3.112 \(\int \frac {\sec ^5(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=100 \[ \frac {2 \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 b^4 d}+\frac {6 \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 b^2 d}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \]

[Out]

2/5*(b*sec(d*x+c))^(5/2)*sin(d*x+c)/b^4/d-6/5*(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))/b/d/cos(d*x+c)^(1/2)/(b*sec(d*x+c))^(1/2)+6/5*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/b^2/d

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Rubi [A]  time = 0.06, antiderivative size = 100, 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} \[ \frac {2 \sin (c+d x) (b \sec (c+d x))^{5/2}}{5 b^4 d}+\frac {6 \sin (c+d x) \sqrt {b \sec (c+d x)}}{5 b^2 d}-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5/(b*Sec[c + d*x])^(3/2),x]

[Out]

(-6*EllipticE[(c + d*x)/2, 2])/(5*b*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Sec[c + d*x]]) + (6*Sqrt[b*Sec[c + d*x]]*Sin[c
+ d*x])/(5*b^2*d) + (2*(b*Sec[c + d*x])^(5/2)*Sin[c + d*x])/(5*b^4*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 \frac {\sec ^5(c+d x)}{(b \sec (c+d x))^{3/2}} \, dx &=\frac {\int (b \sec (c+d x))^{7/2} \, dx}{b^5}\\ &=\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^4 d}+\frac {3 \int (b \sec (c+d x))^{3/2} \, dx}{5 b^3}\\ &=\frac {6 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 b^2 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^4 d}-\frac {3 \int \frac {1}{\sqrt {b \sec (c+d x)}} \, dx}{5 b}\\ &=\frac {6 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 b^2 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^4 d}-\frac {3 \int \sqrt {\cos (c+d x)} \, dx}{5 b \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}\\ &=-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b d \sqrt {\cos (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {6 \sqrt {b \sec (c+d x)} \sin (c+d x)}{5 b^2 d}+\frac {2 (b \sec (c+d x))^{5/2} \sin (c+d x)}{5 b^4 d}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 64, normalized size = 0.64 \[ \frac {2 \tan (c+d x) \left (\sec ^2(c+d x)+3\right )-\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{\sqrt {\cos (c+d x)}}}{5 b d \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5/(b*Sec[c + d*x])^(3/2),x]

[Out]

((-6*EllipticE[(c + d*x)/2, 2])/Sqrt[Cos[c + d*x]] + 2*(3 + Sec[c + d*x]^2)*Tan[c + d*x])/(5*b*d*Sqrt[b*Sec[c
+ d*x]])

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fricas [F]  time = 0.63, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{3}}{b^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{5}}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^5/(b*sec(d*x + c))^(3/2), x)

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maple [C]  time = 0.96, size = 356, normalized size = 3.56 \[ -\frac {2 \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right )^{2} \left (3 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 i \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticF \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )-3 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \EllipticE \left (\frac {i \left (-1+\cos \left (d x +c \right )\right )}{\sin \left (d x +c \right )}, i\right )+3 \left (\cos ^{3}\left (d x +c \right )\right )-2 \left (\cos ^{2}\left (d x +c \right )\right )-1\right )}{5 d \sin \left (d x +c \right )^{5} \cos \left (d x +c \right )^{4} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5/(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)*EllipticF(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)*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*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*cos(d*x+c)^3-2*cos(d*x+c)^2-1)/sin(d*x+c)^5/cos(d*x+c)^4/(b/cos(d*x+c))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{5}}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^5/(b*sec(d*x + c))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^5\,{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^5*(b/cos(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^5*(b/cos(c + d*x))^(3/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{5}{\left (c + d x \right )}}{\left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(b*sec(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**5/(b*sec(c + d*x))**(3/2), x)

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