3.294 \(\int \frac{1}{(d \cos (a+b x))^{5/2} \sqrt{c \sin (a+b x)}} \, dx\)

Optimal. Leaf size=97 \[ \frac{2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{3 b d^2 \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}+\frac{2 \sqrt{c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}} \]

[Out]

(2*Sqrt[c*Sin[a + b*x]])/(3*b*c*d*(d*Cos[a + b*x])^(3/2)) + (2*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b
*x]])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*x]])

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Rubi [A]  time = 0.113712, antiderivative size = 97, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {2571, 2573, 2641} \[ \frac{2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{3 b d^2 \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}+\frac{2 \sqrt{c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((d*Cos[a + b*x])^(5/2)*Sqrt[c*Sin[a + b*x]]),x]

[Out]

(2*Sqrt[c*Sin[a + b*x]])/(3*b*c*d*(d*Cos[a + b*x])^(3/2)) + (2*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[Sin[2*a + 2*b
*x]])/(3*b*d^2*Sqrt[d*Cos[a + b*x]]*Sqrt[c*Sin[a + b*x]])

Rule 2571

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

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{1}{(d \cos (a+b x))^{5/2} \sqrt{c \sin (a+b x)}} \, dx &=\frac{2 \sqrt{c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}+\frac{2 \int \frac{1}{\sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}} \, dx}{3 d^2}\\ &=\frac{2 \sqrt{c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}+\frac{\left (2 \sqrt{\sin (2 a+2 b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx}{3 d^2 \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}\\ &=\frac{2 \sqrt{c \sin (a+b x)}}{3 b c d (d \cos (a+b x))^{3/2}}+\frac{2 F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{\sin (2 a+2 b x)}}{3 b d^2 \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.106237, size = 65, normalized size = 0.67 \[ \frac{2 \cos ^2(a+b x)^{3/4} \sqrt{c \sin (a+b x)} \, _2F_1\left (\frac{1}{4},\frac{7}{4};\frac{5}{4};\sin ^2(a+b x)\right )}{b c d (d \cos (a+b x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((d*Cos[a + b*x])^(5/2)*Sqrt[c*Sin[a + b*x]]),x]

[Out]

(2*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[1/4, 7/4, 5/4, Sin[a + b*x]^2]*Sqrt[c*Sin[a + b*x]])/(b*c*d*(d*Cos
[a + b*x])^(3/2))

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Maple [A]  time = 0.112, size = 184, normalized size = 1.9 \begin{align*} -{\frac{\sqrt{2}\sin \left ( bx+a \right ) \cos \left ( bx+a \right ) }{3\,b \left ( -1+\cos \left ( bx+a \right ) \right ) } \left ( 2\,{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},1/2\,\sqrt{2} \right ) \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{-1+\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sqrt{{\frac{-1+\cos \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}\sin \left ( bx+a \right ) \cos \left ( bx+a \right ) -\cos \left ( bx+a \right ) \sqrt{2}+\sqrt{2} \right ) \left ( d\cos \left ( bx+a \right ) \right ) ^{-{\frac{5}{2}}}{\frac{1}{\sqrt{c\sin \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \cos \left (b x + a\right )\right )^{\frac{5}{2}} \sqrt{c \sin \left (b x + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \cos \left (b x + a\right )} \sqrt{c \sin \left (b x + a\right )}}{c d^{3} \cos \left (b x + a\right )^{3} \sin \left (b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (d \cos \left (b x + a\right )\right )^{\frac{5}{2}} \sqrt{c \sin \left (b x + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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