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

Optimal. Leaf size=98 \[ \frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 b^2 d \sqrt{b \cos (c+d x)}}+\frac{10 \sin (c+d x)}{21 b d (b \cos (c+d x))^{3/2}}+\frac{2 b \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}} \]

[Out]

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

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Rubi [A]  time = 0.0792451, antiderivative size = 98, 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, 2636, 2642, 2641} \[ \frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 b^2 d \sqrt{b \cos (c+d x)}}+\frac{10 \sin (c+d x)}{21 b d (b \cos (c+d x))^{3/2}}+\frac{2 b \sin (c+d x)}{7 d (b \cos (c+d x))^{7/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2636

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

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

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]

Rubi steps

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

Mathematica [A]  time = 0.0709466, size = 66, normalized size = 0.67 \[ \frac{10 \sqrt{\cos (c+d x)} F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 \tan (c+d x) \left (3 \sec ^2(c+d x)+5\right )}{21 b^2 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(10*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + 2*(5 + 3*Sec[c + d*x]^2)*Tan[c + d*x])/(21*b^2*d*Sqrt[b*Cos
[c + d*x]])

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Maple [B]  time = 1.937, size = 398, normalized size = 4.1 \begin{align*} -{\frac{2}{21\,{b}^{2}d} \left ( -40\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+60\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-40\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}\cos \left ( 1/2\,dx+c/2 \right ) -30\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+40\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}\cos \left ( 1/2\,dx+c/2 \right ) +5\,\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -16\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}{\frac{1}{\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2} \right ) }}} \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) ^{-3} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/21*(-40*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)
*sin(1/2*d*x+1/2*c)^6+60*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2
*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-40*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-30*(2*sin(1/2*d*x+1/2*c)^2-1)^(
1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*sin(1/2*d*x+1/2*c)^2+40*sin(1/2*d*x+1/
2*c)^4*cos(1/2*d*x+1/2*c)+5*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))-16*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))/b^2*(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/
2*c)^2)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(2*cos(1/2*d*x+1/2*c)^2-1)^3/sin(1/2*d*
x+1/2*c)/(b*(2*cos(1/2*d*x+1/2*c)^2-1))^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*cos(d*x + c))*sec(d*x + c)^2/(b^3*cos(d*x + c)^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(sec(d*x+c)**2/(b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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