3.873 \(\int \frac{1}{\cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=277 \[ \frac{2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}} \text{EllipticF}\left (\frac{1}{2} (c+d x),\frac{2 a}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}-\frac{2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (a^2-5 b^2\right ) \sin (c+d x)}{3 b d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{8 b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

[Out]

(2*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x
]]*Sqrt[a + b*Sec[c + d*x]]) + (8*b*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c
+ d*x]])/(3*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) - (2*a^2*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*Sqrt
[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)) + (2*a*(a^2 - 5*b^2)*Sin[c + d*x])/(3*b*(a^2 - b^2)^2*d*Sqrt[Cos[c
+ d*x]]*Sqrt[a + b*Sec[c + d*x]])

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Rubi [A]  time = 0.744574, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4264, 3845, 4100, 4035, 3856, 2655, 2653, 3858, 2663, 2661} \[ -\frac{2 a^2 \sin (c+d x)}{3 b d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (a^2-5 b^2\right ) \sin (c+d x)}{3 b d \left (a^2-b^2\right )^2 \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{8 b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

(2*Sqrt[(b + a*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*a)/(a + b)])/(3*(a^2 - b^2)*d*Sqrt[Cos[c + d*x
]]*Sqrt[a + b*Sec[c + d*x]]) + (8*b*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*a)/(a + b)]*Sqrt[a + b*Sec[c
+ d*x]])/(3*(a^2 - b^2)^2*d*Sqrt[(b + a*Cos[c + d*x])/(a + b)]) - (2*a^2*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*Sqrt
[Cos[c + d*x]]*(a + b*Sec[c + d*x])^(3/2)) + (2*a*(a^2 - 5*b^2)*Sin[c + d*x])/(3*b*(a^2 - b^2)^2*d*Sqrt[Cos[c
+ d*x]]*Sqrt[a + b*Sec[c + d*x]])

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]

Rule 3845

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a^2*
d^3*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[d
^3/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 3)*Simp[a^2*(n - 3) + a*b*(
m + 1)*Csc[e + f*x] - (a^2*(n - 2) + b^2*(m + 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && N
eQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

Rule 4100

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 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{1}{\cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \, dx &=\left (\sqrt{\cos (c+d x)} \sqrt{\sec (c+d x)}\right ) \int \frac{\sec ^{\frac{5}{2}}(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\\ &=-\frac{2 a^2 \sin (c+d x)}{3 b \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{a^2}{2}-\frac{3}{2} a b \sec (c+d x)-\frac{1}{2} \left (a^2-3 b^2\right ) \sec ^2(c+d x)}{\sqrt{\sec (c+d x)} (a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (a^2-5 b^2\right ) \sin (c+d x)}{3 b \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{a^2 b^2+\frac{1}{4} a b \left (a^2+3 b^2\right ) \sec (c+d x)}{\sqrt{\sec (c+d x)} \sqrt{a+b \sec (c+d x)}} \, dx}{3 a b \left (a^2-b^2\right )^2}\\ &=-\frac{2 a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (a^2-5 b^2\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (4 b \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 \left (a^2-b^2\right )^2}+\frac{\left (\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 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (a^2-5 b^2\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\sqrt{b+a \cos (c+d x)} \int \frac{1}{\sqrt{b+a \cos (c+d x)}} \, dx}{3 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (4 b \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}\right ) \int \sqrt{b+a \cos (c+d x)} \, dx}{3 \left (a^2-b^2\right )^2 \sqrt{b+a \cos (c+d x)}}\\ &=-\frac{2 a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (a^2-5 b^2\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\sqrt{\frac{b+a \cos (c+d x)}{a+b}} \int \frac{1}{\sqrt{\frac{b}{a+b}+\frac{a \cos (c+d x)}{a+b}}} \, dx}{3 \left (a^2-b^2\right ) \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{\left (4 b \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 \left (a^2-b^2\right )^2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}\\ &=\frac{2 \sqrt{\frac{b+a \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 a}{a+b}\right )}{3 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}+\frac{8 b \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 \left (a^2-b^2\right )^2 d \sqrt{\frac{b+a \cos (c+d x)}{a+b}}}-\frac{2 a^2 \sin (c+d x)}{3 b \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)} (a+b \sec (c+d x))^{3/2}}+\frac{2 a \left (a^2-5 b^2\right ) \sin (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt{\cos (c+d x)} \sqrt{a+b \sec (c+d x)}}\\ \end{align*}

