3.70 \(\int \sec ^4(c+d x) \sqrt {b \sec (c+d x)} \, dx\)

Optimal. Leaf size=97 \[ \frac {2 \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 b^3 d}+\frac {10 \sin (c+d x) (b \sec (c+d x))^{3/2}}{21 b d}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d} \]

[Out]

10/21*(b*sec(d*x+c))^(3/2)*sin(d*x+c)/b/d+2/7*(b*sec(d*x+c))^(7/2)*sin(d*x+c)/b^3/d+10/21*(cos(1/2*d*x+1/2*c)^
2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*(b*sec(d*x+c))^(1/2)/d

________________________________________________________________________________________

Rubi [A]  time = 0.06, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {16, 3768, 3771, 2641} \[ \frac {2 \sin (c+d x) (b \sec (c+d x))^{7/2}}{7 b^3 d}+\frac {10 \sin (c+d x) (b \sec (c+d x))^{3/2}}{21 b d}+\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*Sqrt[b*Sec[c + d*x]],x]

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[b*Sec[c + d*x]])/(21*d) + (10*(b*Sec[c + d*x])^(3/2)*Sin
[c + d*x])/(21*b*d) + (2*(b*Sec[c + d*x])^(7/2)*Sin[c + d*x])/(7*b^3*d)

Rule 16

Int[(u_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Dist[1/b^m, Int[u*(b*v)^(m + n), x], x] /; FreeQ[{b, n}, x
] && IntegerQ[m]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3768

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Csc[c + d*x])^(n - 1))/(d*(n -
1)), x] + Dist[(b^2*(n - 2))/(n - 1), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1
] && IntegerQ[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rubi steps

\begin {align*} \int \sec ^4(c+d x) \sqrt {b \sec (c+d x)} \, dx &=\frac {\int (b \sec (c+d x))^{9/2} \, dx}{b^4}\\ &=\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^3 d}+\frac {5 \int (b \sec (c+d x))^{5/2} \, dx}{7 b^2}\\ &=\frac {10 (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 b d}+\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^3 d}+\frac {5}{21} \int \sqrt {b \sec (c+d x)} \, dx\\ &=\frac {10 (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 b d}+\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^3 d}+\frac {1}{21} \left (5 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {10 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \sec (c+d x)}}{21 d}+\frac {10 (b \sec (c+d x))^{3/2} \sin (c+d x)}{21 b d}+\frac {2 (b \sec (c+d x))^{7/2} \sin (c+d x)}{7 b^3 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 0.22, size = 69, normalized size = 0.71 \[ \frac {\sec ^2(c+d x) \sqrt {b \sec (c+d x)} \left (5 \sin (2 (c+d x))+6 \tan (c+d x)+10 \cos ^{\frac {5}{2}}(c+d x) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{21 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*Sqrt[b*Sec[c + d*x]],x]

[Out]

(Sec[c + d*x]^2*Sqrt[b*Sec[c + d*x]]*(10*Cos[c + d*x]^(5/2)*EllipticF[(c + d*x)/2, 2] + 5*Sin[2*(c + d*x)] + 6
*Tan[c + d*x]))/(21*d)

________________________________________________________________________________________

fricas [F]  time = 0.67, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{4}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(b*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*sec(d*x + c))*sec(d*x + c)^4, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(b*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sec(d*x + c))*sec(d*x + c)^4, x)

________________________________________________________________________________________

maple [C]  time = 1.20, size = 152, normalized size = 1.57 \[ -\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (5 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-5 \left (\cos ^{3}\left (d x +c \right )\right )+5 \left (\cos ^{2}\left (d x +c \right )\right )-3 \cos \left (d x +c \right )+3\right ) \sqrt {\frac {b}{\cos \left (d x +c \right )}}}{21 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(b*sec(d*x+c))^(1/2),x)

[Out]

-2/21/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(5*I*cos(d*x+c)^3*sin(d*x+c)*(1/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+
cos(d*x+c)))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)-5*cos(d*x+c)^3+5*cos(d*x+c)^2-3*cos(d*x+c)+3)*(b/
cos(d*x+c))^(1/2)/sin(d*x+c)^3/cos(d*x+c)^3

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*(b*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sec(d*x + c))*sec(d*x + c)^4, x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{{\cos \left (c+d\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sec {\left (c + d x \right )}} \sec ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(b*sec(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(b*sec(c + d*x))*sec(c + d*x)**4, x)

________________________________________________________________________________________