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

Optimal. Leaf size=399 \[ \frac{2 (a-b) \sqrt{a+b} \left (165 a^2 b+10 a^3-114 a b^2+147 b^3\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{315 b^2 d}-\frac{2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{315 b d}-\frac{4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{315 b d}+\frac{2 (a-b) \sqrt{a+b} \left (-279 a^2 b^2+10 a^4-147 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{315 b^3 d}+\frac{2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}-\frac{4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{63 b d} \]

[Out]

(2*(a - b)*Sqrt[a + b]*(10*a^4 - 279*a^2*b^2 - 147*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/
(315*b^3*d) + (2*(a - b)*Sqrt[a + b]*(10*a^3 + 165*a^2*b - 114*a*b^2 + 147*b^3)*Cot[c + d*x]*EllipticF[ArcSin[
Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Se
c[c + d*x]))/(a - b))])/(315*b^2*d) - (4*a*(5*a^2 - 57*b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(315*b*d) -
(2*(10*a^2 - 49*b^2)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(315*b*d) - (4*a*(a + b*Sec[c + d*x])^(5/2)*Tan
[c + d*x])/(63*b*d) + (2*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d)

________________________________________________________________________________________

Rubi [A]  time = 0.779678, antiderivative size = 399, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3840, 4002, 4005, 3832, 4004} \[ -\frac{2 \left (10 a^2-49 b^2\right ) \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{315 b d}-\frac{4 a \left (5 a^2-57 b^2\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{315 b d}+\frac{2 (a-b) \sqrt{a+b} \left (165 a^2 b+10 a^3-114 a b^2+147 b^3\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{315 b^2 d}+\frac{2 (a-b) \sqrt{a+b} \left (-279 a^2 b^2+10 a^4-147 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{315 b^3 d}+\frac{2 \tan (c+d x) (a+b \sec (c+d x))^{7/2}}{9 b d}-\frac{4 a \tan (c+d x) (a+b \sec (c+d x))^{5/2}}{63 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(a - b)*Sqrt[a + b]*(10*a^4 - 279*a^2*b^2 - 147*b^4)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]
/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/
(315*b^3*d) + (2*(a - b)*Sqrt[a + b]*(10*a^3 + 165*a^2*b - 114*a*b^2 + 147*b^3)*Cot[c + d*x]*EllipticF[ArcSin[
Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Se
c[c + d*x]))/(a - b))])/(315*b^2*d) - (4*a*(5*a^2 - 57*b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(315*b*d) -
(2*(10*a^2 - 49*b^2)*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(315*b*d) - (4*a*(a + b*Sec[c + d*x])^(5/2)*Tan
[c + d*x])/(63*b*d) + (2*(a + b*Sec[c + d*x])^(7/2)*Tan[c + d*x])/(9*b*d)

Rule 3840

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(Cot[e + f*x]*(a
+ b*Csc[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*(m + 2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rule 4002

Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[(B*Cot[e + f*x]*(a + b*Csc[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]

Rule 4005

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, Int[(Csc[e + f*x]*(1 +
Csc[e + f*x]))/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]

Rule 3832

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Csc[e + f*x]))/(a - b))]*EllipticF[ArcSin[Sqrt[a + b*Csc[e +
f*x]]/Rt[a + b, 2]], (a + b)/(a - b)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 4004

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Simp[(-2*(A*b - a*B)*Rt[a + (b*B)/A, 2]*Sqrt[(b*(1 - Csc[e + f*x]))/(a + b)]*Sqrt[-((b*(1 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+b \sec (c+d x))^{5/2} \, dx &=\frac{2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac{2 \int \sec (c+d x) \left (\frac{7 b}{2}-a \sec (c+d x)\right ) (a+b \sec (c+d x))^{5/2} \, dx}{9 b}\\ &=-\frac{4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac{2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac{4 \int \sec (c+d x) (a+b \sec (c+d x))^{3/2} \left (\frac{39 a b}{4}-\frac{1}{4} \left (10 a^2-49 b^2\right ) \sec (c+d x)\right ) \, dx}{63 b}\\ &=-\frac{2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac{4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac{2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac{8 \int \sec (c+d x) \sqrt{a+b \sec (c+d x)} \left (\frac{3}{8} b \left (55 a^2+49 b^2\right )-\frac{3}{4} a \left (5 a^2-57 b^2\right ) \sec (c+d x)\right ) \, dx}{315 b}\\ &=-\frac{4 a \left (5 a^2-57 b^2\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{315 b d}-\frac{2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac{4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac{2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac{16 \int \frac{\sec (c+d x) \left (\frac{3}{16} a b \left (155 a^2+261 b^2\right )-\frac{3}{16} \left (10 a^4-279 a^2 b^2-147 b^4\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{945 b}\\ &=-\frac{4 a \left (5 a^2-57 b^2\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{315 b d}-\frac{2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac{4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac{2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}+\frac{\left ((a-b) \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right )\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{315 b}-\frac{\left (10 a^4-279 a^2 b^2-147 b^4\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{315 b}\\ &=\frac{2 (a-b) \sqrt{a+b} \left (10 a^4-279 a^2 b^2-147 b^4\right ) \cot (c+d x) E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{315 b^3 d}+\frac{2 (a-b) \sqrt{a+b} \left (10 a^3+165 a^2 b-114 a b^2+147 b^3\right ) \cot (c+d x) F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right ) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (1+\sec (c+d x))}{a-b}}}{315 b^2 d}-\frac{4 a \left (5 a^2-57 b^2\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{315 b d}-\frac{2 \left (10 a^2-49 b^2\right ) (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{315 b d}-\frac{4 a (a+b \sec (c+d x))^{5/2} \tan (c+d x)}{63 b d}+\frac{2 (a+b \sec (c+d x))^{7/2} \tan (c+d x)}{9 b d}\\ \end{align*}

