3.147 \(\int \frac {(b \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=25 \[ \frac {b x \sqrt {b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.00, antiderivative size = 25, 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, 8} \[ \frac {b x \sqrt {b \sec (c+d x)}}{\sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 24, normalized size = 0.96 \[ \frac {x (b \sec (c+d x))^{3/2}}{\sec ^{\frac {3}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.79, size = 99, normalized size = 3.96 \[ \left [\frac {\sqrt {-b} b \log \left (-2 \, \sqrt {-b} \sqrt {\frac {b}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )^{\frac {3}{2}} \sin \left (d x + c\right ) + 2 \, b \cos \left (d x + c\right )^{2} - b\right )}{2 \, d}, \frac {b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {\frac {b}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{\sqrt {b} \sqrt {\cos \left (d x + c\right )}}\right )}{d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \sec \left (d x + c\right )\right )^{\frac {3}{2}}}{\sec \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.98, size = 32, normalized size = 1.28 \[ \frac {\left (\frac {b}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (d x +c \right )}{d \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.77, size = 26, normalized size = 1.04 \[ \frac {2 \, b^{\frac {3}{2}} \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.20, size = 25, normalized size = 1.00 \[ \frac {b\,x\,\sqrt {\frac {b}{\cos \left (c+d\,x\right )}}}{\sqrt {\frac {1}{\cos \left (c+d\,x\right )}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 18.08, size = 5, normalized size = 0.20 \[ b^{\frac {3}{2}} x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b**(3/2)*x

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