3.2.60 \(\int \frac {(b \sec (c+d x))^{5/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx\) [160]

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

[Out]

b^2*sin(d*x+c)*(b*sec(d*x+c))^(1/2)/d/sec(d*x+c)^(1/2)-1/3*b^2*sin(d*x+c)^3*(b*sec(d*x+c))^(1/2)/d/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 = {17, 2713} \begin {gather*} \frac {b^2 \sin (c+d x) \sqrt {b \sec (c+d x)}}{d \sqrt {\sec (c+d x)}}-\frac {b^2 \sin ^3(c+d x) \sqrt {b \sec (c+d x)}}{3 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(11/2),x]

[Out]

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

Rule 17

Int[(u_.)*((a_.)*(v_))^(m_)*((b_.)*(v_))^(n_), x_Symbol] :> Dist[a^(m + 1/2)*b^(n - 1/2)*(Sqrt[b*v]/Sqrt[a*v])
, Int[u*v^(m + n), x], x] /; FreeQ[{a, b, m}, x] &&  !IntegerQ[m] && IGtQ[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 {(b \sec (c+d x))^{5/2}}{\sec ^{\frac {11}{2}}(c+d x)} \, dx &=\frac {\left (b^2 \sqrt {b \sec (c+d x)}\right ) \int \cos ^3(c+d x) \, dx}{\sqrt {\sec (c+d x)}}\\ &=-\frac {\left (b^2 \sqrt {b \sec (c+d x)}\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{d \sqrt {\sec (c+d x)}}\\ &=\frac {b^2 \sqrt {b \sec (c+d x)} \sin (c+d x)}{d \sqrt {\sec (c+d x)}}-\frac {b^2 \sqrt {b \sec (c+d x)} \sin ^3(c+d x)}{3 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.18, size = 45, normalized size = 0.59 \begin {gather*} \frac {(5+\cos (2 (c+d x))) (b \sec (c+d x))^{5/2} \sin (c+d x)}{6 d \sec ^{\frac {5}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(b*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(11/2),x]

[Out]

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

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Maple [A]
time = 38.96, size = 52, normalized size = 0.68

method	result	size
default	\(\frac {\left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{3 d \cos \left (d x +c \right )^{3} \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}}}\)	\(52\)
risch	\(-\frac {3 i b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {3 i b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{-i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \sin \left (3 d x +3 c \right )}{12 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(208\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^(5/2)/sec(d*x+c)^(11/2),x,method=_RETURNVERBOSE)

[Out]

1/3/d*(cos(d*x+c)^2+2)*sin(d*x+c)*(b/cos(d*x+c))^(5/2)/cos(d*x+c)^3/(1/cos(d*x+c))^(11/2)

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Maxima [A]
time = 0.71, size = 49, normalized size = 0.64 \begin {gather*} \frac {{\left (b^{2} \sin \left (3 \, d x + 3 \, c\right ) + 9 \, b^{2} \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 )\right )} \sqrt {b}}{12 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(5/2)/sec(d*x+c)^(11/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.09, size = 55, normalized size = 0.72 \begin {gather*} \frac {{\left (b^{2} \cos \left (d x + c\right )^{3} + 2 \, b^{2} \cos \left (d x + c\right )\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{3 \, d \sqrt {\cos \left (d x + c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))^(5/2)/sec(d*x+c)^(11/2),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*sec(d*x+c))**(5/2)/sec(d*x+c)**(11/2),x)

[Out]

Timed out

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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((b*sec(d*x+c))^(5/2)/sec(d*x+c)^(11/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(5/2)/sec(d*x + c)^(11/2), x)

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Mupad [B]
time = 0.50, size = 48, normalized size = 0.63 \begin {gather*} \frac {b^2\,\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\,d\,\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(c + d*x))^(5/2)/(1/cos(c + d*x))^(11/2),x)

[Out]

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

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