3.2.82 \(\int \frac {1}{\sqrt {\sec (c+d x)} (b \sec (c+d x))^{5/2}} \, dx\) [182]

Optimal. Leaf size=76 \[ \frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 d \sqrt {b \sec (c+d x)}}-\frac {\sqrt {\sec (c+d x)} \sin ^3(c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}} \]

[Out]

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

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Rubi [A]
time = 0.01, 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, 2713} \begin {gather*} \frac {\sin (c+d x) \sqrt {\sec (c+d x)}}{b^2 d \sqrt {b \sec (c+d x)}}-\frac {\sin ^3(c+d x) \sqrt {\sec (c+d x)}}{3 b^2 d \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[Sec[c + d*x]]*(b*Sec[c + d*x])^(5/2)),x]

[Out]

(Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(b^2*d*Sqrt[b*Sec[c + d*x]]) - (Sqrt[Sec[c + d*x]]*Sin[c + d*x]^3)/(3*b^2*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 2713

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}{\sqrt {\sec (c+d x)} (b \sec (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\sec (c+d x)} \int \cos ^3(c+d x) \, dx}{b^2 \sqrt {b \sec (c+d x)}}\\ &=-\frac {\sqrt {\sec (c+d x)} \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{b^2 d \sqrt {b \sec (c+d x)}}\\ &=\frac {\sqrt {\sec (c+d x)} \sin (c+d x)}{b^2 d \sqrt {b \sec (c+d x)}}-\frac {\sqrt {\sec (c+d x)} \sin ^3(c+d x)}{3 b^2 d \sqrt {b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.07, size = 48, normalized size = 0.63 \begin {gather*} \frac {(5+\cos (2 (c+d x))) \sqrt {\sec (c+d x)} \sin (c+d x)}{6 b^2 d \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[Sec[c + d*x]]*(b*Sec[c + d*x])^(5/2)),x]

[Out]

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

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Maple [A]
time = 34.75, size = 52, normalized size = 0.68

method	result	size
default	\(\frac {\sin \left (d x +c \right ) \left (2+\cos ^{2}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{3} \sqrt {\frac {1}{\cos \left (d x +c \right )}}\, \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}\)	\(52\)
risch	\(-\frac {i {\mathrm e}^{4 i \left (d x +c \right )}}{24 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}-\frac {3 i {\mathrm e}^{2 i \left (d x +c \right )}}{8 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {3 i}{8 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{24 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) d}\)	\(321\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/3/d*sin(d*x+c)*(cos(d*x+c)^2+2)/cos(d*x+c)^3/(1/cos(d*x+c))^(1/2)/(b/cos(d*x+c))^(5/2)

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Maxima [A]
time = 0.61, size = 42, normalized size = 0.55 \begin {gather*} \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 {5}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),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^(5/2)*d)

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Fricas [A]
time = 3.18, size = 51, normalized size = 0.67 \begin {gather*} \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^{3} d \sqrt {\cos \left (d x + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),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^3*d*sqrt(cos(d*x + c)))

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Sympy [A]
time = 41.97, size = 82, normalized size = 1.08 \begin {gather*} \begin {cases} \frac {2 \tan ^{3}{\left (c + d x \right )}}{3 d \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sqrt {\sec {\left (c + d x \right )}}} + \frac {\tan {\left (c + d x \right )}}{d \left (b \sec {\left (c + d x \right )}\right )^{\frac {5}{2}} \sqrt {\sec {\left (c + d x \right )}}} & \text {for}\: d \neq 0 \\\frac {x}{\left (b \sec {\left (c \right )}\right )^{\frac {5}{2}} \sqrt {\sec {\left (c \right )}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)**(1/2)/(b*sec(d*x+c))**(5/2),x)

[Out]

Piecewise((2*tan(c + d*x)**3/(3*d*(b*sec(c + d*x))**(5/2)*sqrt(sec(c + d*x))) + tan(c + d*x)/(d*(b*sec(c + d*x
))**(5/2)*sqrt(sec(c + d*x))), Ne(d, 0)), (x/((b*sec(c))**(5/2)*sqrt(sec(c))), True))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(1/2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((b*sec(d*x + c))^(5/2)*sqrt(sec(d*x + c))), x)

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Mupad [B]
time = 0.42, size = 48, normalized size = 0.63 \begin {gather*} \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^3\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((b/cos(c + d*x))^(5/2)*(1/cos(c + d*x))^(1/2)),x)

[Out]

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

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