Mathematica [C]  time = 9.08617, size = 311, normalized size = 1.12 \[ \frac{2 (a \cos (c+d x)+b)^2 \left (\frac{a \sin (c+d x) \left (a^2-4 a b \cos (c+d x)-5 b^2\right )}{a \cos (c+d x)+b}+\frac{\sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} \left (-i \left (a^2+4 a b+3 b^2\right ) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} \text{EllipticF}\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{b-a}{a+b}\right )+4 b \tan \left (\frac{1}{2} (c+d x)\right ) \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} (a \cos (c+d x)+b)+4 i b (a+b) \sqrt{\frac{\sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)}{a+b}} E\left (i \sinh ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{b-a}{a+b}\right )\right )}{\sqrt{\sec (c+d x)}}\right )}{3 d \left (a^2-b^2\right )^2 \cos ^{\frac{5}{2}}(c+d x) (a+b \sec (c+d x))^{5/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/(Cos[c + d*x]^(5/2)*(a + b*Sec[c + d*x])^(5/2)),x]

[Out]

(2*(b + a*Cos[c + d*x])^2*((a*(a^2 - 5*b^2 - 4*a*b*Cos[c + d*x])*Sin[c + d*x])/(b + a*Cos[c + d*x]) + (Sqrt[Co
s[(c + d*x)/2]^2*Sec[c + d*x]]*((4*I)*b*(a + b)*EllipticE[I*ArcSinh[Tan[(c + d*x)/2]], (-a + b)/(a + b)]*Sqrt[
((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] - I*(a^2 + 4*a*b + 3*b^2)*EllipticF[I*ArcSinh[Tan[(c + d*x)
/2]], (-a + b)/(a + b)]*Sqrt[((b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2)/(a + b)] + 4*b*(b + a*Cos[c + d*x])*Sqr
t[Sec[(c + d*x)/2]^2]*Tan[(c + d*x)/2]))/Sqrt[Sec[c + d*x]]))/(3*(a^2 - b^2)^2*d*Cos[c + d*x]^(5/2)*(a + b*Sec
[c + d*x])^(5/2))

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Maple [B]  time = 0.303, size = 1333, normalized size = 4.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3/d/(a-b)/(a+b)^2/((a-b)/(a+b))^(1/2)*(cos(d*x+c)^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))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^
2-3*cos(d*x+c)^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))*(1/
(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a*b+4*cos(d*x+c)^2*sin(d*x+c)*(1/(a+b)*(
b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/s
in(d*x+c),(-(a+b)/(a-b))^(1/2))*a*b+cos(d*x+c)*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))*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a^2-2*cos(
d*x+c)*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))*(1/(a+b)*(b+a
*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*a*b-3*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+
c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/sin(d*x+c),(-
(a+b)/(a-b))^(1/2))*b^2+4*cos(d*x+c)*sin(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(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+4*cos(d*x+c)*sin
(d*x+c)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(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))*b^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*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)
-3*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))*((a-b)/(
a+b))^(1/2)/sin(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2*sin(d*x+c)+4*EllipticE((-1+cos(d*x+c))*((a-b)/(a+b))^(1/2)/si
n(d*x+c),(-(a+b)/(a-b))^(1/2))*b^2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)*si
n(d*x+c)+((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a^2-3*((a-b)/(a+b))^(1/2)*cos(d*x+c)^2*a*b+4*((a-b)/(a+b))^(1/2)*cos
(d*x+c)*a*b-4*((a-b)/(a+b))^(1/2)*cos(d*x+c)*b^2-a^2*((a-b)/(a+b))^(1/2)-a*b*((a-b)/(a+b))^(1/2)+4*b^2*((a-b)/
(a+b))^(1/2))*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)/(b+a*cos(d*x+c))^2/sin(d*x+c)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(1/((b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2)), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (d x + c\right ) + a} \sqrt{\cos \left (d x + c\right )}}{b^{3} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right )^{3} \sec \left (d x + c\right ) + a^{3} \cos \left (d x + c\right )^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c) + a)*sqrt(cos(d*x + c))/(b^3*cos(d*x + c)^3*sec(d*x + c)^3 + 3*a*b^2*cos(d*x + c)
^3*sec(d*x + c)^2 + 3*a^2*b*cos(d*x + c)^3*sec(d*x + c) + a^3*cos(d*x + c)^3), 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(1/cos(d*x+c)**(5/2)/(a+b*sec(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cos(d*x+c)^(5/2)/(a+b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c) + a)^(5/2)*cos(d*x + c)^(5/2)), x)