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

[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)

________________________________________________________________________________________

Rubi [A]  time = 1.35008, antiderivative size = 380, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, 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 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]

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{\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*}

Mathematica [C]  time = 19.326, size = 492, normalized size = 1.29 \[ \frac{(a \cos (c+d x)+b) \left (\frac{\sin (c+d x) \left (115 a^2 A+140 a^2 C-112 a b B+96 A b^2\right )}{210 a^3}+\frac{(7 a B-6 A b) \sin (2 (c+d x))}{35 a^2}+\frac{A \sin (3 (c+d x))}{14 a}\right )}{d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 \sqrt{\cos (c+d x)} \left (\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (i a \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (-2 a^2 b (22 A-7 B+35 C)+a^3 (25 A+63 B+35 C)+4 a b^2 (14 B-3 A)-48 A b^3\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{b-a}{a+b}\right )+\tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right )^{3/2} \left (a^2 (44 A b+70 b C)-63 a^3 B-56 a b^2 B+48 A b^3\right ) (a \cos (c+d x)+b)-i (a+b) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \left (-2 a^2 b (22 A+35 C)+63 a^3 B+56 a b^2 B-48 A b^3\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{105 a^4 d \sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \]

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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Maple [B]  time = 0.771, size = 2829, normalized size = 7.4 \begin{align*} \text{output too large to display} \end{align*}

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)*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^3*(25*A*cos(d*x
+c)^2*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^4*(1/(cos(d*x+c)+1))^(3/2)-44*A*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)*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))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^
(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))*(1/(a+b)*(b+
a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^3+44*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+c
os(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*b-44*A*(1/(a+b)*(b+a*cos(d*x+c))/(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+48*A*(1/(a
+b)*(b+a*cos(d*x+c))/(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*b^3-48*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*b^4*(1/(cos(d*x+c)+1))^(3/2)+25*A*cos(d*x+c)*((a-b)/(a+b
))^(1/2)*sin(d*x+c)*a^4*(1/(cos(d*x+c)+1))^(3/2)+35*C*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^4*(1/(cos(d*
x+c)+1))^(3/2)+63*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^4*(1/(cos(d*x+c)+1))^(3/2)+25*A*((a-b)/(a+b))^
(1/2)*sin(d*x+c)*a^3*b*(1/(cos(d*x+c)+1))^(3/2)-44*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(cos(d*x+c)+1))
^(3/2)+24*A*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a*b^3*(1/(cos(d*x+c)+1))^(3/2)+63*B*((a-b)/(a+b))^(1/2)*sin(d*x+c)*
a^3*b*(1/(cos(d*x+c)+1))^(3/2)-28*B*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(cos(d*x+c)+1))^(3/2)+56*B*((a-b
)/(a+b))^(1/2)*sin(d*x+c)*a*b^3*(1/(cos(d*x+c)+1))^(3/2)+35*C*((a-b)/(a+b))^(1/2)*a^3*b*sin(d*x+c)*(1/(cos(d*x
+c)+1))^(3/2)-70*C*((a-b)/(a+b))^(1/2)*a^2*b^2*sin(d*x+c)*(1/(cos(d*x+c)+1))^(3/2)+15*A*cos(d*x+c)^4*((a-b)/(a
+b))^(1/2)*sin(d*x+c)*a^4*(1/(cos(d*x+c)+1))^(3/2)+21*B*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^4*(1/(co
s(d*x+c)+1))^(3/2)+21*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^4*(1/(cos(d*x+c)+1))^(3/2)+15*A*cos(d*x+
c)^3*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^4*(1/(cos(d*x+c)+1))^(3/2)+35*C*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*sin(d*x
+c)*a^4*(1/(cos(d*x+c)+1))^(3/2)+63*B*(1/(a+b)*(b+a*cos(d*x+c))/(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-56*B*(1/(a+b)*(b+a*cos(d*x+c))/(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+56*B*(1/(a+b)*(b+a*
cos(d*x+c))/(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*b^3-14*B*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)*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))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b^2+70*C*(1/(a+b)*(b+a*cos(d*x+c))/(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-70*C*(1/(a+b)*(
b+a*cos(d*x+c))/(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^2-70*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)*a^3*b-3*A*cos(d*x+c)^3*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*b*(1/(cos(d*x+c
)+1))^(3/2)-7*B*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*b*(1/(cos(d*x+c)+1))^(3/2)-19*A*cos(d*x+c)*((a
-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*b*(1/(cos(d*x+c)+1))^(3/2)+6*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b
^2*(1/(cos(d*x+c)+1))^(3/2)-24*A*cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a*b^3*(1/(cos(d*x+c)+1))^(3/2)-7*B*
cos(d*x+c)*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*b*(1/(cos(d*x+c)+1))^(3/2)+28*B*cos(d*x+c)*((a-b)/(a+b))^(1/2)*s
in(d*x+c)*a^2*b^2*(1/(cos(d*x+c)+1))^(3/2)-35*C*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*b*sin(d*x+c)*(1/(cos(d*x+c)
+1))^(3/2)-3*A*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*sin(d*x+c)*a^3*b*(1/(cos(d*x+c)+1))^(3/2)+6*A*cos(d*x+c)^2*((a
-b)/(a+b))^(1/2)*sin(d*x+c)*a^2*b^2*(1/(cos(d*x+c)+1))^(3/2)-63*B*(1/(a+b)*(b+a*cos(d*x+c))/(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+63*B*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)*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))*(1/(a+b)*(b+a*cos(d*x+c
))/(cos(d*x+c)+1))^(1/2)*a^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
))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^4-48*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*El
lipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^4)/a^4/((a-b)/(a+b))^(1/2)/(b+a
*cos(d*x+c))/sin(d*x+c)^6/(1/(cos(d*x+c)+1))^(3/2)

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Maxima [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(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]

Timed out

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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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(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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

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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