3.17 \(\int (c \sec (a+b x))^{7/2} \, dx\)

Optimal. Leaf size=98 \[ \frac{6 c^3 \sin (a+b x) \sqrt{c \sec (a+b x)}}{5 b}-\frac{6 c^4 E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b \sqrt{\cos (a+b x)} \sqrt{c \sec (a+b x)}}+\frac{2 c \sin (a+b x) (c \sec (a+b x))^{5/2}}{5 b} \]

[Out]

(-6*c^4*EllipticE[(a + b*x)/2, 2])/(5*b*Sqrt[Cos[a + b*x]]*Sqrt[c*Sec[a + b*x]]) + (6*c^3*Sqrt[c*Sec[a + b*x]]
*Sin[a + b*x])/(5*b) + (2*c*(c*Sec[a + b*x])^(5/2)*Sin[a + b*x])/(5*b)

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Rubi [A]  time = 0.0558185, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3768, 3771, 2639} \[ \frac{6 c^3 \sin (a+b x) \sqrt{c \sec (a+b x)}}{5 b}-\frac{6 c^4 E\left (\left .\frac{1}{2} (a+b x)\right |2\right )}{5 b \sqrt{\cos (a+b x)} \sqrt{c \sec (a+b x)}}+\frac{2 c \sin (a+b x) (c \sec (a+b x))^{5/2}}{5 b} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sec[a + b*x])^(7/2),x]

[Out]

(-6*c^4*EllipticE[(a + b*x)/2, 2])/(5*b*Sqrt[Cos[a + b*x]]*Sqrt[c*Sec[a + b*x]]) + (6*c^3*Sqrt[c*Sec[a + b*x]]
*Sin[a + b*x])/(5*b) + (2*c*(c*Sec[a + b*x])^(5/2)*Sin[a + b*x])/(5*b)

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

Mathematica [A]  time = 0.165577, size = 62, normalized size = 0.63 \[ \frac{c (c \sec (a+b x))^{5/2} \left (7 \sin (a+b x)+3 \sin (3 (a+b x))-12 \cos ^{\frac{5}{2}}(a+b x) E\left (\left .\frac{1}{2} (a+b x)\right |2\right )\right )}{10 b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sec[a + b*x])^(7/2),x]

[Out]

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

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Maple [C]  time = 0.32, size = 354, normalized size = 3.6 \begin{align*}{\frac{2\, \left ( -1+\cos \left ( bx+a \right ) \right ) ^{2}\cos \left ( bx+a \right ) \left ( \cos \left ( bx+a \right ) +1 \right ) ^{2}}{5\,b \left ( \sin \left ( bx+a \right ) \right ) ^{5}} \left ( 3\,i \left ( \cos \left ( bx+a \right ) \right ) ^{3}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}\sin \left ( bx+a \right ) -3\,i \left ( \cos \left ( bx+a \right ) \right ) ^{3}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}\sin \left ( bx+a \right ) +3\,i\sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}} \left ( \cos \left ( bx+a \right ) \right ) ^{2}{\it EllipticE} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \sin \left ( bx+a \right ) -3\,i\sqrt{ \left ( \cos \left ( bx+a \right ) +1 \right ) ^{-1}}\sqrt{{\frac{\cos \left ( bx+a \right ) }{\cos \left ( bx+a \right ) +1}}}{\it EllipticF} \left ({\frac{i \left ( -1+\cos \left ( bx+a \right ) \right ) }{\sin \left ( bx+a \right ) }},i \right ) \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sin \left ( bx+a \right ) -3\, \left ( \cos \left ( bx+a \right ) \right ) ^{3}+2\, \left ( \cos \left ( bx+a \right ) \right ) ^{2}+1 \right ) \left ({\frac{c}{\cos \left ( bx+a \right ) }} \right ) ^{{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sec(b*x+a))^(7/2),x)

[Out]

2/5/b*(-1+cos(b*x+a))^2*(3*I*cos(b*x+a)^3*EllipticE(I*(-1+cos(b*x+a))/sin(b*x+a),I)*(1/(cos(b*x+a)+1))^(1/2)*(
cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*sin(b*x+a)-3*I*cos(b*x+a)^3*EllipticF(I*(-1+cos(b*x+a))/sin(b*x+a),I)*(1/(cos
(b*x+a)+1))^(1/2)*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*sin(b*x+a)+3*I*(1/(cos(b*x+a)+1))^(1/2)*(cos(b*x+a)/(cos(b
*x+a)+1))^(1/2)*cos(b*x+a)^2*EllipticE(I*(-1+cos(b*x+a))/sin(b*x+a),I)*sin(b*x+a)-3*I*(1/(cos(b*x+a)+1))^(1/2)
*(cos(b*x+a)/(cos(b*x+a)+1))^(1/2)*EllipticF(I*(-1+cos(b*x+a))/sin(b*x+a),I)*cos(b*x+a)^2*sin(b*x+a)-3*cos(b*x
+a)^3+2*cos(b*x+a)^2+1)*cos(b*x+a)*(cos(b*x+a)+1)^2*(c/cos(b*x+a))^(7/2)/sin(b*x+a)^5

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sec \left (b x + a\right )\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sec(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

integrate((c*sec(b*x + a))^(7/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sec(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sec(b*x+a))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (c \sec \left (b x + a\right )\right )^{\frac{7}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sec(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate((c*sec(b*x + a))^(7/2), x)