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

Optimal. Leaf size=32 \[ \frac {\sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]

[Out]

sin(d*x+c)/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 = 32, 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} \[ \frac {\sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [A]  time = 0.02, size = 32, normalized size = 1.00 \[ \frac {\sin (c+d x)}{d \sqrt {\cos (c+d x)} \sqrt {b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.66, size = 31, normalized size = 0.97 \[ \frac {\sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{b d \cos \left (d x + c\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.12, size = 29, normalized size = 0.91 \[ \frac {\sin \left (d x +c \right )}{d \sqrt {\cos \left (d x +c \right )}\, \sqrt {b \cos \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 1.59, size = 59, normalized size = 1.84 \[ \frac {2 \, \sqrt {b} \sin \left (2 \, d x + 2 \, c\right )}{{\left (b \cos \left (2 \, d x + 2 \, c\right )^{2} + b \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, b \cos \left (2 \, d x + 2 \, c\right ) + b\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.54, size = 62, normalized size = 1.94 \[ \frac {\sqrt {b\,\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}+\sin \left (2\,c+2\,d\,x\right )+1{}\mathrm {i}\right )}{b\,d\,\sqrt {\cos \left (c+d\,x\right )}\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((b*cos(c + d*x))^(1/2)*(cos(2*c + 2*d*x)*1i + sin(2*c + 2*d*x) + 1i))/(b*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 \[ \int \frac {1}{\sqrt {b \cos {\left (c + d x \right )}} \cos ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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