3.1350 \(\int \cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)
Optimal. Leaf size=441 \[ \frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 (5 A+7 C)+56 a b B+15 A b^2\right ) \sqrt {a+b \sec (c+d x)}}{105 d}+\frac {2 \sqrt {\cos (c+d x)} \left (63 a^3 B+5 a^2 b (29 A+49 C)+161 a b^2 B+15 A b^3\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \left (-5 a^4 (5 A+7 C)-56 a^3 b B+10 a^2 b^2 (A-7 C)+56 a b^3 B+15 A b^4\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 (7 a B+5 A b) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{35 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}+\frac {2 b^3 C \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]
[Out]
2/35*(5*A*b+7*B*a)*cos(d*x+c)^(3/2)*(a+b*sec(d*x+c))^(3/2)*sin(d*x+c)/d+2/7*A*cos(d*x+c)^(5/2)*(a+b*sec(d*x+c)
)^(5/2)*sin(d*x+c)/d-2/105*(15*A*b^4-56*a^3*b*B+56*a*b^3*B+10*a^2*b^2*(A-7*C)-5*a^4*(5*A+7*C))*(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*b^3*C*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*El
lipticPi(sin(1/2*d*x+1/2*c),2,2^(1/2)*(a/(a+b))^(1/2))*((b+a*cos(d*x+c))/(a+b))^(1/2)/d/cos(d*x+c)^(1/2)/(a+b*
sec(d*x+c))^(1/2)+2/105*(15*A*b^2+56*a*b*B+5*a^2*(5*A+7*C))*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)
/d+2/105*(15*A*b^3+63*a^3*B+161*a*b^2*B+5*a^2*b*(29*A+49*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*E
llipticE(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)
________________________________________________________________________________________
Rubi [A] time = 1.82, antiderivative size = 441, normalized size of antiderivative = 1.00,
number of steps used = 15, number of rules used = 13, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.289, Rules
used = {4265, 4094, 4108, 3859, 2807, 2805, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ \frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 (5 A+7 C)+56 a b B+15 A b^2\right ) \sqrt {a+b \sec (c+d x)}}{105 d}-\frac {2 \left (10 a^2 b^2 (A-7 C)-5 a^4 (5 A+7 C)-56 a^3 b B+56 a b^3 B+15 A b^4\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \left (5 a^2 b (29 A+49 C)+63 a^3 B+161 a b^2 B+15 A b^3\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 (7 a B+5 A b) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2}}{35 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}}{7 d}+\frac {2 b^3 C \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]
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] + C*Sec[c + d*x]^2),x]
[Out]
(-2*(15*A*b^4 - 56*a^3*b*B + 56*a*b^3*B + 10*a^2*b^2*(A - 7*C) - 5*a^4*(5*A + 7*C))*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*b^3
*C*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticPi[2, (c + d*x)/2, (2*a)/(a + b)])/(d*Sqrt[Cos[c + d*x]]*Sqrt[a
+ b*Sec[c + d*x]]) + (2*(15*A*b^3 + 63*a^3*B + 161*a*b^2*B + 5*a^2*b*(29*A + 49*C))*Sqrt[Cos[c + d*x]]*Ellipti
cE[(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*(15
*A*b^2 + 56*a*b*B + 5*a^2*(5*A + 7*C))*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(105*d) + (2*
(5*A*b + 7*a*B)*Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(3/2)*Sin[c + d*x])/(35*d) + (2*A*Cos[c + d*x]^(5/2)*(
a + b*Sec[c + d*x])^(5/2)*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 2805
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Rule 2807
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] && !GtQ[c + d, 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 3859
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(d*Sqr
t[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]])/Sqrt[a + b*Csc[e + f*x]], Int[1/(Sin[e + f*x]*Sqrt[b + a*Sin[e + f
*x]]), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 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 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 4108
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d
_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]), x_Symbol] :> Dist[C/d^2, Int[(d*Csc[e + f*x])^(3/2)/Sqrt[a +
b*Csc[e + f*x]], x], x] + Int[(A + B*Csc[e + f*x])/(Sqrt[d*Csc[e + f*x]]*Sqrt[a + b*Csc[e + f*x]]), x] /; Fre
eQ[{a, b, d, e, f, A, B, C}, x] && NeQ[a^2 - b^2, 0]
Rule 4265
Int[(cos[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cos[a + b*x])^m*(c*Sec[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Sec[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] && !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]
Rubi steps
