3.188 \(\int \frac{1}{\sqrt{\cos (c+d x)} (b \cos (c+d x))^{3/2}} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.01258, antiderivative size = 35, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {18, 3767, 8} \[ \frac{\sin (c+d x)}{b d \sqrt{\cos (c+d x)} \sqrt{b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

Sin[c + d*x]/(b*d*Sqrt[Cos[c + d*x]]*Sqrt[b*Cos[c + d*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 3767

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]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.0150371, size = 32, normalized size = 0.91 \[ \frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{d (b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.233, size = 29, normalized size = 0.8 \begin{align*}{\frac{\sin \left ( dx+c \right ) }{d}\sqrt{\cos \left ( dx+c \right ) } \left ( b\cos \left ( dx+c \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.76311, size = 90, normalized size = 2.57 \begin{align*} \frac{2 \, \sqrt{b} \sin \left (2 \, d x + 2 \, c\right )}{{\left (b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.85864, size = 84, normalized size = 2.4 \begin{align*} \frac{\sqrt{b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{2} d \cos \left (d x + c\right )^{\frac{3}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (b \cos \left (d x + c\right )\right )^{\frac{3}{2}} \sqrt{\cos \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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