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

Optimal. Leaf size=399 \[ -\frac{2 (4 a+3 b) \left (4 a^2+b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{5 b^4 d \sqrt{a+b}}-\frac{2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt{a+b \sec (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}-\frac{2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{5 b^3 d \left (a^2-b^2\right )}-\frac{2 \left (-8 a^2 b^2+16 a^4-3 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{5 b^5 d \sqrt{a+b}} \]

[Out]

(-2*(16*a^4 - 8*a^2*b^2 - 3*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))])/(5*b^5*Sqrt[a + b]*d) -
(2*(4*a + 3*b)*(4*a^2 + b^2)*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 + Sec[c + d*x]))/(a - b))])/(5*b^4*Sqrt[a + b]*d) - (2
*a^2*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (2*a*(8*a^2 - 3*b^2)*Sqrt[a + b
*Sec[c + d*x]]*Tan[c + d*x])/(5*b^3*(a^2 - b^2)*d) + (2*(6*a^2 - b^2)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Ta
n[c + d*x])/(5*b^2*(a^2 - b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 0.774744, antiderivative size = 399, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3845, 4092, 4082, 4005, 3832, 4004} \[ -\frac{2 a^2 \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (6 a^2-b^2\right ) \tan (c+d x) \sec (c+d x) \sqrt{a+b \sec (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}-\frac{2 a \left (8 a^2-3 b^2\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{5 b^3 d \left (a^2-b^2\right )}-\frac{2 (4 a+3 b) \left (4 a^2+b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{5 b^4 d \sqrt{a+b}}-\frac{2 \left (-8 a^2 b^2+16 a^4-3 b^4\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{5 b^5 d \sqrt{a+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(16*a^4 - 8*a^2*b^2 - 3*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))])/(5*b^5*Sqrt[a + b]*d) -
(2*(4*a + 3*b)*(4*a^2 + b^2)*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 + Sec[c + d*x]))/(a - b))])/(5*b^4*Sqrt[a + b]*d) - (2
*a^2*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) - (2*a*(8*a^2 - 3*b^2)*Sqrt[a + b
*Sec[c + d*x]]*Tan[c + d*x])/(5*b^3*(a^2 - b^2)*d) + (2*(6*a^2 - b^2)*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Ta
n[c + d*x])/(5*b^2*(a^2 - b^2)*d)

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 4092

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

Rule 4082

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_
.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(C*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*Simp[b*A*(m + 2) + b*C*(m + 1) + (b*B*
(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

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 \frac{\sec ^5(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx &=-\frac{2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 \int \frac{\sec ^2(c+d x) \left (2 a^2-\frac{1}{2} a b \sec (c+d x)-\frac{1}{2} \left (6 a^2-b^2\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{4 \int \frac{\sec (c+d x) \left (-\frac{1}{2} a \left (6 a^2-b^2\right )+\frac{1}{4} b \left (2 a^2+3 b^2\right ) \sec (c+d x)+\frac{3}{4} a \left (8 a^2-3 b^2\right ) \sec ^2(c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{5 b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 a \left (8 a^2-3 b^2\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{8 \int \frac{\sec (c+d x) \left (-\frac{3}{8} a b \left (4 a^2+b^2\right )-\frac{3}{8} \left (16 a^4-8 a^2 b^2-3 b^4\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 a \left (8 a^2-3 b^2\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac{\left ((4 a+3 b) \left (4 a^2+b^2\right )\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{5 b^3 (a+b)}+\frac{\left (16 a^4-8 a^2 b^2-3 b^4\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{5 b^3 \left (a^2-b^2\right )}\\ &=-\frac{2 \left (16 a^4-8 a^2 b^2-3 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}}}{5 b^5 \sqrt{a+b} d}-\frac{2 (4 a+3 b) \left (4 a^2+b^2\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}}}{5 b^4 \sqrt{a+b} d}-\frac{2 a^2 \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt{a+b \sec (c+d x)}}-\frac{2 a \left (8 a^2-3 b^2\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac{2 \left (6 a^2-b^2\right ) \sec (c+d x) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}\\ \end{align*}

