3.609 \(\int \cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx\)
Optimal. Leaf size=340 \[ \frac {2 \left (25 a^2 A+77 a b B+45 A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{105 d}+\frac {2 \left (a^2-b^2\right ) \left (25 a^2 A+56 a b B+15 A b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (63 a^3 B+145 a^2 A b+161 a b^2 B+15 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 a (7 a B+10 A b) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{35 d}+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{7 d} \]
[Out]
2/7*a*A*cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/105*(a^2-b^2)*(25*A*a^2+15*A*b^2+56*B*a*b)*(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)/a/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/35*a*(10*A*b+7*B*a)*cos(d*x+c)^(3/2)*sin(d*x+
c)*(a+b*sec(d*x+c))^(1/2)/d+2/105*(25*A*a^2+45*A*b^2+77*B*a*b)*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1
/2)/d+2/105*(145*A*a^2*b+15*A*b^3+63*B*a^3+161*B*a*b^2)*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt
icE(sin(1/2*d*x+1/2*c),2^(1/2)*(a/(a+b))^(1/2))*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a/d/((b+a*cos(d*x+c))/
(a+b))^(1/2)
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Rubi [A] time = 1.32, antiderivative size = 340, normalized size of antiderivative =
1.00, number of steps used = 11, number of rules used = 11, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) =
0.314, Rules used = {2955, 4025, 4094, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ \frac {2 \left (25 a^2 A+77 a b B+45 A b^2\right ) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{105 d}+\frac {2 \left (a^2-b^2\right ) \left (25 a^2 A+56 a b B+15 A b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (145 a^2 A b+63 a^3 B+161 a b^2 B+15 A b^3\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 a (7 a B+10 A b) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{35 d}+\frac {2 a A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{7 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),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)])/(105*a*d*Sqrt[Cos[c + d*x]]*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)*Sqrt[Cos[c + d*x]]*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)]) + (2*(25*a^2*A + 45*A*b^2 + 77*a*b*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]
*Sin[c + d*x])/(105*d) + (2*a*(10*A*b + 7*a*B)*Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(35*d
) + (2*a*A*Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(7*d)
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 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 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 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 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 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]
Rubi steps
\begin {align*} \int \cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d}-\frac {1}{7} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d}-\frac {1}{35} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 a (10 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \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 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 a (10 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d}+\frac {\left (\left (a^2-b^2\right ) \left (25 a^2 A+15 A b^2+56 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}{105 a}+\frac {\left (\left (145 a^2 A b+15 A b^3+63 a^3 B+161 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}{105 a}\\ &=\frac {2 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 a (10 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d}+\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)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{105 a \sqrt {\cos (c+d x)} \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 {\cos (c+d x)} \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)}}\\ &=\frac {2 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 a (10 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d}+\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}}\right ) \int \frac {1}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{105 a \sqrt {\cos (c+d x)} \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 {\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}{105 a \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=\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 )}{105 a d \sqrt {\cos (c+d x)} \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 ) \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)}}{105 a d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 \left (25 a^2 A+45 A b^2+77 a b B\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 a (10 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 d}+\frac {2 a A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{7 d}\\ \end {align*}
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Mathematica [C] time = 19.41, size = 470, normalized size = 1.38 \[ \frac {\cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac {1}{210} \left (115 a^2 A+308 a b B+180 A b^2\right ) \sin (c+d x)+\frac {1}{14} a^2 A \sin (3 (c+d x))+\frac {1}{35} a (7 a B+15 A b) \sin (2 (c+d x))\right )}{d (a \cos (c+d x)+b)^2}-\frac {2 \cos ^{\frac {3}{2}}(c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} (a+b \sec (c+d x))^{5/2} \left (i a (a+b) \left (a^2 (25 A+63 B)+8 a b (15 A+7 B)+15 b^2 (A+7 B)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-\left (63 a^3 B+145 a^2 A b+161 a b^2 B+15 A b^3\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i (a+b) \left (63 a^3 B+145 a^2 A b+161 a b^2 B+15 A b^3\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{105 a d \sec ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+b)^3} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(7/2)*(a + b*Sec[c + d*x])^(5/2)*(A + B*Sec[c + d*x]),x]
[Out]
