3.174 \(\int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=76 \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{b d \sqrt {b \sec (c+d x)}}-\frac {\sin ^3(c+d x) \sqrt {\sec (c+d x)}}{3 b d \sqrt {b \sec (c+d x)}} \]

[Out]

sin(d*x+c)*sec(d*x+c)^(1/2)/b/d/(b*sec(d*x+c))^(1/2)-1/3*sin(d*x+c)^3*sec(d*x+c)^(1/2)/b/d/(b*sec(d*x+c))^(1/2
)

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Rubi [A]  time = 0.02, antiderivative size = 76, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {18, 2633} \[ \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{b d \sqrt {b \sec (c+d x)}}-\frac {\sin ^3(c+d x) \sqrt {\sec (c+d x)}}{3 b d \sqrt {b \sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(3/2)),x]

[Out]

(Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b*d*Sqrt[b*Sec[c + d*x]]) - (Sqrt[Sec[c + d*x]]*Sin[c + d*x]^3)/(3*b*d*Sqrt
[b*Sec[c + d*x]])

Rule 18

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[(a^(m - 1/2)*b^(n + 1/2)*Sqrt[a*v])/Sqrt[b*v]
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && ILtQ[n - 1/2, 0] && IntegerQ[m + n]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{3/2}} \, dx &=\frac {\sqrt {\sec (c+d x)} \int \cos ^3(c+d x) \, dx}{b \sqrt {b \sec (c+d x)}}\\ &=-\frac {\sqrt {\sec (c+d x)} \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{b d \sqrt {b \sec (c+d x)}}\\ &=\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{b d \sqrt {b \sec (c+d x)}}-\frac {\sqrt {\sec (c+d x)} \sin ^3(c+d x)}{3 b d \sqrt {b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.10, size = 45, normalized size = 0.59 \[ \frac {\sin (c+d x) (\cos (2 (c+d x))+5) \sec ^{\frac {3}{2}}(c+d x)}{6 d (b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(3/2)),x]

[Out]

((5 + Cos[2*(c + d*x)])*Sec[c + d*x]^(3/2)*Sin[c + d*x])/(6*d*(b*Sec[c + d*x])^(3/2))

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fricas [A]  time = 0.71, size = 51, normalized size = 0.67 \[ \frac {{\left (\cos \left (d x + c\right )^{3} + 2 \, \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, b^{2} d \sqrt {\cos \left (d x + c\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/3*(cos(d*x + c)^3 + 2*cos(d*x + c))*sqrt(b/cos(d*x + c))*sin(d*x + c)/(b^2*d*sqrt(cos(d*x + c)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}} \sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c))^(3/2)*sec(d*x + c)^(3/2)), x)

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maple [A]  time = 1.02, size = 52, normalized size = 0.68 \[ \frac {\sin \left (d x +c \right ) \left (2+\cos ^{2}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3} \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3/d*sin(d*x+c)*(2+cos(d*x+c)^2)/cos(d*x+c)^3/(1/cos(d*x+c))^(3/2)/(b/cos(d*x+c))^(3/2)

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maxima [A]  time = 0.72, size = 42, normalized size = 0.55 \[ \frac {\sin \left (3 \, d x + 3 \, c\right ) + 9 \, \sin \left (\frac {1}{3} \, \arctan \left (\sin \left (3 \, d x + 3 \, c\right ), \cos \left (3 \, d x + 3 \, c\right )\right )\right )}{12 \, b^{\frac {3}{2}} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/12*(sin(3*d*x + 3*c) + 9*sin(1/3*arctan2(sin(3*d*x + 3*c), cos(3*d*x + 3*c))))/(b^(3/2)*d)

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mupad [B]  time = 0.33, size = 48, normalized size = 0.63 \[ \frac {\left (9\,\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{12\,b^2\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((9*sin(c + d*x) + sin(3*c + 3*d*x))*(b/cos(c + d*x))^(1/2))/(12*b^2*d*(1/cos(c + d*x))^(1/2))

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sympy [A]  time = 121.59, size = 65, normalized size = 0.86 \[ \begin {cases} \frac {2 \tan ^{3}{\left (c + d x \right )}}{3 b^{\frac {3}{2}} d \sec ^{3}{\left (c + d x \right )}} + \frac {\tan {\left (c + d x \right )}}{b^{\frac {3}{2}} d \sec ^{3}{\left (c + d x \right )}} & \text {for}\: d \neq 0 \\\frac {x}{\left (b \sec {\relax (c )}\right )^{\frac {3}{2}} \sec ^{\frac {3}{2}}{\relax (c )}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)**(3/2)/(b*sec(d*x+c))**(3/2),x)

[Out]

Piecewise((2*tan(c + d*x)**3/(3*b**(3/2)*d*sec(c + d*x)**3) + tan(c + d*x)/(b**(3/2)*d*sec(c + d*x)**3), Ne(d,
0)), (x/((b*sec(c))**(3/2)*sec(c)**(3/2)), True))

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