3.2.55 \(\int \frac {(b \sec (c+d x))^{5/2}}{\sqrt {\sec (c+d x)}} \, dx\) [155]

Optimal. Leaf size=35 \[ \frac {b^2 \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)} \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 35, 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, 3852, 8} \begin {gather*} \frac {b^2 \sin (c+d x) \sqrt {\sec (c+d x)} \sqrt {b \sec (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 3852

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

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 32, normalized size = 0.91 \begin {gather*} \frac {(b \sec (c+d x))^{5/2} \sin (c+d x)}{d \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 35.53, size = 39, normalized size = 1.11

method	result	size
default	\(\frac {\left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {5}{2}} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{d \sqrt {\frac {1}{\cos \left (d x +c \right )}}}\)	\(39\)
risch	\(\frac {2 i b^{2} \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}}{\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.62, size = 54, normalized size = 1.54 \begin {gather*} \frac {2 \, b^{\frac {5}{2}} \sin \left (2 \, d x + 2 \, c\right )}{{\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} 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)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 3.30, size = 33, normalized size = 0.94 \begin {gather*} \frac {b^{2} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{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)^(1/2),x, algorithm="fricas")

[Out]

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

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3005 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((b*sec(d*x+c))^(5/2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.72, size = 66, normalized size = 1.89 \begin {gather*} \frac {b^2\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+1{}\mathrm {i}\right )}{d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )\,\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))^(1/2),x)

[Out]

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

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