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

Optimal. Leaf size=35 \[ \frac {2 \sqrt {c \sin (a+b x)}}{b c d \sqrt {d \cos (a+b x)}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 35, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.040, Rules used = {2563} \[ \frac {2 \sqrt {c \sin (a+b x)}}{b c d \sqrt {d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2563

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Simp[((a*Sin[e +
f*x])^(m + 1)*(b*Cos[e + f*x])^(n + 1))/(a*b*f*(m + 1)), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2,
0] && NeQ[m, -1]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 36, normalized size = 1.03 \[ \frac {\sin (2 (a+b x))}{b \sqrt {c \sin (a+b x)} (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.51, size = 39, normalized size = 1.11 \[ \frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )}}{b c d^{2} \cos \left (b x + a\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.10, size = 38, normalized size = 1.09 \[ \frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right )}{b \left (d \cos \left (b x +a \right )\right )^{\frac {3}{2}} \sqrt {c \sin \left (b x +a \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.81, size = 31, normalized size = 0.89 \[ \frac {2\,\sqrt {c\,\sin \left (a+b\,x\right )}}{b\,c\,d\,\sqrt {d\,\cos \left (a+b\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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