3.1356 \(\int \frac {\cos ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=380 \[ -\frac {2 (6 A b-7 a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{35 a^2 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 (5 A+7 C)-28 a b B+24 A b^2\right ) \sqrt {a+b \sec (c+d x)}}{105 a^3 d}-\frac {2 \sqrt {\cos (c+d x)} \left (-63 a^3 B+a^2 (44 A b+70 b C)-56 a b^2 B+48 A b^3\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a^4 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 \left (5 a^4 (5 A+7 C)-49 a^3 b B+2 a^2 b^2 (16 A+35 C)-56 a b^3 B+48 A b^4\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a^4 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{7 a d} \]

[Out]

2/105*(48*A*b^4-49*a^3*b*B-56*a*b^3*B+5*a^4*(5*A+7*C)+2*a^2*b^2*(16*A+35*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^4/d/cos(
d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)-2/35*(6*A*b-7*B*a)*cos(d*x+c)^(3/2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a^2/
d+2/7*A*cos(d*x+c)^(5/2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a/d+2/105*(24*A*b^2-28*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)/a^3/d-2/105*(48*A*b^3-63*a^3*B-56*a*b^2*B+a^2*(44*A*b+70*C*b))*(
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))*cos(d*x+c
)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^4/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

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Rubi [A]  time = 1.35, antiderivative size = 380, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 9, integrand size = 45, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {4265, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (5 a^2 (5 A+7 C)-28 a b B+24 A b^2\right ) \sqrt {a+b \sec (c+d x)}}{105 a^3 d}+\frac {2 \left (2 a^2 b^2 (16 A+35 C)+5 a^4 (5 A+7 C)-49 a^3 b B-56 a b^3 B+48 A b^4\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a^4 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \left (a^2 (44 A b+70 b C)-63 a^3 B-56 a b^2 B+48 A b^3\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{105 a^4 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 (6 A b-7 a B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{35 a^2 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{7 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(7/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(2*(48*A*b^4 - 49*a^3*b*B - 56*a*b^3*B + 5*a^4*(5*A + 7*C) + 2*a^2*b^2*(16*A + 35*C))*Sqrt[(b + a*Cos[c + d*x]
)/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(105*a^4*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) - (2
*(48*A*b^3 - 63*a^3*B - 56*a*b^2*B + a^2*(44*A*b + 70*b*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a
+ b)]*Sqrt[a + b*Sec[c + d*x]])/(105*a^4*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (2*(24*A*b^2 - 28*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*a^3*d) - (2*(6*A*b - 7*a*B)*Co
s[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(35*a^2*d) + (2*A*Cos[c + d*x]^(5/2)*Sqrt[a + b*Sec[c
+ d*x]]*Sin[c + d*x])/(7*a*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 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 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]

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 \frac {\cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{\sqrt {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)+C \sec ^2(c+d x)}{\sec ^{\frac {7}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 a d}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} (6 A b-7 a B)-\frac {1}{2} a (5 A+7 C) \sec (c+d x)-2 A b \sec ^2(c+d x)}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{7 a}\\ &=-\frac {2 (6 A b-7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 a^2 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 a d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (24 A b^2-28 a b B+5 a^2 (5 A+7 C)\right )+\frac {1}{4} a (2 A b+21 a B) \sec (c+d x)-\frac {1}{2} b (6 A b-7 a B) \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{35 a^2}\\ &=\frac {2 \left (24 A b^2-28 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 a^3 d}-\frac {2 (6 A b-7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 a^2 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 a d}-\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{8} \left (48 A b^3-63 a^3 B-56 a b^2 B+2 a^2 b (22 A+35 C)\right )+\frac {1}{8} a \left (12 A b^2-14 a b B-5 a^2 (5 A+7 C)\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{105 a^3}\\ &=\frac {2 \left (24 A b^2-28 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 a^3 d}-\frac {2 (6 A b-7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 a^2 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 a d}-\frac {\left (\left (48 A b^3-63 a^3 B-56 a b^2 B+2 a^2 b (22 A+35 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^4}+\frac {\left (8 \left (-\frac {1}{8} a^2 \left (12 A b^2-14 a b B-5 a^2 (5 A+7 C)\right )+\frac {1}{8} b \left (48 A b^3-63 a^3 B-56 a b^2 B+2 a^2 b (22 A+35 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^4}\\ &=\frac {2 \left (24 A b^2-28 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 a^3 d}-\frac {2 (6 A b-7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 a^2 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 a d}+\frac {\left (8 \left (-\frac {1}{8} a^2 \left (12 A b^2-14 a b B-5 a^2 (5 A+7 C)\right )+\frac {1}{8} b \left (48 A b^3-63 a^3 B-56 a b^2 B+2 a^2 b (22 A+35 C)\right )\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{105 a^4 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (48 A b^3-63 a^3 B-56 a b^2 B+2 a^2 b (22 A+35 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^4 \sqrt {b+a \cos (c+d x)}}\\ &=\frac {2 \left (24 A b^2-28 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 a^3 d}-\frac {2 (6 A b-7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 a^2 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 a d}+\frac {\left (8 \left (-\frac {1}{8} a^2 \left (12 A b^2-14 a b B-5 a^2 (5 A+7 C)\right )+\frac {1}{8} b \left (48 A b^3-63 a^3 B-56 a b^2 B+2 a^2 b (22 A+35 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^4 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (\left (48 A b^3-63 a^3 B-56 a b^2 B+2 a^2 b (22 A+35 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^4 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=\frac {2 \left (48 A b^4-49 a^3 b B-56 a b^3 B+5 a^4 (5 A+7 C)+2 a^2 b^2 (16 A+35 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^4 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \left (48 A b^3-63 a^3 B-56 a b^2 B+a^2 (44 A b+70 b 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^4 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 \left (24 A b^2-28 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 a^3 d}-\frac {2 (6 A b-7 a B) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{35 a^2 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{7 a d}\\ \end {align*}

