3.563 \(\int \frac{\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=325 \[ \frac{2 (2 a+b) (4 a+b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{3 b^3 d \sqrt{a+b}}-\frac{2 a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (4 a^2-b^2\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{3 b^2 d \left (a^2-b^2\right )}+\frac{2 a \left (8 a^2-5 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^4 d \sqrt{a+b}} \]
[Out]
(2*a*(8*a^2 - 5*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqr
t[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^4*Sqrt[a + b]*d) + (2*(2*a + b
)*(4*a + b)*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*b^3*Sqrt[a + b]*d) - (2*a^2*Sec[c + d*x]*
Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(4*a^2 - b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d
*x])/(3*b^2*(a^2 - b^2)*d)
________________________________________________________________________________________
Rubi [A] time = 0.503577, antiderivative size = 325, 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 =
{3845, 4082, 4005, 3832, 4004} \[ -\frac{2 a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) \sqrt{a+b \sec (c+d x)}}+\frac{2 \left (4 a^2-b^2\right ) \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{3 b^2 d \left (a^2-b^2\right )}+\frac{2 a \left (8 a^2-5 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^4 d \sqrt{a+b}}+\frac{2 (2 a+b) (4 a+b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{3 b^3 d \sqrt{a+b}} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^4/(a + b*Sec[c + d*x])^(3/2),x]
[Out]
(2*a*(8*a^2 - 5*b^2)*Cot[c + d*x]*EllipticE[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqr
t[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(3*b^4*Sqrt[a + b]*d) + (2*(2*a + b
)*(4*a + b)*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*b^3*Sqrt[a + b]*d) - (2*a^2*Sec[c + d*x]*
Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Sec[c + d*x]]) + (2*(4*a^2 - b^2)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d
*x])/(3*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 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 ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx &=-\frac{2 a^2 \sec (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 (c+d x) \left (a^2-\frac{1}{2} a b \sec (c+d x)-\frac{1}{2} \left (4 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 (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 (4 a^2-b^2\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}-\frac{4 \int \frac{\sec (c+d x) \left (\frac{1}{4} b \left (2 a^2+b^2\right )+\frac{1}{4} a \left (8 a^2-5 b^2\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 \sec (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 (4 a^2-b^2\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}+\frac{((2 a+b) (4 a+b)) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{3 b^2 (a+b)}-\frac{\left (a \left (8 a^2-5 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^2 \left (a^2-b^2\right )}\\ &=\frac{2 a \left (8 a^2-5 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 b^4 \sqrt{a+b} d}+\frac{2 (2 a+b) (4 a+b) \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 b^3 \sqrt{a+b} d}-\frac{2 a^2 \sec (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 (4 a^2-b^2\right ) \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d}\\ \end{align*}
Mathematica [A] time = 14.0506, size = 470, normalized size = 1.45 \[ \frac{\sec ^2(c+d x) (a \cos (c+d x)+b)^2 \left (-\frac{2 a \left (5 b^2-8 a^2\right ) \sin (c+d x)}{3 b^3 \left (b^2-a^2\right )}-\frac{2 a^3 \sin (c+d x)}{b^2 \left (b^2-a^2\right ) (a \cos (c+d x)+b)}+\frac{2 \tan (c+d x)}{3 b^2}\right )}{d (a+b \sec (c+d x))^{3/2}}-\frac{2 \sec ^{\frac{3}{2}}(c+d x) \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} (a \cos (c+d x)+b) \left (-2 b \left (2 a^2 b+8 a^3-5 a b^2+b^3\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+a \left (8 a^2-5 b^2\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)+2 a \left (8 a^2 b+8 a^3-5 a b^2-5 b^3\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{3 b^3 d \left (b^2-a^2\right ) \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{3/2}} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]^4/(a + b*Sec[c + d*x])^(3/2),x]
[Out]
(-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*a*(8*a^3 + 8*a^2*b - 5*a*
b^2 - 5*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]))]*Ell
ipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] - 2*b*(8*a^3 + 2*a^2*b - 5*a*b^2 + 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*(8*a^2 - 5*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(
3*b^3*(-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*a*(-8*a^2 + 5*b^2)*Sin[c + d*x])/(3*b^3*(-a^2 + b^2)) - (2*a^3*Sin[c + d*x])/(b^2*(-a^2 + b^2)*(b +
a*Cos[c + d*x])) + (2*Tan[c + d*x])/(3*b^2)))/(d*(a + b*Sec[c + d*x])^(3/2))
________________________________________________________________________________________
Maple [B] time = 0.474, size = 1789, normalized size = 5.5 \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)^4/(a+b*sec(d*x+c))^(3/2),x)
[Out]
1/3/d/(a-b)/(a+b)/b^3*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(a^2*b^2-8*cos(d*x+c)^2*a^4-8*EllipticE((-1+
cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*cos(d*x+c)^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+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)*b^4-4*cos(d*x+
c)^3*a^3*b-5*cos(d*x+c)^3*a^2*b^2-b^4-8*cos(d*x+c)^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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3*b+5*cos(d*x+
c)^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)*EllipticE((-1+cos(d*x+c
))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b^2+5*cos(d*x+c)^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)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*
a*b^3-8*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^4+cos(d*x+c)^2*b^4+8*cos(d*x+c)^3*a^4+cos(d*x+c)
^3*a*b^3+4*cos(d*x+c)^2*a^2*b^2+4*cos(d*x+c)*a*b^3+8*cos(d*x+c)^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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^3
*b+2*cos(d*x+c)^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)*EllipticF(
(-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*a^2*b^2-5*cos(d*x+c)^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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2)
)*sin(d*x+c)*a*b^3+cos(d*x+c)^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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*sin(d*x+c)*b^4+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))*s
in(d*x+c)*cos(d*x+c)*a*b^3+8*(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))*sin(d*x+c)*cos(d*x+c)*a^3*b+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)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b)
)^(1/2))*sin(d*x+c)*cos(d*x+c)*a^2*b^2-5*(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))*sin(d*x+c)*cos(d*x+c)*a*b^3-8*(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))*sin(d*x+c)*cos(d*x+c)*a^3*b+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))*sin(d*x+c)*cos(d*x+c)*a^2*b^2
+8*cos(d*x+c)^2*a^3*b-5*cos(d*x+c)^2*a*b^3-4*cos(d*x+c)*a^3*b)/(b+a*cos(d*x+c))/sin(d*x+c)/cos(d*x+c)
________________________________________________________________________________________
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)^4/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Timed out
________________________________________________________________________________________
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 )^{4}}{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)^4/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^4/(b^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
________________________________________________________________________________________
Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{4}{\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)**4/(a+b*sec(d*x+c))**(3/2),x)
[Out]
Integral(sec(c + d*x)**4/(a + b*sec(c + d*x))**(3/2), x)
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{4}}{{\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)^4/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate(sec(d*x + c)^4/(b*sec(d*x + c) + a)^(3/2), x)