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

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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 = {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 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 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]

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*}

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Mathematica [A]  time = 0.01, 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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fricas [A]  time = 0.62, size = 31, normalized size = 0.89 \[ \frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{b^{2} d \cos \left (d x + c\right )^{\frac {3}{2}}} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}} \sqrt {\cos \left (d x + c\right )}}\,{d x} \]

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)

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maple [A]  time = 0.12, size = 29, normalized size = 0.83 \[ \frac {\left (\sqrt {\cos }\left (d x +c \right )\right ) \sin \left (d x +c \right )}{d \left (b \cos \left (d x +c \right )\right )^{\frac {3}{2}}} \]

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 = 1.00, size = 67, normalized size = 1.91 \[ \frac {2 \, \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{{\left (b^{2} \cos \left (2 \, d x + 2 \, c\right )^{2} + b^{2} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b^{2} \cos \left (2 \, d x + 2 \, c\right ) + b^{2}\right )} d} \]

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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mupad [B]  time = 0.51, size = 62, normalized size = 1.77 \[ \frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+1{}\mathrm {i}\right )}{b^2\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (b \cos {\left (c + d x \right )}\right )^{\frac {3}{2}} \sqrt {\cos {\left (c + d x \right )}}}\, dx \]

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]

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

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