3.264 \(\int \frac {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=37 \[ \frac {2 (c \sin (a+b x))^{3/2}}{3 b c d (d \cos (a+b x))^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.05, antiderivative size = 37, 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} \[ \frac {2 (c \sin (a+b x))^{3/2}}{3 b c d (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(5/2),x]

[Out]

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

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 {\sqrt {c \sin (a+b x)}}{(d \cos (a+b x))^{5/2}} \, dx &=\frac {2 (c \sin (a+b x))^{3/2}}{3 b c d (d \cos (a+b x))^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 37, normalized size = 1.00 \[ \frac {2 (c \sin (a+b x))^{3/2}}{3 b c d (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*Sin[a + b*x]]/(d*Cos[a + b*x])^(5/2),x]

[Out]

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

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fricas [A]  time = 0.46, size = 42, normalized size = 1.14 \[ \frac {2 \, \sqrt {d \cos \left (b x + a\right )} \sqrt {c \sin \left (b x + a\right )} \sin \left (b x + a\right )}{3 \, b d^{3} \cos \left (b x + a\right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(5/2),x, algorithm="giac")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(5/2), x)

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maple [A]  time = 0.14, size = 38, normalized size = 1.03 \[ \frac {2 \sin \left (b x +a \right ) \cos \left (b x +a \right ) \sqrt {c \sin \left (b x +a \right )}}{3 b \left (d \cos \left (b x +a \right )\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(5/2),x)

[Out]

2/3/b*sin(b*x+a)*cos(b*x+a)*(c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(5/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {c \sin \left (b x + a\right )}}{\left (d \cos \left (b x + a\right )\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))^(1/2)/(d*cos(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c*sin(b*x + a))/(d*cos(b*x + a))^(5/2), x)

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mupad [B]  time = 0.99, size = 50, normalized size = 1.35 \[ \frac {2\,\sin \left (2\,a+2\,b\,x\right )\,\sqrt {c\,\sin \left (a+b\,x\right )}}{3\,b\,d^2\,\left (\cos \left (2\,a+2\,b\,x\right )+1\right )\,\sqrt {d\,\cos \left (a+b\,x\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sin(a + b*x))^(1/2)/(d*cos(a + b*x))^(5/2),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sin(b*x+a))**(1/2)/(d*cos(b*x+a))**(5/2),x)

[Out]

Timed out

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