\(\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{5/2}} \, dx\) [870]
Optimal result
Integrand size = 25, antiderivative size = 317 \[
\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 b \left (9 a^2-8 b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{3 a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^4-15 a^2 b^2+8 b^4\right ) \sqrt {\cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right ) \sqrt {a+b \sec (c+d x)}}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}
\]
[Out]
2/3*b^2*sin(d*x+c)/a/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)/cos(d*x+c)^(1/2)+8/3*b^2*(2*a^2-b^2)*sin(d*x+c)/a^2/(a
^2-b^2)^2/d/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)-2/3*b*(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)*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/(a^2-b^2)/d
/cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(1/2)+2/3*(3*a^4-15*a^2*b^2+8*b^4)*(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/(a^2
-b^2)^2/d/((b+a*cos(d*x+c))/(a+b))^(1/2)
Rubi [A] (verified)
Time = 1.03 (sec) , antiderivative size = 317, normalized size of antiderivative = 1.00, number of
steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {4349, 3932, 4185, 4120,
3941, 2734, 2732, 3943, 2742, 2740} \[
\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{3 a^2 d \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 b^2 \sin (c+d x)}{3 a d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac {2 b \left (9 a^2-8 b^2\right ) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^4-15 a^2 b^2+8 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{3 a^3 d \left (a^2-b^2\right )^2 \sqrt {\frac {a \cos (c+d x)+b}{a+b}}}
\]
[In]
Int[Sqrt[Cos[c + d*x]]/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(-2*b*(9*a^2 - 8*b^2)*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*a^3*(a^2 -
b^2)*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]) + (2*(3*a^4 - 15*a^2*b^2 + 8*b^4)*Sqrt[Cos[c + d*x]]*Ellip
ticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c + d*x]])/(3*a^3*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a
+ b)]) + (2*b^2*Sin[c + d*x])/(3*a*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)) + (8*b^2*(2*a
^2 - b^2)*Sin[c + d*x])/(3*a^2*(a^2 - b^2)^2*d*Sqrt[Cos[c + d*x]]*Sqrt[a + b*Sec[c + d*x]])
Rule 2732
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2734
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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] && !GtQ[a + b, 0]
Rule 2740
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
Rule 2742
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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && !GtQ[a + b, 0]
Rule 3932
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b^2*Co
t[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)
*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*(a^2*(m + 1) - b^2*(m + n + 1) - a*b*(m + 1
)*Csc[e + f*x] + b^2*(m + n + 2)*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, n}, x] && NeQ[a^2 - b^2, 0]
&& LtQ[m, -1] && IntegersQ[2*m, 2*n]
Rule 3941
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 3943
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[Sqrt[d*C
sc[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 4120
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 4185
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*b^2 - a*b*B + a^2*C)*Cot[e + f*x]*(a +
b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^n/(a*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(a*(m + 1)*(a^2 - b^2)), I
nt[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[a*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C)*
(m + n + 1) - a*(A*b - a*B + b*C)*(m + 1)*Csc[e + f*x] + (A*b^2 - a*b*B + a^2*C)*(m + n + 2)*Csc[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, d, e, f, A, B, C, n}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && !(ILtQ[m + 1/2, 0] &
& ILtQ[n, 0])
Rule 4349
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*}
