3.81 \(\int \sec ^2(c+d x) (b \sec (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=98 \[ -\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 b d}+\frac {6 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 d} \]

[Out]

2/5*(b*sec(d*x+c))^(5/2)*sin(d*x+c)/b/d-6/5*b^2*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(
1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+6/5*b*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.06, antiderivative size = 98, 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, 2639} \[ -\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {2 \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 b d}+\frac {6 b \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*b^2*EllipticE[(c + d*x)/2, 2])/(5*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (6*b*Sqrt[b*Sec[c + d*x]]*S
in[c + d*x])/(5*d) + (2*(b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*b*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 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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 ^2(c+d x) (b \sec (c+d x))^{3/2} \, dx &=\frac {\int (b \sec (c+d x))^{7/2} \, dx}{b^2}\\ &=\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d}+\frac {3}{5} \int (b \sec (c+d x))^{3/2} \, dx\\ &=\frac {6 b \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d}-\frac {1}{5} \left (3 b^2\right ) \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx\\ &=\frac {6 b \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d}-\frac {\left (3 b^2\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\\ &=-\frac {6 b^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {6 b \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b d}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 64, normalized size = 0.65 \[ \frac {(b \sec (c+d x))^{5/2} \left (7 \sin (c+d x)+3 \sin (3 (c+d x))-12 \cos ^{\frac {5}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{10 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((b*Sec[c + d*x])^(5/2)*(-12*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + 7*Sin[c + d*x] + 3*Sin[3*(c + d*x)
]))/(10*b*d)

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fricas [F]  time = 0.66, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\sqrt {b \sec \left (d x + c\right )} b \sec \left (d x + c\right )^{3}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*sec(d*x + c))*b*sec(d*x + c)^3, x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \sec \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.76, size = 356, normalized size = 3.63 \[ \frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (3 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 \left (\cos ^{3}\left (d x +c \right )\right )+2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}}{5 d \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*(b*sec(d*x+c))^(3/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \sec \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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