Mathematica [A]  time = 15.0895, size = 498, normalized size = 1.25 \[ \frac{2 \left (4 a^2+b^2\right ) \sec ^{\frac{3}{2}}(c+d x) \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} (a \cos (c+d x)+b) \left (2 b \left (-4 a^2-a b+3 b^2\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+\left (4 a^2-3 b^2\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)+2 \left (4 a^2 b+4 a^3-3 a b^2-3 b^3\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{5 b^4 d \left (b^2-a^2\right ) \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{3/2}}+\frac{\sec ^2(c+d x) (a \cos (c+d x)+b)^2 \left (\frac{2 \left (8 a^2 b^2-16 a^4+3 b^4\right ) \sin (c+d x)}{5 b^4 \left (b^2-a^2\right )}+\frac{2 a^4 \sin (c+d x)}{b^3 \left (b^2-a^2\right ) (a \cos (c+d x)+b)}-\frac{6 a \tan (c+d x)}{5 b^3}+\frac{2 \tan (c+d x) \sec (c+d x)}{5 b^2}\right )}{d (a+b \sec (c+d x))^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*(4*a^2 + b^2)*(b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(4*a^3 + 4*a
^2*b - 3*a*b^2 - 3*b^3)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c +
d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(-4*a^2 - a*b + 3*b^2)*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)] + (4*a^2 - 3*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(5*
b^4*(-a^2 + b^2)*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(3/2)) + ((b + a*Cos[c + d*x])^2*Sec[c + d*x]
^2*((2*(-16*a^4 + 8*a^2*b^2 + 3*b^4)*Sin[c + d*x])/(5*b^4*(-a^2 + b^2)) + (2*a^4*Sin[c + d*x])/(b^3*(-a^2 + b^
2)*(b + a*Cos[c + d*x])) - (6*a*Tan[c + d*x])/(5*b^3) + (2*Sec[c + d*x]*Tan[c + d*x])/(5*b^2)))/(d*(a + b*Sec[
c + d*x])^(3/2))

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Maple [B]  time = 0.851, size = 2477, normalized size = 6.2 \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)^5/(a+b*sec(d*x+c))^(3/2),x)

[Out]

-1/5/d/(a-b)/(a+b)/b^4*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-a^2*b^3-8*cos(d*x+c)^4*a^4*b+3*cos(d*x+c)
^4*a^2*b^3+6*cos(d*x+c)^3*a^3*b^2+5*cos(d*x+c)^3*a*b^4+6*cos(d*x+c)^2*a^2*b^3-2*cos(d*x+c)*a*b^4-3*sin(d*x+c)*
cos(d*x+c)^3*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*b^5-16*sin(d*x+c)*cos(d*x+c)^3*EllipticE((-1+cos(d*x+c))/sin(d*x+c)
,((a-b)/(a+b))^(1/2))*(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)*a^5+b^
5+16*cos(d*x+c)^4*a^5-16*cos(d*x+c)^3*a^5-3*cos(d*x+c)^3*b^5+2*cos(d*x+c)^2*b^5-8*cos(d*x+c)^4*a^3*b^2-3*cos(d
*x+c)^4*a*b^4+16*cos(d*x+c)^3*a^4*b-8*cos(d*x+c)^3*a^2*b^3-8*cos(d*x+c)^2*a^4*b+2*cos(d*x+c)*a^3*b^2+3*sin(d*x
+c)*cos(d*x+c)^3*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*b^5-3*sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x
+c),((a-b)/(a+b))^(1/2))*(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)*b^5
-16*sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^5+3*sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+
c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*b^5+16*sin(d*x+c)*cos(d*x+c)^3*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^4*b+4*sin(d*x+c)*cos(d*x+c)^3*EllipticF
((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^3*b^2-8*sin(d*x+c)*cos(d*x+c)^3*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))*(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)*a^2*b^3+sin(d*x+c)*cos(d*
x+c)^3*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a*b^4-16*sin(d*x+c)*cos(d*x+c)^3*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a
-b)/(a+b))^(1/2))*(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)*a^4*b+8*si
n(d*x+c)*cos(d*x+c)^3*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^3*b^2+8*sin(d*x+c)*cos(d*x+c)^3*EllipticE((-1+cos(d*x+c)
)/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^2*b^3+3*sin(d*x+c)*cos(d*x+c)^3*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a*b^4+16*sin(d*x+c)*cos(d*x+c)^2*Ellipti
cF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^4*b+4*sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))*(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)*a^3*b^2-8*sin(d*x+c)*cos(
d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^2*b^3+sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((
a-b)/(a+b))^(1/2))*(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)*a*b^4-16*
sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^4*b+8*sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c)
)/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^3*b^2+8*sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a^2*b^3+3*sin(d*x+c)*cos(d*x+c)^2*Ellipt
icE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(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)*a*b^4)/(b+a*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)^2

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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