3.2.46 \(\int \frac {(b \sec (c+d x))^{3/2}}{\sqrt {\sec (c+d x)}} \, dx\) [146]

Optimal. Leaf size=34 \[ \frac {b \tanh ^{-1}(\sin (c+d x)) \sqrt {b \sec (c+d x)}}{d \sqrt {\sec (c+d x)}} \]

[Out]

b*arctanh(sin(d*x+c))*(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 = 34, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {17, 3855} \begin {gather*} \frac {b \sqrt {b \sec (c+d x)} \tanh ^{-1}(\sin (c+d x))}{d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*ArcTanh[Sin[c + d*x]]*Sqrt[b*Sec[c + d*x]])/(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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 33, normalized size = 0.97 \begin {gather*} \frac {\tanh ^{-1}(\sin (c+d x)) (b \sec (c+d x))^{3/2}}{d \sec ^{\frac {3}{2}}(c+d x)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 33.52, size = 52, normalized size = 1.53

method	result	size
default	\(-\frac {2 \arctanh \left (\frac {\cos \left (d x +c \right )-1}{\sin \left (d x +c \right )}\right ) \left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \cos \left (d x +c \right )}{d \sqrt {\frac {1}{\cos \left (d x +c \right )}}}\)	\(52\)
risch	\(-\frac {b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}+\frac {b \sqrt {\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{\sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}}{{\mathrm e}^{2 i \left (d x +c \right )}+1}}\, d}\)	\(141\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 68 vs. \(2 (30) = 60\).
time = 0.62, size = 68, normalized size = 2.00 \begin {gather*} \frac {{\left (b \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) + 1\right ) - b \log \left (\cos \left (d x + c\right )^{2} + \sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )\right )} \sqrt {b}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(b*log(cos(d*x + c)^2 + sin(d*x + c)^2 + 2*sin(d*x + c) + 1) - b*log(cos(d*x + c)^2 + sin(d*x + c)^2 - 2*s
in(d*x + c) + 1))*sqrt(b)/d

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Fricas [A]
time = 3.01, size = 112, normalized size = 3.29 \begin {gather*} \left [\frac {b^{\frac {3}{2}} \log \left (-\frac {b \cos \left (d x + c\right )^{2} - 2 \, \sqrt {b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 2 \, b}{\cos \left (d x + c\right )^{2}}\right )}{2 \, d}, -\frac {\sqrt {-b} b \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )}{b}\right )}{d}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}{\sqrt {\sec {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((b*sec(c + d*x))**(3/2)/sqrt(sec(c + d*x)), x)

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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))^(3/2)/sec(d*x+c)^(1/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\left (\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b/cos(c + d*x))^(3/2)/(1/cos(c + d*x))^(1/2),x)

[Out]

int((b/cos(c + d*x))^(3/2)/(1/cos(c + d*x))^(1/2), x)

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