\text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{5/2}} \, dx \\ & = \frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}-\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {3 a^2}{2}+2 b^2+\frac {3}{2} a b \sec (c+d x)-b^2 \sec ^2(c+d x)}{\sqrt {\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )} \\ & = \frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {1}{4} \left (3 a^4-15 a^2 b^2+8 b^4\right )-\frac {1}{2} a b \left (3 a^2-b^2\right ) \sec (c+d x)}{\sqrt {\sec (c+d x)} \sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = \frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^4-15 a^2 b^2+8 b^4\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}{3 a^3 \left (a^2-b^2\right )^2}-\frac {\left (4 \left (\frac {1}{2} a^2 b \left (3 a^2-b^2\right )+\frac {1}{4} b \left (3 a^4-15 a^2 b^2+8 b^4\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}{3 a^3 \left (a^2-b^2\right )^2} \\ & = \frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (4 \left (\frac {1}{2} a^2 b \left (3 a^2-b^2\right )+\frac {1}{4} b \left (3 a^4-15 a^2 b^2+8 b^4\right )\right ) \sqrt {b+a \cos (c+d x)}\right ) \int \frac {1}{\sqrt {b+a \cos (c+d x)}} \, dx}{3 a^3 \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^4-15 a^2 b^2+8 b^4\right ) \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}\right ) \int \sqrt {b+a \cos (c+d x)} \, dx}{3 a^3 \left (a^2-b^2\right )^2 \sqrt {b+a \cos (c+d x)}} \\ & = \frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}-\frac {\left (4 \left (\frac {1}{2} a^2 b \left (3 a^2-b^2\right )+\frac {1}{4} b \left (3 a^4-15 a^2 b^2+8 b^4\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}{3 a^3 \left (a^2-b^2\right )^2 \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {\left (\left (3 a^4-15 a^2 b^2+8 b^4\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}{3 a^3 \left (a^2-b^2\right )^2 \sqrt {\frac {b+a \cos (c+d x)}{a+b}}} \\ & = -\frac {2 b \left (9 a^2-8 b^2\right ) \sqrt {\frac {b+a \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 a}{a+b}\right )}{3 a^3 \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (3 a^4-15 a^2 b^2+8 b^4\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)}}{3 a^3 \left (a^2-b^2\right )^2 d \sqrt {\frac {b+a \cos (c+d x)}{a+b}}}+\frac {2 b^2 \sin (c+d x)}{3 a \left (a^2-b^2\right ) d \sqrt {\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac {8 b^2 \left (2 a^2-b^2\right ) \sin (c+d x)}{3 a^2 \left (a^2-b^2\right )^2 d \sqrt {\cos (c+d x)} \sqrt {a+b \sec (c+d x)}} \\
\end{align*}
Mathematica [C] (warning: unable to verify)
Result contains complex when optimal does not.
Time = 18.25 (sec) , antiderivative size = 507, normalized size of antiderivative = 1.60
\[
\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \left (-\frac {2 b^3 \sin (c+d x)}{3 a^2 \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}-\frac {2 \left (-9 a^2 b^2 \sin (c+d x)+5 b^4 \sin (c+d x)\right )}{3 a^2 \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d \cos ^{\frac {5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}}-\frac {2 \cos ^{\frac {3}{2}}(c+d x) (b+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \left (\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )^{3/2} \left (-i \left (3 a^5+3 a^4 b-15 a^3 b^2-15 a^2 b^3+8 a b^4+8 b^5\right ) E\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}+i a \left (3 a^4-6 a^3 b-15 a^2 b^2+2 a b^3+8 b^4\right ) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {-a+b}{a+b}\right ) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {(b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}}-\left (3 a^4-15 a^2 b^2+8 b^4\right ) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right )^{3/2} \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 a \left (a^3-a b^2\right )^2 d (a+b \sec (c+d x))^{5/2}}
\]
[In]
Integrate[Sqrt[Cos[c + d*x]]/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
((b + a*Cos[c + d*x])^3*((-2*b^3*Sin[c + d*x])/(3*a^2*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) - (2*(-9*a^2*b^2*Sin
[c + d*x] + 5*b^4*Sin[c + d*x]))/(3*a^2*(a^2 - b^2)^2*(b + a*Cos[c + d*x]))))/(d*Cos[c + d*x]^(5/2)*(a + b*Sec
[c + d*x])^(5/2)) - (2*Cos[c + d*x]^(3/2)*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)*(Cos[(c + d*x)/2]^2*Sec[c
+ d*x])^(3/2)*((-I)*(3*a^5 + 3*a^4*b - 15*a^3*b^2 - 15*a^2*b^3 + 8*a*b^4 + 8*b^5)*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^4 - 6*a^3*b - 15*a^2*b^2 + 2*a*b^3 + 8*b^4)*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)] - (3*a^4 - 15*a^2*b^2 + 8*b^4)*(b + a*C
os[c + d*x])*(Sec[(c + d*x)/2]^2)^(3/2)*Tan[(c + d*x)/2]))/(3*a*(a^3 - a*b^2)^2*d*(a + b*Sec[c + d*x])^(5/2))
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(4119\) vs. \(2(347)=694\).