(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2)*(((115*a^2*A + 180*A*b^2 + 308*a*b*B)*Sin[c + d*x])/210 + (a*(1
5*A*b + 7*a*B)*Sin[2*(c + d*x)])/35 + (a^2*A*Sin[3*(c + d*x)])/14))/(d*(b + a*Cos[c + d*x])^2) - (2*Cos[c + d*
x]^(3/2)*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*(a + b*Sec[c + d*x])^(5/2)*((-I)*(a + b)*(145*a^2*A*b + 15*A*
b^3 + 63*a^3*B + 161*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)*(15*b^2*(A + 7*B) + 8*a*b*(15*A + 7*B) + a^2
*(25*A + 63*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)] - (145*a^2*A*b + 15*A*b^3 + 63*a^3*B + 161*a*b^2*B)*(b + a*Cos[c + d*x])
*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(105*a*d*(b + a*Cos[c + d*x])^3*Sec[c + d*x]^(5/2))
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fricas [F] time = 1.09, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{3} + {\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(7/2)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="fricas")
[Out]
integral((B*b^2*cos(d*x + c)^3*sec(d*x + c)^3 + A*a^2*cos(d*x + c)^3 + (2*B*a*b + A*b^2)*cos(d*x + c)^3*sec(d*
x + c)^2 + (B*a^2 + 2*A*a*b)*cos(d*x + c)^3*sec(d*x + c))*sqrt(b*sec(d*x + c) + a)*sqrt(cos(d*x + c)), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(7/2)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(7/2), x)
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maple [B] time = 3.03, size = 2450, normalized size = 7.21 \[ \text {Expression too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(7/2)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x)
[Out]
2/105/d*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)*(-1+cos(d*x+c))*(1+cos(d*x+c))*(60*A*((a-b)/(a+b)
)^(1/2)*cos(d*x+c)^4*a^3*b*(1/(1+cos(d*x+c)))^(1/2)+161*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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))*sin(d*x+c)*a^2*b^2-15*A*((a-b)/(a+b))
^(1/2)*b^4*(1/(1+cos(d*x+c)))^(1/2)-161*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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*b^3+119*B*EllipticF((-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)*sin(d*x+c)*a^3*
b-161*B*EllipticF((-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+co
s(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2+90*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^2*b^2*(1/(1+cos(d*x+c)))^(1/
2)-145*A*EllipticF((-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+c
os(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*a^3*b+135*A*EllipticF((-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)*sin(d*x+c)*a^2*b^2-15*A*EllipticF((-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)*sin(d*x+c
)*a*b^3+145*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^3*b-145*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^2*b^2+15*A*((b+a*cos(d*x+c))/(
1+cos(d*x+c))/(a+b))^(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*b^3+98*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^3*b*(1/(1+cos(d*x+c)))^(1/2)+110*A*((a-b)/(a+b))^(1/2)*co
s(d*x+c)^2*a^3*b*(1/(1+cos(d*x+c)))^(1/2)+60*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a*b^3*(1/(1+cos(d*x+c)))^(1/2)
+238*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^2*b^2*(1/(1+cos(d*x+c)))^(1/2)-145*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*
a^3*b*(1/(1+cos(d*x+c)))^(1/2)+55*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(1/2)-15*A*((a-b
)/(a+b))^(1/2)*cos(d*x+c)*a*b^3*(1/(1+cos(d*x+c)))^(1/2)-35*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^3*b*(1/(1+cos(d
*x+c)))^(1/2)-161*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(1/2)+161*B*((a-b)/(a+b))^(1/2)*
cos(d*x+c)*a*b^3*(1/(1+cos(d*x+c)))^(1/2)-63*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^3*b+105*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))*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)
*a*b^3+25*A*EllipticF((-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)*sin(d*x+c)*a^4-15*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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)*b^4+21*B*((a-b)/(a+b))^(1/2)*cos(d*x+
c)^4*a^4*(1/(1+cos(d*x+c)))^(1/2)+42*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^4*(1/(1+cos(d*x+c)))^(1/2)+15*A*((a-
b)/(a+b))^(1/2)*cos(d*x+c)*b^4*(1/(1+cos(d*x+c)))^(1/2)-63*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*a^4*(1/(1+cos(d*x+
c)))^(1/2)-25*A*((a-b)/(a+b))^(1/2)*a^3*b*(1/(1+cos(d*x+c)))^(1/2)-145*A*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(1+cos
(d*x+c)))^(1/2)-45*A*((a-b)/(a+b))^(1/2)*a*b^3*(1/(1+cos(d*x+c)))^(1/2)-63*B*((a-b)/(a+b))^(1/2)*a^3*b*(1/(1+c
os(d*x+c)))^(1/2)-77*B*((a-b)/(a+b))^(1/2)*a^2*b^2*(1/(1+cos(d*x+c)))^(1/2)-161*B*((a-b)/(a+b))^(1/2)*a*b^3*(1
/(1+cos(d*x+c)))^(1/2)+10*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*a^4*(1/(1+cos(d*x+c)))^(1/2)-25*A*((a-b)/(a+b))^(
1/2)*cos(d*x+c)*a^4*(1/(1+cos(d*x+c)))^(1/2)+63*B*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(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^4-63*B*sin(d*x+c)*((b+a*cos(d*x+
c))/(1+cos(d*x+c))/(a+b))^(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+15*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^5*a^4*(1/(1+cos(d*x+c)))^(1/2))/a/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c)
)/(1/(1+cos(d*x+c)))^(1/2)/sin(d*x+c)^3
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(7/2)*(a+b*sec(d*x+c))^(5/2)*(A+B*sec(d*x+c)),x, algorithm="maxima")
[Out]
integrate((B*sec(d*x + c) + A)*(b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(7/2), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^{7/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}\right )\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^(7/2)*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^(5/2),x)
[Out]
int(cos(c + d*x)^(7/2)*(A + B/cos(c + d*x))*(a + b/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(cos(d*x+c)**(7/2)*(a+b*sec(d*x+c))**(5/2)*(A+B*sec(d*x+c)),x)
[Out]
Timed out
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