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Mathematica [C]  time = 19.04, size = 492, normalized size = 1.29 \[ \frac {(a \cos (c+d x)+b) \left (\frac {(7 a B-6 A b) \sin (2 (c+d x))}{35 a^2}+\frac {\sin (c+d x) \left (115 a^2 A+140 a^2 C-112 a b B+96 A b^2\right )}{210 a^3}+\frac {A \sin (3 (c+d x))}{14 a}\right )}{d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \left (-63 a^3 B+a^2 (44 A b+70 b C)-56 a b^2 B+48 A b^3\right ) (a \cos (c+d x)+b)+i a \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (a^3 (25 A+63 B+35 C)-2 a^2 b (22 A-7 B+35 C)+4 a b^2 (14 B-3 A)-48 A b^3\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-i (a+b) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (63 a^3 B-2 a^2 b (22 A+35 C)+56 a b^2 B-48 A b^3\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{105 a^4 d \sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]^(7/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + b*Sec[c + d*x]],x]

[Out]

((b + a*Cos[c + d*x])*(((115*a^2*A + 96*A*b^2 - 112*a*b*B + 140*a^2*C)*Sin[c + d*x])/(210*a^3) + ((-6*A*b + 7*
a*B)*Sin[2*(c + d*x)])/(35*a^2) + (A*Sin[3*(c + d*x)])/(14*a)))/(d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]
) - (2*Sqrt[Cos[c + d*x]]*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*((-I)*(a + b)*(-48*A*b^3 + 63*a^3*B + 56*a*b
^2*B - 2*a^2*b*(22*A + 35*C))*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*(-48*A*b^3 + 4*a*b^2*(-3*A + 14*B) - 2*a^2*b*(22*A -
7*B + 35*C) + a^3*(25*A + 63*B + 35*C))*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)] + (48*A*b^3 - 63*a^3*B - 56*a*b^2*B + a^2*(44*A
*b + 70*b*C))*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(105*a^4*d*Sqrt[Sec[c + d*x]]
*Sqrt[a + b*Sec[c + d*x]])

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + B \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{3}\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \sec \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^3*sec(d*x + c)^2 + B*cos(d*x + c)^3*sec(d*x + c) + A*cos(d*x + c)^3)*sqrt(cos(d*x + c
))/sqrt(b*sec(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(7/2)/sqrt(b*sec(d*x + c) + a), x)

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maple [B]  time = 2.56, size = 2829, normalized size = 7.44 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),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^4-48*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)-28*B*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(3/2)+56*B*((a-b)/(a+b))^(1/2)*s
in(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)-70
*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)-44*A*((a-b
)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(3/2)+24*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^4-7*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)-19*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)+6*A*((a-b)/(a+b))^(1/2)*cos(d*x+c)*si
n(d*x+c)*a^2*b^2*(1/(1+cos(d*x+c)))^(3/2)-24*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)-7*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)+28*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)-3*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)-3*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)+6*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)-35*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)-48*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*b^4*(1/(1+cos(d*x+c)))^(3/2)-1
4*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+56*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)*Ellipt
icE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*b-56*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+56*
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^3-70*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+70*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-70*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^2-44*A*El
lipticF((-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+12*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*c
os(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*a^2*b^2-48*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+44*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-44*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+48*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*b^3)/a^4/((a-b)/(a+b))^(1/2)/(b+a*cos(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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {7}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(7/2)/sqrt(b*sec(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{7/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(7/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^(1/2),x)

[Out]

int((cos(c + d*x)^(7/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + b/cos(c + d*x))^(1/2), 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)+C*sec(d*x+c)**2)/(a+b*sec(d*x+c))**(1/2),x)

[Out]

Timed out

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