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

Optimal. Leaf size=125 \[ \frac{2 b^2 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 \sin (c+d x)}{15 b^2 d \sqrt{b \cos (c+d x)}}-\frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 b^3 d \sqrt{\cos (c+d x)}}+\frac{14 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}} \]

[Out]

(-14*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*b^3*d*Sqrt[Cos[c + d*x]]) + (2*b^2*Sin[c + d*x])/(9*d
*(b*Cos[c + d*x])^(9/2)) + (14*Sin[c + d*x])/(45*d*(b*Cos[c + d*x])^(5/2)) + (14*Sin[c + d*x])/(15*b^2*d*Sqrt[
b*Cos[c + d*x]])

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Rubi [A]  time = 0.0979237, antiderivative size = 125, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {16, 2636, 2640, 2639} \[ \frac{2 b^2 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 \sin (c+d x)}{15 b^2 d \sqrt{b \cos (c+d x)}}-\frac{14 E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{b \cos (c+d x)}}{15 b^3 d \sqrt{\cos (c+d x)}}+\frac{14 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-14*Sqrt[b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2])/(15*b^3*d*Sqrt[Cos[c + d*x]]) + (2*b^2*Sin[c + d*x])/(9*d
*(b*Cos[c + d*x])^(9/2)) + (14*Sin[c + d*x])/(45*d*(b*Cos[c + d*x])^(5/2)) + (14*Sin[c + d*x])/(15*b^2*d*Sqrt[
b*Cos[c + d*x]])

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 2640

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

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 ^3(c+d x)}{(b \cos (c+d x))^{5/2}} \, dx &=b^3 \int \frac{1}{(b \cos (c+d x))^{11/2}} \, dx\\ &=\frac{2 b^2 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{1}{9} (7 b) \int \frac{1}{(b \cos (c+d x))^{7/2}} \, dx\\ &=\frac{2 b^2 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{7 \int \frac{1}{(b \cos (c+d x))^{3/2}} \, dx}{15 b}\\ &=\frac{2 b^2 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 b^2 d \sqrt{b \cos (c+d x)}}-\frac{7 \int \sqrt{b \cos (c+d x)} \, dx}{15 b^3}\\ &=\frac{2 b^2 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 b^2 d \sqrt{b \cos (c+d x)}}-\frac{\left (7 \sqrt{b \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{15 b^3 \sqrt{\cos (c+d x)}}\\ &=-\frac{14 \sqrt{b \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 b^3 d \sqrt{\cos (c+d x)}}+\frac{2 b^2 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac{14 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac{14 \sin (c+d x)}{15 b^2 d \sqrt{b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.088816, size = 80, normalized size = 0.64 \[ \frac{42 \sin (c+d x)-42 \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+2 \tan (c+d x) \sec (c+d x) \left (5 \sec ^2(c+d x)+7\right )}{45 b^2 d \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-42*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 42*Sin[c + d*x] + 2*Sec[c + d*x]*(7 + 5*Sec[c + d*x]^2)*Ta
n[c + d*x])/(45*b^2*d*Sqrt[b*Cos[c + d*x]])

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Maple [B]  time = 3.252, size = 414, normalized size = 3.3 \begin{align*} -2\,{\frac{\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}{{b}^{2}\sin \left ( 1/2\,dx+c/2 \right ) \sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) }d} \left ( -{\frac{\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}{144\,b \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{5}}}-{\frac{7\,\cos \left ( 1/2\,dx+c/2 \right ) \sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}{180\,b \left ( \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1/2 \right ) ^{3}}}-{\frac{14\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}\cos \left ( 1/2\,dx+c/2 \right ) }{15\,\sqrt{b \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}}}+{\frac{7\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) }{15\,\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}}-{\frac{7\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1} \left ({\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \right ) }{15\,\sqrt{-b \left ( 2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}- \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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*(-1/144*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/2*d
*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/
2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*
x+1/2*c)/(b*(2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1
/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c)
,2^(1/2))-7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(
1/2*d*x+1/2*c)^2))^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))/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 )^{3}}{\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)^3/(b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

[Out]

integral(sqrt(b*cos(d*x + c))*sec(d*x + c)^3/(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)**3/(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 )^{3}}{\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)^3/(b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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