\begin {align*} \int \cos ^{\frac {7}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \sec (c+d x))^{5/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sec ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{7} \left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \sec (c+d x))^{3/2} \left (\frac {1}{2} (5 A b+7 a B)+\frac {1}{2} (5 a A+7 b B+7 a C) \sec (c+d x)+\frac {7}{2} b C \sec ^2(c+d x)\right )}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\\ &=\frac {2 (5 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{35} \left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {a+b \sec (c+d x)} \left (\frac {1}{4} \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right )+\frac {1}{4} \left (40 a A b+21 a^2 B+35 b^2 B+70 a b C\right ) \sec (c+d x)+\frac {35}{4} b^2 C \sec ^2(c+d x)\right )}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (5 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{105} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right )+\frac {1}{8} \left (119 a^2 b B+105 b^3 B+45 a b^2 (3 A+7 C)+5 a^3 (5 A+7 C)\right ) \sec (c+d x)+\frac {105}{8} b^3 C \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (5 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {1}{105} \left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right )+\frac {1}{8} \left (119 a^2 b B+105 b^3 B+45 a b^2 (3 A+7 C)+5 a^3 (5 A+7 C)\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx+\left (b^3 C \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (5 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}+\frac {\left (\left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\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 {\left (8 \left (-\frac {1}{8} a \left (119 a^2 b B+105 b^3 B+45 a b^2 (3 A+7 C)+5 a^3 (5 A+7 C)\right )+\frac {1}{8} b \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right )\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 (b^3 C \sqrt {b+a \cos (c+d x)}\right ) \int \frac {\sec (c+d x)}{\sqrt {b+a \cos (c+d x)}} \, dx}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}\\ &=\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (5 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}-\frac {\left (8 \left (-\frac {1}{8} a \left (119 a^2 b B+105 b^3 B+45 a b^2 (3 A+7 C)+5 a^3 (5 A+7 C)\right )+\frac {1}{8} b \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right )\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 (b^3 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}}\right ) \int \frac {\sec (c+d x)}{\sqrt {\frac {b}{a+b}+\frac {a \cos (c+d x)}{a+b}}} \, dx}{\sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\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 b^3 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (5 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}-\frac {\left (8 \left (-\frac {1}{8} a \left (119 a^2 b B+105 b^3 B+45 a b^2 (3 A+7 C)+5 a^3 (5 A+7 C)\right )+\frac {1}{8} b \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\right )\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 (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\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 (15 A b^4-56 a^3 b B+56 a b^3 B+10 a^2 b^2 (A-7 C)-5 a^4 (5 A+7 C)\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 b^3 C \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \Pi \left (2;\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (15 A b^3+63 a^3 B+161 a b^2 B+5 a^2 b (29 A+49 C)\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 (15 A b^2+56 a b B+5 a^2 (5 A+7 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{105 d}+\frac {2 (5 A b+7 a B) \cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \sin (c+d x)}{35 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2} \sin (c+d x)}{7 d}\\ \end {align*}
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Mathematica [C] time = 35.24, size = 64878, normalized size = 147.12 \[ \text {Result too large to show} \]
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] + C*Sec[c + d*x]^2),x]
[Out]
Result too large to show
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fricas [F] time = 2.84, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{4} + {\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{3} + A a^{2} \cos \left (d x + c\right )^{3} + {\left (C a^{2} + 2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + {\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}, x\right ) \]
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)+C*sec(d*x+c)^2),x, algorithm="fricas")
[Out]
integral((C*b^2*cos(d*x + c)^3*sec(d*x + c)^4 + (2*C*a*b + B*b^2)*cos(d*x + c)^3*sec(d*x + c)^3 + A*a^2*cos(d*
x + c)^3 + (C*a^2 + 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 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]
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)+C*sec(d*x+c)^2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + 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 [C] time = 2.44, size = 3164, normalized size = 7.17 \[ \text {output too large to display} \]
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)+C*sec(d*x+c)^2),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))^2*(-1+cos(d*x+c))^3*(15*A*((a-b)/
