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*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d
*x+c)+48*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*((b+a*cos(d*x+
c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+25*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*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)
))^(1/2)*sin(d*x+c)-40*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*
((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-30*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*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1
/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-40*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*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-24*A*cos(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))*((b+a*cos(d*x+c))/(1+cos(d*
x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*a^2*b^2+12*A*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c)
)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b)
)^(1/2))*sin(d*x+c)*a^3*b+36*A*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/
2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a^2*b^2+48*A*cos(
d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)
/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a*b^3+25*B*cos(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))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c
)))^(1/2)*sin(d*x+c)*a^3*b-40*B*cos(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))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)*a*b^3-30*B*cos(d
*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/
(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*sin(d*x+c)*a^3*b-40*B*cos(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c)
)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b)
)^(1/2))*sin(d*x+c)*a^2*b^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^4*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+6*A*cos(d*x+c)^3*((a-b)
/(a+b))^(1/2)*a^2*b^2-5*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^3*b-6*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3*b-24
*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^3+20*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^2*b^2-6*A*cos(d*x+c)*((a-b)/
(a+b))^(1/2)*a^3*b+18*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b^2-20*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b-3*A*c
os(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3*b+40*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^3)*((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))/a^4/(a+b)/((a-b)/(a+b))^(1/2)
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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((A+B*sec(d*x+c))/sec(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Timed out
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {A+\frac {B}{\cos \left (c+d\,x\right )}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((A + B/cos(c + d*x))/((a + b/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(5/2)),x)
[Out]
int((A + B/cos(c + d*x))/((a + b/cos(c + d*x))^(3/2)*(1/cos(c + d*x))^(5/2)), x)
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sympy [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((A+B*sec(d*x+c))/sec(d*x+c)**(5/2)/(a+b*sec(d*x+c))**(3/2),x)
[Out]
Timed out
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