\(\int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [563]

Optimal result

Integrand size = 23, antiderivative size = 325 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 a \left (8 a^2-5 b^2\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^4 \sqrt {a+b} d}+\frac {2 (2 a+b) (4 a+b) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 b^3 \sqrt {a+b} d}-\frac {2 a^2 \sec (c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (4 a^2-b^2\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{3 b^2 \left (a^2-b^2\right ) d} \]

[Out]

2/3*a*(8*a^2-5*b^2)*cot(d*x+c)*EllipticE((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x
+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/d/(a+b)^(1/2)+2/3*(2*a+b)*(4*a+b)*cot(d*x+c)*EllipticF((
a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b)
)^(1/2)/b^3/d/(a+b)^(1/2)-2*a^2*sec(d*x+c)*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)+2/3*(4*a^2-b^2)*(a+
b*sec(d*x+c))^(1/2)*tan(d*x+c)/b^2/(a^2-b^2)/d

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 325, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3930, 4167, 4090, 3917, 4089} \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 a \left (8 a^2-5 b^2\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^4 d \sqrt {a+b}}-\frac {2 a^2 \tan (c+d x) \sec (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (4 a^2-b^2\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b^2 d \left (a^2-b^2\right )}+\frac {2 (2 a+b) (4 a+b) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 b^3 d \sqrt {a+b}} \]

[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 3917

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

Rule 3930

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] &&
NeQ[a^2 - b^2, 0] && LtQ[m, -1] && (IGtQ[n, 3] || (IntegersQ[n + 1/2, 2*m] && GtQ[n, 2]))

Rule 4089

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 + C
sc[e + f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + b*(B/A), 2]], (a
*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]

Rule 4090

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 4167

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]

Rubi steps \begin{align*} \text {integral}& = -\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 (\arcsin \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) \operatorname {EllipticF}\left (\arcsin \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] (warning: unable to verify)

Time = 11.67 (sec) , antiderivative size = 470, normalized size of antiderivative = 1.45 \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=-\frac {2 (b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 a \left (8 a^3+8 a^2 b-5 a b^2-5 b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )-2 b \left (8 a^3+2 a^2 b-5 a b^2+b^3\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+a \left (8 a^2-5 b^2\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b^3 \left (-a^2+b^2\right ) d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{3/2}}+\frac {(b+a \cos (c+d x))^2 \sec ^2(c+d x) \left (-\frac {2 a \left (-8 a^2+5 b^2\right ) \sin (c+d x)}{3 b^3 \left (-a^2+b^2\right )}-\frac {2 a^3 \sin (c+d x)}{b^2 \left (-a^2+b^2\right ) (b+a \cos (c+d x))}+\frac {2 \tan (c+d x)}{3 b^2}\right )}{d (a+b \sec (c+d x))^{3/2}} \]

[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] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2285\) vs. \(2(297)=594\).

Time = 12.33 (sec) , antiderivative size = 2286, normalized size of antiderivative = 7.03

method	result	size
default	\(\text {Expression too large to display}\)	\(2286\)

[In]

int(sec(d*x+c)^4/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/d/(a-b)/(a+b)/b^3*(a+b*sec(d*x+c))^(1/2)/(b+a*cos(d*x+c))/(cos(d*x+c)+1)*(4*a^3*b*sin(d*x+c)-4*a*b^3*sin(
d*x+c)+b^4*sin(d*x+c)-a^2*b^2*sin(d*x+c)+8*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c
)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b-5*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x
+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2-5
*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x
+c)/(cos(d*x+c)+1))^(1/2)*a*b^3-EllipticF(cot(d*x+c)-csc(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)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^4*cos(d*x+c)^2+8*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x
+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^4*cos(d
*x+c)^2-2*EllipticF(cot(d*x+c)-csc(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)
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*b^4*cos(d*x+c)+16*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+
c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^4*cos(d*x+c)-8*(1/(a+b)*(b+a*c
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*a^3*b-2*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b)
)^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2+5*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF
(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^3+a*b^3*cos(d*x+c)*sin(d*x+c
)+8*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*a^4-4*a^3*b*cos(d*x+c)*sin(d*x+c)-5*a^2*b^2*cos(d*x+c)*sin(d*x+c)-(1/(a+b)*(b+a*c
os(d*x+c))/(cos(d*x+c)+1))^(1/2)*EllipticF(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*b^4+tan(d*x+c)*b^4-5*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/
2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b^2*cos(d*x+c)^2-5*(1/(a+b)*(b+a*cos(d*x+c))/(cos(
d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3*
cos(d*x+c)^2-16*EllipticF(cot(d*x+c)-csc(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)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b*cos(d*x+c)-4*EllipticF(cot(d*x+c)-csc(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)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2*cos(d*x+c)+10*Ell
ipticF(cot(d*x+c)-csc(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)*(cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*a*b^3*cos(d*x+c)+16*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x
+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a^3*b*cos(d*x+c)-10*(1/(a+b)*(b+a*cos(d*x+c
))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2)
)*a^2*b^2*cos(d*x+c)-10*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*Elli
pticE(cot(d*x+c)-csc(d*x+c),((a-b)/(a+b))^(1/2))*a*b^3*cos(d*x+c)-8*EllipticF(cot(d*x+c)-csc(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)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^3*b*cos(d*x+c)^
2-2*EllipticF(cot(d*x+c)-csc(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)*(cos(
d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b^2*cos(d*x+c)^2+5*EllipticF(cot(d*x+c)-csc(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)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^3*cos(d*x+c)^2+8*(1/(a+b)*(b+
a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticE(cot(d*x+c)-csc(d*x+c),((a-b)/(
a+b))^(1/2))*a^3*b*cos(d*x+c)^2-a^2*b^2*tan(d*x+c)+8*a^4*cos(d*x+c)*sin(d*x+c))

Fricas [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[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]

\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

[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)

Maxima [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[In]

integrate(sec(d*x+c)^4/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^4/(b*sec(d*x + c) + a)^(3/2), x)

Giac [F]

\[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {\sec \left (d x + c\right )^{4}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]

[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)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sec ^4(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

[In]

int(1/(cos(c + d*x)^4*(a + b/cos(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + b/cos(c + d*x))^(3/2)), x)