3.843 \(\int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=240 \[ \frac {2 b \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{5 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{5 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{5 d}+\frac {4 b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{5 d} \]

[Out]

2/5*b*(a^2-b^2)*(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)+2/5*a*cos(d*x+c)^(3/2)*sin(
d*x+c)*(a+b*sec(d*x+c))^(1/2)/d+4/5*b*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/d+2/5*(3*a^2+b^2)*(co
s(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/d/((b+a*cos(d*x+c))/(a+b))^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 0.68, antiderivative size = 240, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4264, 3864, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ \frac {2 b \left (a^2-b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{5 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{5 a d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 a \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{5 d}+\frac {4 b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(2*b*(a^2 - b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(5*a*d*Sqrt[Cos[c +
d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(3*a^2 + b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*S
qrt[a + b*Sec[c + d*x]])/(5*a*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) + (4*b*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c
+ d*x]]*Sin[c + d*x])/(5*d) + (2*a*Cos[c + d*x]^(3/2)*Sqrt[a + b*Sec[c + d*x]]*Sin[c + d*x])/(5*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 3864

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(3/2), x_Symbol] :> Simp[(a*Co
t[e + f*x]*Sqrt[a + b*Csc[e + f*x]]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(2*d*n), Int[((d*Csc[e + f*x])^(n +
1)*Simp[a*b*(2*n - 1) + 2*(b^2*n + a^2*(n + 1))*Csc[e + f*x] + a*b*(2*n + 3)*Csc[e + f*x]^2, x])/Sqrt[a + b*C
sc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LeQ[n, -1] && IntegersQ[2*n]

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 4264

Int[(u_)*((c_.)*sin[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Csc[a + b*x])^m*(c*Sin[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Csc[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSecantIntegrandQ[
u, x]

Rubi steps

\begin {align*} \int \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {(a+b \sec (c+d x))^{3/2}}{\sec ^{\frac {5}{2}}(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 d}-\frac {1}{5} \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-6 a b-\left (3 a^2+5 b^2\right ) \sec (c+d x)-2 a b \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {4 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{2} a \left (3 a^2+b^2\right )+6 a^2 b \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{15 a}\\ &=\frac {4 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {\left (b \left (a^2-b^2\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}{5 a}+\frac {\left (\left (3 a^2+b^2\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}{5 a}\\ &=\frac {4 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {\left (b \left (a^2-b^2\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{5 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^2+b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{5 a \sqrt {b+a \cos (c+d x)}}\\ &=\frac {4 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {\left (b \left (a^2-b^2\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}{5 a \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^2+b^2\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}{5 a \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=\frac {2 b \left (a^2-b^2\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 )}{5 a d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^2+b^2\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)}}{5 a d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {4 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 a \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 9.26, size = 344, normalized size = 1.43 \[ \frac {\cos ^{\frac {3}{2}}(c+d x) (a+b \sec (c+d x))^{3/2} \left (2 \sin (c+d x) (a \cos (c+d x)+b) (a \cos (c+d x)+2 b)-\frac {2 \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-\left (3 a^2+b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)+i a \left (3 a^2+4 a b+b^2\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )-i \left (3 a^3+3 a^2 b+a b^2+b^3\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{a \sec ^{\frac {3}{2}}(c+d x)}\right )}{5 d (a \cos (c+d x)+b)^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(Cos[c + d*x]^(3/2)*(a + b*Sec[c + d*x])^(3/2)*(2*(b + a*Cos[c + d*x])*(2*b + a*Cos[c + d*x])*Sin[c + d*x] - (
2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*((-I)*(3*a^3 + 3*a^2*b + a*b^2 + b^3)*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*(3
*a^2 + 4*a*b + b^2)*EllipticF[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sec[(c + d*x)/2]^2*Sqrt[((b + a*C
os[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - (3*a^2 + b^2)*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[
(c + d*x)/2]))/(a*Sec[c + d*x]^(3/2))))/(5*d*(b + a*Cos[c + d*x])^2)

________________________________________________________________________________________

fricas [F]  time = 2.35, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (d x + c\right )^{2} \sec \left (d x + c\right ) + a \cos \left (d x + c\right )^{2}\right )} \sqrt {b \sec \left (d x + c\right ) + a} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c)^2*sec(d*x + c) + a*cos(d*x + c)^2)*sqrt(b*sec(d*x + c) + a)*sqrt(cos(d*x + c)), x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c) + a)^(3/2)*cos(d*x + c)^(5/2), x)

________________________________________________________________________________________

maple [B]  time = 1.49, size = 1695, normalized size = 7.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(3/2),x)

[Out]

2/5/d*cos(d*x+c)^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(3*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d
*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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^3-4*cos(d*x+c)*sin(d*x+c)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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^2*b+cos(d*x+c)*sin(d*x+c)*
((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(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^2-3*cos(d*x+c)*sin(d*x+c)*EllipticE((-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)*(1/(1+cos(d*x+c)))^(1/2)*
a^3+3*cos(d*x+c)*sin(d*x+c)*EllipticE((-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)*(1/(1+cos(d*x+c)))^(1/2)*a^2*b-cos(d*x+c)*sin(d*x+c)*EllipticE((-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)
*(1/(1+cos(d*x+c)))^(1/2)*a*b^2+cos(d*x+c)*sin(d*x+c)*EllipticE((-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)*(1/(1+cos(d*x+c)))^(1/2)*b^3-cos(d*x+c)^4
*((a-b)/(a+b))^(1/2)*a^3+3*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^3*
((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-4*EllipticF((-1+cos(d*x+c))*
((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1
+cos(d*x+c)))^(1/2)*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))*
a*b^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-3*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+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(
1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+3*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*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-EllipticE((-1+co
s(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b^2*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(
1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b
))^(1/2))*b^3*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)-3*cos(d*x+c)^3
*((a-b)/(a+b))^(1/2)*a^2*b-2*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3-3*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^2+3*c
os(d*x+c)*((a-b)/(a+b))^(1/2)*a^3+cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2-cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^3+3*a^
2*b*((a-b)/(a+b))^(1/2)+2*a*b^2*((a-b)/(a+b))^(1/2)+b^3*((a-b)/(a+b))^(1/2))/(b+a*cos(d*x+c))/sin(d*x+c)/a/((a
-b)/(a+b))^(1/2)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{\frac {5}{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(5/2)*(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sec(d*x + c) + a)^(3/2)*cos(d*x + c)^(5/2), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^{5/2}\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^(5/2)*(a + b/cos(c + d*x))^(3/2), x)

________________________________________________________________________________________

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))**(3/2),x)

[Out]

Timed out

________________________________________________________________________________________