3.2.83 \(\int \frac {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx\) [183]

Optimal. Leaf size=107 \[ \frac {3 x \sqrt {\sec (c+d x)}}{8 b^2 \sqrt {b \sec (c+d x)}}+\frac {\sin (c+d x)}{4 b^2 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)}}+\frac {3 \sin (c+d x)}{8 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}} \]

[Out]

1/4*sin(d*x+c)/b^2/d/sec(d*x+c)^(5/2)/(b*sec(d*x+c))^(1/2)+3/8*sin(d*x+c)/b^2/d/sec(d*x+c)^(1/2)/(b*sec(d*x+c)
)^(1/2)+3/8*x*sec(d*x+c)^(1/2)/b^2/(b*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {18, 2715, 8} \begin {gather*} \frac {3 x \sqrt {\sec (c+d x)}}{8 b^2 \sqrt {b \sec (c+d x)}}+\frac {\sin (c+d x)}{4 b^2 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)}}+\frac {3 \sin (c+d x)}{8 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(5/2)),x]

[Out]

(3*x*Sqrt[Sec[c + d*x]])/(8*b^2*Sqrt[b*Sec[c + d*x]]) + Sin[c + d*x]/(4*b^2*d*Sec[c + d*x]^(5/2)*Sqrt[b*Sec[c
+ d*x]]) + (3*Sin[c + d*x])/(8*b^2*d*Sqrt[Sec[c + d*x]]*Sqrt[b*Sec[c + d*x]])

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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 {1}{\sec ^{\frac {3}{2}}(c+d x) (b \sec (c+d x))^{5/2}} \, dx &=\frac {\sqrt {\sec (c+d x)} \int \cos ^4(c+d x) \, dx}{b^2 \sqrt {b \sec (c+d x)}}\\ &=\frac {\sin (c+d x)}{4 b^2 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)}}+\frac {\left (3 \sqrt {\sec (c+d x)}\right ) \int \cos ^2(c+d x) \, dx}{4 b^2 \sqrt {b \sec (c+d x)}}\\ &=\frac {\sin (c+d x)}{4 b^2 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)}}+\frac {3 \sin (c+d x)}{8 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}}+\frac {\left (3 \sqrt {\sec (c+d x)}\right ) \int 1 \, dx}{8 b^2 \sqrt {b \sec (c+d x)}}\\ &=\frac {3 x \sqrt {\sec (c+d x)}}{8 b^2 \sqrt {b \sec (c+d x)}}+\frac {\sin (c+d x)}{4 b^2 d \sec ^{\frac {5}{2}}(c+d x) \sqrt {b \sec (c+d x)}}+\frac {3 \sin (c+d x)}{8 b^2 d \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 58, normalized size = 0.54 \begin {gather*} \frac {\sqrt {\sec (c+d x)} (12 (c+d x)+8 \sin (2 (c+d x))+\sin (4 (c+d x)))}{32 b^2 d \sqrt {b \sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sec[c + d*x]^(3/2)*(b*Sec[c + d*x])^(5/2)),x]

[Out]

(Sqrt[Sec[c + d*x]]*(12*(c + d*x) + 8*Sin[2*(c + d*x)] + Sin[4*(c + d*x)]))/(32*b^2*d*Sqrt[b*Sec[c + d*x]])

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Maple [A]
time = 35.69, size = 74, normalized size = 0.69

method	result	size
default	\(\frac {2 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 \sin \left (d x +c \right ) \cos \left (d x +c \right )+3 d x +3 c}{8 d \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right )^{4}}\)	\(74\)
risch	\(\frac {3 \,{\mathrm e}^{i \left (d x +c \right )} x}{8 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}-\frac {i {\mathrm e}^{5 i \left (d x +c \right )}}{64 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )}}{8 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}-\frac {7 i \cos \left (3 d x +3 c \right )}{64 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {9 \sin \left (3 d x +3 c \right )}{64 b^{2} \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(408\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/8/d*(2*cos(d*x+c)^3*sin(d*x+c)+3*sin(d*x+c)*cos(d*x+c)+3*d*x+3*c)/(1/cos(d*x+c))^(3/2)/(b/cos(d*x+c))^(5/2)/
cos(d*x+c)^4

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Maxima [A]
time = 0.66, size = 49, normalized size = 0.46 \begin {gather*} \frac {12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (4 \, d x + 4 \, c\right ), \cos \left (4 \, d x + 4 \, c\right )\right )\right )}{32 \, b^{\frac {5}{2}} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

1/32*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(1/2*arctan2(sin(4*d*x + 4*c), cos(4*d*x + 4*c))))/(b^(5/2)*d)

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Fricas [A]
time = 3.62, size = 208, normalized size = 1.94 \begin {gather*} \left [\frac {\frac {2 \, {\left (2 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} - 3 \, \sqrt {-b} \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 )}{16 \, b^{3} d}, \frac {\frac {{\left (2 \, \cos \left (d x + c\right )^{4} + 3 \, \cos \left (d x + c\right )^{2}\right )} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}} + 3 \, \sqrt {b} \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 )}{8 \, b^{3} d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/sec(d*x+c)^(3/2)/(b*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[1/16*(2*(2*cos(d*x + c)^4 + 3*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) - 3*sqrt(-
b)*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))/(b^3*d), 1/8
*((2*cos(d*x + c)^4 + 3*cos(d*x + c)^2)*sqrt(b/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 3*sqrt(b)*arcta
n(sqrt(b/cos(d*x + c))*sin(d*x + c)/(sqrt(b)*sqrt(cos(d*x + c)))))/(b^3*d)]

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3006 deep

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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)^(3/2)/(b*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.46, size = 55, normalized size = 0.51 \begin {gather*} \frac {\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\left (8\,\sin \left (2\,c+2\,d\,x\right )+\sin \left (4\,c+4\,d\,x\right )+12\,d\,x\right )}{32\,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))^(3/2)),x)

[Out]

((b/cos(c + d*x))^(1/2)*(8*sin(2*c + 2*d*x) + sin(4*c + 4*d*x) + 12*d*x))/(32*b^3*d*(1/cos(c + d*x))^(1/2))

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