3.855 \(\int \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx\)
Optimal. Leaf size=249 \[ -\frac {8 b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{15 a^2 d}-\frac {2 b \left (7 a^2+8 b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (9 a^2+8 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d} \]
[Out]
-2/15*b*(7*a^2+8*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^3/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/5*cos(d*x+c)^(3/2
)*sin(d*x+c)*(a+b*sec(d*x+c))^(1/2)/a/d-8/15*b*sin(d*x+c)*cos(d*x+c)^(1/2)*(a+b*sec(d*x+c))^(1/2)/a^2/d+2/15*(
9*a^2+8*b^2)*(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 = 0.65, antiderivative size = 249, 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, 3863, 4104, 4035, 3856, 2655, 2653, 3858, 2663, 2661}
\[ -\frac {2 b \left (7 a^2+8 b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (9 a^2+8 b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{15 a^3 d \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}-\frac {8 b \sin (c+d x) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}{15 a^2 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}}{5 a d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^(5/2)/Sqrt[a + b*Sec[c + d*x]],x]
[Out]
(-2*b*(7*a^2 + 8*b^2)*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*(9*a^2 + 8*b^2)*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)]) - (8*b*Sqrt[Cos[c + d*x]]*Sqr
t[a + b*Sec[c + d*x]]*Sin[c + d*x])/(15*a^2*d) + (2*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 3863
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(Cos[e
+ f*x]*(d*Csc[e + f*x])^(n + 1)*Sqrt[a + b*Csc[e + f*x]])/(a*d*f*n), x] + Dist[1/(2*a*d*n), Int[((d*Csc[e + f*
x])^(n + 1)*Simp[-(b*(2*n + 1)) + 2*a*(n + 1)*Csc[e + f*x] + b*(2*n + 3)*Csc[e + f*x]^2, x])/Sqrt[a + b*Csc[e
+ f*x]], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[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 \frac {\cos ^{\frac {5}{2}}(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx\\ &=\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {4 b-3 a \sec (c+d x)-2 b \sec ^2(c+d x)}{\sec ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)}} \, dx}{5 a}\\ &=-\frac {8 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 \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} \left (9 a^2+8 b^2\right )+a b \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{15 a^2}\\ &=-\frac {8 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}+\frac {\left (\left (9 a^2+8 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}{15 a^3}-\frac {\left (2 \left (-a^2 b+\frac {1}{2} b \left (9 a^2+8 b^2\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 {8 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {\left (2 \left (-a^2 b+\frac {1}{2} b \left (9 a^2+8 b^2\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 (9 a^2+8 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}{15 a^3 \sqrt {b+a \cos (c+d x)}}\\ &=-\frac {8 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 \cos ^{\frac {3}{2}}(c+d x) \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{5 a d}-\frac {\left (2 \left (-a^2 b+\frac {1}{2} b \left (9 a^2+8 b^2\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 (9 a^2+8 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}{15 a^3 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}\\ &=-\frac {2 b \left (7 a^2+8 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 )}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (9 a^2+8 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)}}{15 a^3 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}-\frac {8 b \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} \sin (c+d x)}{15 a^2 d}+\frac {2 \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 = 9.68, size = 340, normalized size = 1.37 \[ \frac {2 a \sin (c+d x) (a \cos (c+d x)+b) (3 a \cos (c+d x)-4 b)+\frac {2 \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (\left (9 a^2+8 b^2\right ) \tan \left (\frac {1}{2} (c+d x)\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} (a \cos (c+d x)+b)-i a \left (9 a^2+2 a b+8 b^2\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} F\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )+i \left (9 a^3+9 a^2 b+8 a b^2+8 b^3\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\sec ^2\left (\frac {1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {b-a}{a+b}\right )\right )}{\sec ^{\frac {3}{2}}(c+d x)}}{15 a^3 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cos[c + d*x]^(5/2)/Sqrt[a + b*Sec[c + d*x]],x]
[Out]
(2*a*(b + a*Cos[c + d*x])*(-4*b + 3*a*Cos[c + d*x])*Sin[c + d*x] + (2*(Cos[(c + d*x)/2]^2*Sec[c + d*x])^(3/2)*
(I*(9*a^3 + 9*a^2*b + 8*a*b^2 + 8*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*(9*a^2 + 2*a*b + 8*b^2)*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
)] + (9*a^2 + 8*b^2)*(b + a*Cos[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/Sec[c + d*x]^(3/2))/(1
5*a^3*d*Sqrt[Cos[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 \[ {\rm integral}\left (\frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral(cos(d*x + c)^(5/2)/sqrt(b*sec(d*x + c) + a), x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(cos(d*x + c)^(5/2)/sqrt(b*sec(d*x + c) + a), x)
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maple [B] time = 1.29, size = 1726, normalized size = 6.93 \[ \text {result too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x)
[Out]
-2/15/d*(-9*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)+8*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*((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)+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*((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)-8*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)+9*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^3-8*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+co
s(d*x+c))/(a+b))^(1/2)*(1/(1+cos(d*x+c)))^(1/2)*b^3-9*a^2*b*((a-b)/(a+b))^(1/2)+4*a*b^2*((a-b)/(a+b))^(1/2)-9*
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+6*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a^3-9*cos
(d*x+c)*((a-b)/(a+b))^(1/2)*a^3+8*cos(d*x+c)*((a-b)/(a+b))^(1/2)*b^3+3*cos(d*x+c)^4*((a-b)/(a+b))^(1/2)*a^3-co
s(d*x+c)^3*((a-b)/(a+b))^(1/2)*a^2*b+4*cos(d*x+c)^2*((a-b)/(a+b))^(1/2)*a*b^2+10*cos(d*x+c)*((a-b)/(a+b))^(1/2
)*a^2*b-8*cos(d*x+c)*((a-b)/(a+b))^(1/2)*a*b^2-9*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+8*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+2*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)/si
n(d*x+c),(-(a+b)/(a-b))^(1/2))*a^2*b-8*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^2-8*b
^3*((a-b)/(a+b))^(1/2)+9*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)-8*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)-9*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))*((b+a*cos(d*x+c))/cos(d*x
+c))^(1/2)*cos(d*x+c)^(1/2)/(b+a*cos(d*x+c))/sin(d*x+c)/a^3/((a-b)/(a+b))^(1/2)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cos \left (d x + c\right )^{\frac {5}{2}}}{\sqrt {b \sec \left (d x + c\right ) + a}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(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 \[ \int \frac {{\cos \left (c+d\,x\right )}^{5/2}}{\sqrt {a+\frac {b}{\cos \left (c+d\,x\right )}}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(c + d*x)^(5/2)/(a + b/cos(c + d*x))^(1/2),x)
[Out]
int(cos(c + d*x)^(5/2)/(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))**(1/2),x)
[Out]
Timed out
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