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]

(-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)

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Rubi [A]  time = 0.0576487, antiderivative size = 100, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, 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 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]

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]

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*}

Mathematica [A]  time = 0.0534337, 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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Maple [C]  time = 0.223, size = 356, normalized size = 3.6 \begin{align*}{\frac{2\, \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2} \left ( -1+\cos \left ( dx+c \right ) \right ) ^{2}}{5\,d \left ( \sin \left ( dx+c \right ) \right ) ^{5} \left ( \cos \left ( dx+c \right ) \right ) ^{4}} \left ( 3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{3}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) -3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \left ( \cos \left ( dx+c \right ) \right ) ^{2}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \sin \left ( dx+c \right ) -3\,i\sqrt{ \left ( \cos \left ( dx+c \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }},i \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) -3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}+2\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}+1 \right ) \left ({\frac{b}{\cos \left ( dx+c \right ) }} \right ) ^{-{\frac{3}{2}}}} \end{align*}

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*(cos(d*x+c)+1)^2*(-1+cos(d*x+c))^2*(3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d
*x+c)^3*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-3*I*(1/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x
+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)^3*sin(d*x+c)+3*I*(1/(cos(d*x+c)+1))^(1/2)*(
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)^2*EllipticE(I*(-1+cos(d*x+c))/sin(d*x+c),I)*sin(d*x+c)-3*I*(1/(cos
(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*EllipticF(I*(-1+cos(d*x+c))/sin(d*x+c),I)*cos(d*x+c)^2*sin
(d*x+c)-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., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{5}}{\left (b \sec \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b \sec \left (d x + c\right )} \sec \left (d x + c\right )^{3}}{b^{2}}, x\right ) \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (c + d x \right )}}{\left (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)**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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{5}}{\left (b \sec \left (d x + c\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

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)