3.572 \(\int \frac{\sec ^3(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=337 \[ \frac{2 \left (2 a^2+3 a b-3 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3 b^2 d (a-b) (a+b)^{3/2}}-\frac{2 a^2 \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{4 a \left (a^2-3 b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{4 a \left (a^2-3 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^3 d (a-b) (a+b)^{3/2}} \]
[Out]
(4*a*(a^2 - 3*b^2)*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))])/(3*(a - b)*b^3*(a + b)^(3/2)*d) + (2*
(2*a^2 + 3*a*b - 3*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))])/(3*(a - b)*b^2*(a + b)^(3/2)*d)
- (2*a^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (4*a*(a^2 - 3*b^2)*Tan[c + d*x])/(3*b*
(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])
________________________________________________________________________________________
Rubi [A] time = 0.501503, antiderivative size = 337, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used =
{3839, 4003, 4005, 3832, 4004} \[ -\frac{2 a^2 \tan (c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac{4 a \left (a^2-3 b^2\right ) \tan (c+d x)}{3 b d \left (a^2-b^2\right )^2 \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (2 a^2+3 a b-3 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^2 d (a-b) (a+b)^{3/2}}+\frac{4 a \left (a^2-3 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^3 d (a-b) (a+b)^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^3/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
(4*a*(a^2 - 3*b^2)*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))])/(3*(a - b)*b^3*(a + b)^(3/2)*d) + (2*
(2*a^2 + 3*a*b - 3*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))])/(3*(a - b)*b^2*(a + b)^(3/2)*d)
- (2*a^2*Tan[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Sec[c + d*x])^(3/2)) + (4*a*(a^2 - 3*b^2)*Tan[c + d*x])/(3*b*
(a^2 - b^2)^2*d*Sqrt[a + b*Sec[c + d*x]])
Rule 3839
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(a^2*Cot[e + f*x
]*(a + b*Csc[e + f*x])^(m + 1))/(b*f*(m + 1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e + f*
x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[a*b*(m + 1) - (a^2 + b^2*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b
, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]
Rule 4003
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_))
, x_Symbol] :> -Simp[((A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(a^2 - b^2)), Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[(a*A - b*B)*(m + 1) - (A*b - a*B
)*(m + 2)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] &
& 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 ^3(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx &=-\frac{2 a^2 \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac{2 \int \frac{\sec (c+d x) \left (-\frac{3 a b}{2}-\frac{1}{2} \left (2 a^2-3 b^2\right ) \sec (c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{4 a \left (a^2-3 b^2\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{4 \int \frac{\sec (c+d x) \left (\frac{1}{4} b \left (a^2+3 b^2\right )-\frac{1}{2} a \left (a^2-3 b^2\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b \left (a^2-b^2\right )^2}\\ &=-\frac{2 a^2 \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{4 a \left (a^2-3 b^2\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}+\frac{\left (2 a^2+3 a b-3 b^2\right ) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 (a-b) b (a+b)^2}-\frac{\left (2 a \left (a^2-3 b^2\right )\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b \left (a^2-b^2\right )^2}\\ &=\frac{4 a \left (a^2-3 b^2\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}}}{3 (a-b) b^3 (a+b)^{3/2} d}+\frac{2 \left (2 a^2+3 a b-3 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}}}{3 (a-b) b^2 (a+b)^{3/2} d}-\frac{2 a^2 \tan (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}+\frac{4 a \left (a^2-3 b^2\right ) \tan (c+d x)}{3 b \left (a^2-b^2\right )^2 d \sqrt{a+b \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 14.0179, size = 503, normalized size = 1.49 \[ \frac{4 \sec ^{\frac{5}{2}}(c+d x) \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} (a \cos (c+d x)+b)^2 \left (b \left (a^2 b-2 a^3+6 a b^2+3 b^3\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+a \left (a^2-3 b^2\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)+2 a \left (a^2 b+a^3-3 a b^2-3 b^3\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{3 d \left (b^3-a^2 b\right )^2 \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}}+\frac{\sec ^3(c+d x) (a \cos (c+d x)+b)^3 \left (\frac{4 a \left (3 b^2-a^2\right ) \sin (c+d x)}{3 b^2 \left (b^2-a^2\right )^2}-\frac{2 a \sin (c+d x)}{3 \left (b^2-a^2\right ) (a \cos (c+d x)+b)^2}-\frac{2 \left (5 a b^2 \sin (c+d x)-a^3 \sin (c+d x)\right )}{3 b \left (b^2-a^2\right )^2 (a \cos (c+d x)+b)}\right )}{d (a+b \sec (c+d x))^{5/2}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]^3/(a + b*Sec[c + d*x])^(5/2),x]