Mathematica [A]  time = 16.1654, size = 552, normalized size = 1.38 \[ \frac{2 \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} (a+b \sec (c+d x))^{5/2} \left (2 b \left (279 a^2 b^2+155 a^3 b-10 a^4+261 a b^3+147 b^4\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+\left (-279 a^2 b^2+10 a^4-147 b^4\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)+2 \left (-279 a^3 b^2-279 a^2 b^3+10 a^4 b+10 a^5-147 a b^4-147 b^5\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{315 b^2 d \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} \sec ^{\frac{5}{2}}(c+d x) (a \cos (c+d x)+b)^3}+\frac{\cos ^2(c+d x) (a+b \sec (c+d x))^{5/2} \left (\frac{2 \left (279 a^2 b^2-10 a^4+147 b^4\right ) \sin (c+d x)}{315 b^2}+\frac{2}{315} \sec ^2(c+d x) \left (75 a^2 \sin (c+d x)+49 b^2 \sin (c+d x)\right )+\frac{2 \sec (c+d x) \left (5 a^3 \sin (c+d x)+163 a b^2 \sin (c+d x)\right )}{315 b}+\frac{38}{63} a b \tan (c+d x) \sec ^2(c+d x)+\frac{2}{9} b^2 \tan (c+d x) \sec ^3(c+d x)\right )}{d (a \cos (c+d x)+b)^2} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(a + b*Sec[c + d*x])^(5/2)*(2*(10*a^5 + 10*a^4*b - 279*a^3*b^2 - 279*
a^2*b^3 - 147*a*b^4 - 147*b^5)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + C
os[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(-10*a^4 + 155*a^3*b + 279*a^2*b^2 +
261*a*b^3 + 147*b^4)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c + d*
x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (10*a^4 - 279*a^2*b^2 - 147*b^4)*Cos[c + d*x]*(b
+ a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(315*b^2*d*(b + a*Cos[c + d*x])^3*Sqrt[Sec[(c + d*x)/2
]^2]*Sec[c + d*x]^(5/2)) + (Cos[c + d*x]^2*(a + b*Sec[c + d*x])^(5/2)*((2*(-10*a^4 + 279*a^2*b^2 + 147*b^4)*Si
n[c + d*x])/(315*b^2) + (2*Sec[c + d*x]^2*(75*a^2*Sin[c + d*x] + 49*b^2*Sin[c + d*x]))/315 + (2*Sec[c + d*x]*(
5*a^3*Sin[c + d*x] + 163*a*b^2*Sin[c + d*x]))/(315*b) + (38*a*b*Sec[c + d*x]^2*Tan[c + d*x])/63 + (2*b^2*Sec[c
+ d*x]^3*Tan[c + d*x])/9))/(d*(b + a*Cos[c + d*x])^2)

________________________________________________________________________________________

Maple [B]  time = 1.017, size = 2523, normalized size = 6.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/315/d/b^2*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(-279*cos(d*x+c)^6*a^3*b^2-
147*cos(d*x+c)^6*a*b^4+10*cos(d*x+c)^5*a^4*b+199*cos(d*x+c)^5*a^3*b^2-279*cos(d*x+c)^5*a^2*b^3-65*cos(d*x+c)^5
*a*b^4-5*cos(d*x+c)^4*a^4*b+272*cos(d*x+c)^4*a^2*b^3+80*cos(d*x+c)^3*a^3*b^2+82*cos(d*x+c)^3*a*b^4+170*cos(d*x
+c)^2*a^2*b^3+130*cos(d*x+c)*a*b^4-5*cos(d*x+c)^6*a^4*b-10*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))
^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))*a^5-147*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))
^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^5-163*cos(d*x+c)^6*a^2*b^3+35*b^5+10*cos(d*
x+c)^6*a^5-10*cos(d*x+c)^5*a^5-147*cos(d*x+c)^5*b^5+98*cos(d*x+c)^4*b^5+14*cos(d*x+c)^2*b^5+147*sin(d*x+c)*cos
(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(
d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^5-147*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(
a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^5-10*s
in(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Ellip
ticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^5+147*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(cos(d*x+c)+1
))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1
/2))*b^5+10*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1)
)^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b-155*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),(
(a-b)/(a+b))^(1/2))*a^3*b^2-279*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*
x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^3-261*sin(d*x+c)*c
os(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+co
s(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^4-10*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(
1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b
+279*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)
*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b^2+279*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(co
s(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)
/(a+b))^(1/2))*a^2*b^3+147*sin(d*x+c)*cos(d*x+c)^5*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))
/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^4+10*sin(d*x+c)*cos(d*x+c
)^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c)
)/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b-155*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)
*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b^2-279*
sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*Elli
pticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^3-261*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(cos(d*x
+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b
))^(1/2))*a*b^4-10*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*
x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^4*b+279*sin(d*x+c)*cos(d*x+c)^4*(co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d
*x+c),((a-b)/(a+b))^(1/2))*a^3*b^2+279*sin(d*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a
*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^3+147*sin(d
*x+c)*cos(d*x+c)^4*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE
((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^4)/(b+a*cos(d*x+c))/cos(d*x+c)^4/sin(d*x+c)^5

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

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b^{2} \sec \left (d x + c\right )^{5} + 2 \, a b \sec \left (d x + c\right )^{4} + a^{2} \sec \left (d x + c\right )^{3}\right )} \sqrt{b \sec \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b^2*sec(d*x + c)^5 + 2*a*b*sec(d*x + c)^4 + a^2*sec(d*x + c)^3)*sqrt(b*sec(d*x + c) + a), 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(sec(d*x+c)**3*(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{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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