(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^3*a^4*(1/(1+cos(d*x+c)))^(3/2)+63*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)*a^4-63*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))*
a^4-35*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+c
os(d*x+c))/(a+b))^(1/2)*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))*b^4-25*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)*a^4+25*A*((a-b)/(a+b))^(1/2)*sin(d*x+c
)*cos(d*x+c)^2*(1/(1+cos(d*x+c)))^(3/2)*a^4+35*C*((a-b)/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^2*a^4*(1/(1+cos(d*x
+c)))^(3/2)+21*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*a^4*(1/(1+cos(d*x+c)))^(3/2)+21*B*((a-b)/(a+b))^(
1/2)*cos(d*x+c)^2*sin(d*x+c)*a^4*(1/(1+cos(d*x+c)))^(3/2)+63*B*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*b*(1/(1+cos(
d*x+c)))^(3/2)+77*B*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(3/2)+161*B*((a-b)/(a+b))^(1/2)*
sin(d*x+c)*a*b^3*(1/(1+cos(d*x+c)))^(3/2)+35*C*((a-b)/(a+b))^(1/2)*a^3*b*sin(d*x+c)*(1/(1+cos(d*x+c)))^(3/2)+2
45*C*((a-b)/(a+b))^(1/2)*a^2*b^2*sin(d*x+c)*(1/(1+cos(d*x+c)))^(3/2)+63*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*sin(d
*x+c)*a^4*(1/(1+cos(d*x+c)))^(3/2)+25*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*b*(1/(1+cos(d*x+c)))^(3/2)+145*A*((
a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(3/2)+45*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a*b^3*(1/(1+
cos(d*x+c)))^(3/2)+25*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)*a^4*(1/(1+cos(d*x+c)))^(3/2)+35*C*((a-b)/(a+
b))^(1/2)*sin(d*x+c)*cos(d*x+c)*a^4*(1/(1+cos(d*x+c)))^(3/2)+15*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*cos(d*x+c)^4*
(1/(1+cos(d*x+c)))^(3/2)*a^4+98*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^3*b*(1/(1+cos(d*x+c)))^(3/2)+1
70*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*sin(d*x+c)*a^3*b*(1/(1+cos(d*x+c)))^(3/2)+90*A*((a-b)/(a+b))^(1/2)*cos(d*x
+c)*sin(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(3/2)+60*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*sin(d*x+c)*a*b^3*(1/(1+cos
(d*x+c)))^(3/2)+98*B*((a-b)/(a+b))^(1/2)*cos(d*x+c)*sin(d*x+c)*a^3*b*(1/(1+cos(d*x+c)))^(3/2)+238*B*((a-b)/(a+
b))^(1/2)*cos(d*x+c)*sin(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(3/2)+60*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^3*sin(d*x
+c)*a^3*b*(1/(1+cos(d*x+c)))^(3/2)+60*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^3*b*(1/(1+cos(d*x+c)))^(
3/2)+90*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(3/2)+280*C*((a-b)/(a+b))^(1/
2)*cos(d*x+c)*a^3*b*sin(d*x+c)*(1/(1+cos(d*x+c)))^(3/2)-315*C*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/si
n(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^2*b^2-105*B*((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*b^
3+105*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+co
s(d*x+c))/(a+b))^(1/2)*a*b^3-210*C*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticPi((-1+cos(d*x+c))*((
a-b)/(a+b))^(1/2)/sin(d*x+c),(a+b)/(a-b),I/((a-b)/(a+b))^(1/2))*a*b^3+15*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*b^4*
(1/(1+cos(d*x+c)))^(3/2)-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)*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+cos(d*x+c))/(a+b))^(1/2)*a^2*b^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))*a^3*b-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))*a^2*b^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))*a*b^3+245*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^3*b-245*C*((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))*a^3*b+245
*C*((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))*a^2*b^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+cos(d*x+c))/(a+b))^(1/2)*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)*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)*a*b^3-14
5*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))*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))*a^2*b^2-15*A*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellip
ticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^3)/a/((a-b)/(a+b))^(1/2)/(b+a*co
s(d*x+c))/(1/(1+cos(d*x+c)))^(3/2)/sin(d*x+c)^6
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{\frac {7}{2}}\,{d x} \]
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)+C*sec(d*x+c)^2),x, algorithm="maxima")
[Out]
integrate((C*sec(d*x + c)^2 + 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 \[ \int {\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right ) \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
[Out]
int(cos(c + d*x)^(7/2)*(a + b/cos(c + d*x))^(5/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^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)+C*sec(d*x+c)**2),x)
[Out]
Timed out
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