[Out]
((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((4*a*(-a^2 + 3*b^2)*Sin[c + d*x])/(3*b^2*(-a^2 + b^2)^2) - (2*a*Sin[c
+ d*x])/(3*(-a^2 + b^2)*(b + a*Cos[c + d*x])^2) - (2*(-(a^3*Sin[c + d*x]) + 5*a*b^2*Sin[c + d*x]))/(3*b*(-a^2
+ b^2)^2*(b + a*Cos[c + d*x]))))/(d*(a + b*Sec[c + d*x])^(5/2)) + (4*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)
*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*a*(a^3 + 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
)] + b*(-2*a^3 + a^2*b + 6*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]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + a*(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]))/(3*(-(a^2*b) + b^3)^2*d*Sqrt[Sec[(c + d*x)/2]^2]*(a
+ b*Sec[c + d*x])^(5/2))
________________________________________________________________________________________
Maple [B] time = 0.329, size = 2731, normalized size = 8.1 \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]
-1/3/d/(a-b)^2/(a+b)^2/b^2*4^(1/2)*(6*cos(d*x+c)^3*a^3*b^2+12*cos(d*x+c)^2*a^2*b^3+6*cos(d*x+c)*a*b^4+3*Ellipt
icF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*(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+2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*
sin(d*x+c)*(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^2-6*Ellipti
cE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*(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^3-6*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+
c)*(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^4-2*EllipticF((-1+cos
(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*(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+EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*(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+6*EllipticF((-1+cos(d*x+c))/s
in(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*(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+2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*(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^5-4*cos(d*x+c)^2*a^3*b^2-6*cos(d*x
+c)^2*a*b^4+3*cos(d*x+c)*a^4*b-7*cos(d*x+c)*a^2*b^3+2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*sin(d*x+c)*(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-2*cos(d*x
+c)^3*a^5+cos(d*x+c)^3*a^4*b-5*cos(d*x+c)^3*a^2*b^3-4*cos(d*x+c)^2*a^4*b-2*cos(d*x+c)*a^3*b^2+2*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+2*cos(d*x+c)^2*a^5-4*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(
a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*(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^2-6*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*(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^4-2*EllipticF((-1+cos(d*x+c))/sin(d*x
+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(co
s(d*x+c)+1))^(1/2)*a^4*b-EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*(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+7*EllipticF((-1+cos(d*x+c
))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)*(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+9*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*co
s(d*x+c)*(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+4*sin(d*x+c)*
cos(d*x+c)*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-12*sin(d*x+c)*cos(d*x+c)*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*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*(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^5-2*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^4*b+s
in(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+6*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+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+2*sin(d*x+c)*cos(d*x+c)^2*Ellipti
cE((-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-6*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^2-6*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)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)/sin(d*x+c)/(b+a*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)^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 (\frac{\sqrt{b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3}}{b^{3} \sec \left (d x + c\right )^{3} + 3 \, a b^{2} \sec \left (d x + c\right )^{2} + 3 \, a^{2} b \sec \left (d x + c\right ) + a^{3}}, 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(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^3/(b^3*sec(d*x + c)^3 + 3*a*b^2*sec(d*x + c)^2 + 3*a^2*b*sec(d*
x + c) + a^3), x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec{\left (c + d x \right )}\right )^{\frac{5}{2}}}\, dx \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]
Integral(sec(c + d*x)**3/(a + b*sec(c + d*x))**(5/2), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{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(sec(d*x + c)^3/(b*sec(d*x + c) + a)^(5/2), x)