Time = 9.19 (sec) , antiderivative size = 4120, normalized size of antiderivative =
13.00
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\(4120\) |
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[In]
int(cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
[Out]
2/3/d/((a-b)/(a+b))^(1/2)/(a+b)^2/(a-b)/a^3*(-((1-cos(d*x+c))^2*csc(d*x+c)^2-1)/((1-cos(d*x+c))^2*csc(d*x+c)^2
+1))^(1/2)*((a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/((1-cos(d*x+c))^2*csc(d*x+c)
^2-1))^(1/2)*(3*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d
*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^
5*(1-cos(d*x+c))^2*csc(d*x+c)^2+3*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b
))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b
)/(a-b))^(1/2))*a^4*b*(1-cos(d*x+c))^2*csc(d*x+c)^2+15*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*c
sc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+
c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*b^2*(1-cos(d*x+c))^2*csc(d*x+c)^2-15*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)
^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/
(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b^3*(1-cos(d*x+c))^2*csc(d*x+c)^2-8*(-(a*(1-co
s(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(
1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^4*(1-cos(d*x+c))^2*csc(d
*x+c)^2-15*((a-b)/(a+b))^(1/2)*a^3*b^2*(1-cos(d*x+c))^5*csc(d*x+c)^5+15*((a-b)/(a+b))^(1/2)*a^2*b^3*(1-cos(d*x
+c))^5*csc(d*x+c)^5+8*((a-b)/(a+b))^(1/2)*a*b^4*(1-cos(d*x+c))^5*csc(d*x+c)^5-6*((a-b)/(a+b))^(1/2)*a^4*b*(1-c
os(d*x+c))^3*csc(d*x+c)^3+12*((a-b)/(a+b))^(1/2)*a^3*b^2*(1-cos(d*x+c))^3*csc(d*x+c)^3+28*((a-b)/(a+b))^(1/2)*
a^2*b^3*(1-cos(d*x+c))^3*csc(d*x+c)^3-8*((a-b)/(a+b))^(1/2)*a*b^4*(1-cos(d*x+c))^3*csc(d*x+c)^3+3*((a-b)/(a+b)
)^(1/2)*a^5*(-cot(d*x+c)+csc(d*x+c))-8*((a-b)/(a+b))^(1/2)*b^5*(-cot(d*x+c)+csc(d*x+c))+((a-b)/(a+b))^(1/2)*a^
2*b^3*(-cot(d*x+c)+csc(d*x+c))-16*((a-b)/(a+b))^(1/2)*a*b^4*(-cot(d*x+c)+csc(d*x+c))+9*((a-b)/(a+b))^(1/2)*a^4
*b*(-cot(d*x+c)+csc(d*x+c))+6*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(
1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a
-b))^(1/2))*a^4*b*(1-cos(d*x+c))^2*csc(d*x+c)^2-15*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d
*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+c
sc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*b^2*(1-cos(d*x+c))^2*csc(d*x+c)^2-2*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*
(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticF(((a-b)/(a+b)
)^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b^3*(1-cos(d*x+c))^2*csc(d*x+c)^2+8*(-(a*(1-cos(d*x
+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*
EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^4*(1-cos(d*x+c))^2*csc(d*x+c)
^2-12*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*c
sc(d*x+c)^2+1)^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^4*b-3*(-(a
*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^
2+1)^(1/2)*EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*b^2+14*(-(a*(1-cos
(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1
/2)*EllipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b^3+8*(-(a*(1-cos(d*x+c))
^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*Elli
pticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^4+3*(-(a*(1-cos(d*x+c))^2*csc(d*x
+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-
b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^4*b-15*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(
1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/(a+b))
^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^3*b^2-15*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d
*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*
(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^2*b^3+8*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2
*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*
x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a*b^4-3*((a-b)/(a+b))^(1/2)*a^4*b*(1-cos(d*x+c))^5*csc(d*x+c)^5-3*(-(a*
