3.2.59 \(\int \frac {(b \sec (c+d x))^{5/2}}{\sec ^{\frac {9}{2}}(c+d x)} \, dx\) [159]

Optimal. Leaf size=69 \[ \frac {b^2 x \sqrt {b \sec (c+d x)}}{2 \sqrt {\sec (c+d x)}}+\frac {b^2 \sqrt {b \sec (c+d x)} \sin (c+d x)}{2 d \sec ^{\frac {3}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {17, 2715, 8} \begin {gather*} \frac {b^2 x \sqrt {b \sec (c+d x)}}{2 \sqrt {\sec (c+d x)}}+\frac {b^2 \sin (c+d x) \sqrt {b \sec (c+d x)}}{2 d \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*Sec[c + d*x])^(5/2)/Sec[c + d*x]^(9/2),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2715

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

Rubi steps

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

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Mathematica [A]
time = 0.11, size = 45, normalized size = 0.65 \begin {gather*} \frac {(b \sec (c+d x))^{5/2} (2 (c+d x)+\sin (2 (c+d x)))}{4 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]^(9/2),x]

[Out]

((b*Sec[c + d*x])^(5/2)*(2*(c + d*x) + Sin[2*(c + d*x)]))/(4*d*Sec[c + d*x]^(5/2))

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Maple [A]
time = 35.36, size = 54, normalized size = 0.78

method	result	size
default	\(\frac {\left (\sin \left (d x +c \right ) \cos \left (d x +c \right )+d x +c \right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}}}{2 d \cos \left (d x +c \right )^{2} \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {9}{2}}}\)	\(54\)
risch	\(\frac {b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, x}{2 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{2 i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, {\mathrm e}^{-2 i \left (d x +c \right )}}{8 \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(197\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*sec(d*x+c))^(5/2)/sec(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

1/2/d*(sin(d*x+c)*cos(d*x+c)+d*x+c)*(b/cos(d*x+c))^(5/2)/cos(d*x+c)^2/(1/cos(d*x+c))^(9/2)

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Maxima [A]
time = 0.62, size = 32, normalized size = 0.46 \begin {gather*} \frac {{\left (2 \, {\left (d x + c\right )} b^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )\right )} \sqrt {b}}{4 \, 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)^(9/2),x, algorithm="maxima")

[Out]

1/4*(2*(d*x + c)*b^2 + b^2*sin(2*d*x + 2*c))*sqrt(b)/d

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Fricas [A]
time = 3.30, size = 167, normalized size = 2.42 \begin {gather*} \left [\frac {2 \, b^{2} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + \sqrt {-b} b^{2} \log \left (-2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + 2 \, b \cos \left (d x + c\right )^{2} - b\right )}{4 \, d}, \frac {b^{2} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + b^{\frac {5}{2}} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {b} \sqrt {\cos \left (d x + c\right )}}\right )}{2 \, d}\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)^(9/2),x, algorithm="fricas")

[Out]

[1/4*(2*b^2*sqrt(b/cos(d*x + c))*cos(d*x + c)^(3/2)*sin(d*x + c) + sqrt(-b)*b^2*log(-2*sqrt(-b)*sqrt(b/cos(d*x
+ c))*cos(d*x + c)^(3/2)*sin(d*x + c) + 2*b*cos(d*x + c)^2 - b))/d, 1/2*(b^2*sqrt(b/cos(d*x + c))*cos(d*x + c
)^(3/2)*sin(d*x + c) + b^(5/2)*arctan(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(cos(d*x + c)))))/d]

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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)**(9/2),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)^(9/2),x, algorithm="giac")

[Out]

integrate((b*sec(d*x + c))^(5/2)/sec(d*x + c)^(9/2), x)

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Mupad [B]
time = 0.37, size = 44, normalized size = 0.64 \begin {gather*} \frac {b^2\,\left (\sin \left (2\,c+2\,d\,x\right )+2\,d\,x\right )\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{4\,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))^(9/2),x)

[Out]

(b^2*(sin(2*c + 2*d*x) + 2*d*x)*(b/cos(c + d*x))^(1/2))/(4*d*(1/cos(c + d*x))^(1/2))

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