3.1357 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\)

Optimal. Leaf size=291 \[ -\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{15 a^2 d}+\frac {2 \sqrt {\cos (c+d x)} \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 \left (-5 a^3 B+a^2 b (7 A+15 C)-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d} \]

[Out]

-2/15*(8*A*b^3-5*a^3*B-10*a*b^2*B+a^2*b*(7*A+15*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^3/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x
+c))^(1/2)+2/5*A*cos(d*x+c)^(3/2)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a/d-2/15*(4*A*b-5*B*a)*sin(d*x+c)*cos(d*x+
c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^2/d+2/15*(8*A*b^2-10*a*b*B+3*a^2*(3*A+5*C))*(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^3/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

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Rubi [A]  time = 0.98, antiderivative size = 291, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ -\frac {2 \left (a^2 b (7 A+15 C)-5 a^3 B-10 a b^2 B+8 A b^3\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {2 (4 A b-5 a B) \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{15 a^2 d}+\frac {2 A \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(-2*(8*A*b^3 - 5*a^3*B - 10*a*b^2*B + a^2*b*(7*A + 15*C))*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*
x)/2, (2*a)/(a + b)])/(15*a^3*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(8*A*b^2 - 10*a*b*B + 3*a^2*
(3*A + 5*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(15*a^3*d*Sqrt
[(b + a*Cos[c + d*x])/(a + b)]) - (2*(4*A*b - 5*a*B)*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])
/(15*a^2*d) + (2*A*Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*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 {5}{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 {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{2} (4 A b-5 a B)-\frac {1}{2} a (3 A+5 C) \sec (c+d x)-A b \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{5 a}\\ &=-\frac {2 (4 A b-5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (8 A b^2-10 a b B+3 a^2 (3 A+5 C)\right )+\frac {1}{4} a (2 A b+5 a B) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {2 (4 A b-5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}+\frac {\left (\left (8 A b^2-10 a b B+3 a^2 (3 A+5 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}{15 a^3}-\frac {\left (4 \left (-\frac {1}{4} a^2 (2 A b+5 a B)+\frac {1}{4} b \left (8 A b^2-10 a b B+3 a^2 (3 A+5 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}{15 a^3}\\ &=-\frac {2 (4 A b-5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {\left (4 \left (-\frac {1}{4} a^2 (2 A b+5 a B)+\frac {1}{4} b \left (8 A b^2-10 a b B+3 a^2 (3 A+5 C)\right )\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{15 a^3 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (8 A b^2-10 a b B+3 a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{15 a^3 \sqrt {b+a \cos (c+d x)}}\\ &=-\frac {2 (4 A b-5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {\left (4 \left (-\frac {1}{4} a^2 (2 A b+5 a B)+\frac {1}{4} b \left (8 A b^2-10 a b B+3 a^2 (3 A+5 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}{15 a^3 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (8 A b^2-10 a b B+3 a^2 (3 A+5 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}{15 a^3 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=-\frac {2 \left (8 A b^3-5 a^3 B-10 a b^2 B+a^2 b (7 A+15 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 )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (8 A b^2-10 a b B+3 a^2 (3 A+5 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)}}{15 a^3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}-\frac {2 (4 A b-5 a B) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 A \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}\\ \end {align*}

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Mathematica [C]  time = 17.07, size = 379, normalized size = 1.30 \[ \frac {2 a \sin (c+d x) (a \cos (c+d x)+b) (3 a A \cos (c+d x)+5 a B-4 A b)+\frac {2 \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (\tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) (a \cos (c+d x)+b)-i a \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (a^2 (9 A+5 (B+3 C))+2 a b (A-5 B)+8 A b^2\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+i (a+b) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (3 a^2 (3 A+5 C)-10 a b B+8 A b^2\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{\sec ^{\frac {3}{2}}(c+d x)}}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(Cos[c + d*x]^(5/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/Sqrt[a + b*Sec[c + d*x]],x]

[Out]

(2*a*(b + a*Cos[c + d*x])*(-4*A*b + 5*a*B + 3*a*A*Cos[c + d*x])*Sin[c + d*x] + (2*(Cos[(c + d*x)/2]^2*Sec[c +
d*x])^(3/2)*(I*(a + b)*(8*A*b^2 - 10*a*b*B + 3*a^2*(3*A + 5*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*(8*A*b^2 + 2*a*b*(
A - 5*B) + a^2*(9*A + 5*(B + 3*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)] + (8*A*b^2 - 10*a*b*B + 3*a^2*(3*A + 5*C))*(b + a*Co
s[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/Sec[c + d*x]^(3/2))/(15*a^3*d*Sqrt[Cos[c + d*x]]*Sqr
t[a + b*Sec[c + d*x]])

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fricas [F]  time = 0.55, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{2} \sec \left (d x + c\right )^{2} + B \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + A \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )}}{\sqrt {b \sec \left (d x + c\right ) + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(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)^2*sec(d*x + c)^2 + B*cos(d*x + c)^2*sec(d*x + c) + A*cos(d*x + c)^2)*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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(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)^(5/2)/sqrt(b*sec(d*x + c) + a), x)

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maple [B]  time = 2.32, size = 1887, normalized size = 6.48 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(1/2),x)

[Out]

-2/15/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*(9*A*sin(d*x+c
)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(3/2)-4*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)
))^(3/2)+9*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*((b+a*cos(d*x+c))/
(1+cos(d*x+c))/(a+b))^(1/2)*a^3-9*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+8*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))*b^3-5*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^3-15*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+15*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^3+5*B*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(3/2)-10*B*
sin(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(3/2)+15*C*((a-b)/(a+b))^(1/2)*a^2*b*sin(d*x+c)*(1/(1+
cos(d*x+c)))^(3/2)+3*A*cos(d*x+c)^2*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(1+cos(d*x+c)))^(3/2)+9*A*cos(d*x+c)
*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*(1/(1+cos(d*x+c)))^(3/2)+15*C*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a^3*sin(d*x+c
)*(1/(1+cos(d*x+c)))^(3/2)+5*B*cos(d*x+c)^2*sin(d*x+c)*((a-b)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(3/2)*a^3+5*B*co
s(d*x+c)*sin(d*x+c)*((a-b)/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(3/2)*a^3+3*A*cos(d*x+c)^3*sin(d*x+c)*((a-b)/(a+b))
^(1/2)*a^3*(1/(1+cos(d*x+c)))^(3/2)+8*A*sin(d*x+c)*((a-b)/(a+b))^(1/2)*b^3*(1/(1+cos(d*x+c)))^(3/2)+9*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-8*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^2+10*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-10*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^2-10*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+15*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*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+8*A*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/s
in(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^2-A*cos(d*x+c)^2*sin(d*x+c)*
((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(3/2)-A*cos(d*x+c)*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos
(d*x+c)))^(3/2)+4*A*cos(d*x+c)*sin(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2*(1/(1+cos(d*x+c)))^(3/2)-5*B*cos(d*x+c)*si
n(d*x+c)*((a-b)/(a+b))^(1/2)*a^2*b*(1/(1+cos(d*x+c)))^(3/2))/a^3/((a-b)/(a+b))^(1/2)/(b+a*cos(d*x+c))/(1/(1+co
s(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 \[ \int \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(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)^(5/2)/sqrt(b*sec(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^{5/2}\,\left (A+\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}\right )}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^(5/2)*(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)^(5/2)*(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)**(5/2)*(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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