(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2
+1)^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^5*(1-cos(d*x+c))^2*cs
c(d*x+c)^2+8*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+
c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b^5*(
1-cos(d*x+c))^2*csc(d*x+c)^2+15*((a-b)/(a+b))^(1/2)*a^3*b^2*(-cot(d*x+c)+csc(d*x+c))+3*((a-b)/(a+b))^(1/2)*a^5
*(1-cos(d*x+c))^5*csc(d*x+c)^5-8*((a-b)/(a+b))^(1/2)*b^5*(1-cos(d*x+c))^5*csc(d*x+c)^5-6*((a-b)/(a+b))^(1/2)*a
^5*(1-cos(d*x+c))^3*csc(d*x+c)^3-16*((a-b)/(a+b))^(1/2)*b^5*(1-cos(d*x+c))^3*csc(d*x+c)^3-3*(-(a*(1-cos(d*x+c)
)^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*Ell
ipticF(((a-b)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^5+3*(-(a*(1-cos(d*x+c))^2*csc(d*x+
c)^2-b*(1-cos(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b
)/(a+b))^(1/2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*a^5+8*(-(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-co
s(d*x+c))^2*csc(d*x+c)^2-a-b)/(a+b))^(1/2)*((1-cos(d*x+c))^2*csc(d*x+c)^2+1)^(1/2)*EllipticE(((a-b)/(a+b))^(1/
2)*(-cot(d*x+c)+csc(d*x+c)),(-(a+b)/(a-b))^(1/2))*b^5)/(a*(1-cos(d*x+c))^2*csc(d*x+c)^2-b*(1-cos(d*x+c))^2*csc
(d*x+c)^2-a-b)^2
Fricas [C] (verification not implemented)
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.16 (sec) , antiderivative size = 884, normalized size of antiderivative = 2.79
\[
\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Too large to display}
\]
[In]
integrate(cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
1/9*(6*(8*a^4*b^3 - 4*a^2*b^5 + (9*a^5*b^2 - 5*a^3*b^4)*cos(d*x + c))*sqrt((a*cos(d*x + c) + b)/cos(d*x + c))*
sqrt(cos(d*x + c))*sin(d*x + c) - 4*(sqrt(2)*(-6*I*a^6*b + 9*I*a^4*b^3 - 4*I*a^2*b^5)*cos(d*x + c)^2 + 2*sqrt(
2)*(-6*I*a^5*b^2 + 9*I*a^3*b^4 - 4*I*a*b^6)*cos(d*x + c) + sqrt(2)*(-6*I*a^4*b^3 + 9*I*a^2*b^5 - 4*I*b^7))*sqr
t(a)*weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*s
in(d*x + c) + 2*b)/a) - 4*(sqrt(2)*(6*I*a^6*b - 9*I*a^4*b^3 + 4*I*a^2*b^5)*cos(d*x + c)^2 + 2*sqrt(2)*(6*I*a^5
*b^2 - 9*I*a^3*b^4 + 4*I*a*b^6)*cos(d*x + c) + sqrt(2)*(6*I*a^4*b^3 - 9*I*a^2*b^5 + 4*I*b^7))*sqrt(a)*weierstr
assPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*sin(d*x + c) +
2*b)/a) - 3*(sqrt(2)*(-3*I*a^7 + 15*I*a^5*b^2 - 8*I*a^3*b^4)*cos(d*x + c)^2 + 2*sqrt(2)*(-3*I*a^6*b + 15*I*a^
4*b^3 - 8*I*a^2*b^5)*cos(d*x + c) + sqrt(2)*(-3*I*a^5*b^2 + 15*I*a^3*b^4 - 8*I*a*b^6))*sqrt(a)*weierstrassZeta
(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a
^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) + 3*I*a*sin(d*x + c) + 2*b)/a)) - 3*(sqrt(2)*(3*I*a^7 - 15*I*a^5*b^2
+ 8*I*a^3*b^4)*cos(d*x + c)^2 + 2*sqrt(2)*(3*I*a^6*b - 15*I*a^4*b^3 + 8*I*a^2*b^5)*cos(d*x + c) + sqrt(2)*(3*I
*a^5*b^2 - 15*I*a^3*b^4 + 8*I*a*b^6))*sqrt(a)*weierstrassZeta(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)
/a^3, weierstrassPInverse(-4/3*(3*a^2 - 4*b^2)/a^2, 8/27*(9*a^2*b - 8*b^3)/a^3, 1/3*(3*a*cos(d*x + c) - 3*I*a*
sin(d*x + c) + 2*b)/a)))/((a^10 - 2*a^8*b^2 + a^6*b^4)*d*cos(d*x + c)^2 + 2*(a^9*b - 2*a^7*b^3 + a^5*b^5)*d*co
s(d*x + c) + (a^8*b^2 - 2*a^6*b^4 + a^4*b^6)*d)
Sympy [F(-1)]
Timed out. \[
\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{5/2}} \, dx=\text {Timed out}
\]
[In]
integrate(cos(d*x+c)**(1/2)/(a+b*sec(d*x+c))**(5/2),x)
[Out]
Timed out
Maxima [F]
\[
\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x }
\]
[In]
integrate(cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate(sqrt(cos(d*x + c))/(b*sec(d*x + c) + a)^(5/2), x)
Giac [F]
\[
\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{5/2}} \, dx=\int { \frac {\sqrt {\cos \left (d x + c\right )}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x }
\]
[In]
integrate(cos(d*x+c)^(1/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate(sqrt(cos(d*x + c))/(b*sec(d*x + c) + a)^(5/2), x)
Mupad [F(-1)]
Timed out. \[
\int \frac {\sqrt {\cos (c+d x)}}{(a+b \sec (c+d x))^{5/2}} \, dx=\int \frac {\sqrt {\cos \left (c+d\,x\right )}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x
\]
[In]
int(cos(c + d*x)^(1/2)/(a + b/cos(c + d*x))^(5/2),x)
[Out]
int(cos(c + d*x)^(1/2)/(a + b/cos(c + d*x))^(5/2), x)