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

Optimal. Leaf size=93 \[ \frac{c^2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}-\frac{c \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}{b d} \]

[Out]

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

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Rubi [A]  time = 0.116628, antiderivative size = 93, 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 = {2568, 2573, 2641} \[ \frac{c^2 \sqrt{\sin (2 a+2 b x)} F\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{2 b \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}-\frac{c \sqrt{c \sin (a+b x)} \sqrt{d \cos (a+b x)}}{b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2568

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(a*(b*Cos[e
+ f*x])^(n + 1)*(a*Sin[e + f*x])^(m - 1))/(b*f*(m + n)), x] + Dist[(a^2*(m - 1))/(m + n), Int[(b*Cos[e + f*x])
^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && 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{(c \sin (a+b x))^{3/2}}{\sqrt{d \cos (a+b x)}} \, dx &=-\frac{c \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}{b d}+\frac{1}{2} c^2 \int \frac{1}{\sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}} \, dx\\ &=-\frac{c \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}{b d}+\frac{\left (c^2 \sqrt{\sin (2 a+2 b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx}{2 \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}\\ &=-\frac{c \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}{b d}+\frac{c^2 F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{\sin (2 a+2 b x)}}{2 b \sqrt{d \cos (a+b x)} \sqrt{c \sin (a+b x)}}\\ \end{align*}

Mathematica [C]  time = 0.0806252, size = 67, normalized size = 0.72 \[ \frac{2 \cos ^2(a+b x)^{3/4} \tan (a+b x) (c \sin (a+b x))^{3/2} \, _2F_1\left (\frac{3}{4},\frac{5}{4};\frac{9}{4};\sin ^2(a+b x)\right )}{5 b \sqrt{d \cos (a+b x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(Cos[a + b*x]^2)^(3/4)*Hypergeometric2F1[3/4, 5/4, 9/4, Sin[a + b*x]^2]*(c*Sin[a + b*x])^(3/2)*Tan[a + b*x]
)/(5*b*Sqrt[d*Cos[a + b*x]])

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Maple [A]  time = 0.118, size = 182, normalized size = 2. \begin{align*} -{\frac{\sqrt{2}}{2\,b \left ( -1+\cos \left ( bx+a \right ) \right ) \sin \left ( bx+a \right ) } \left ( \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 ) }}}\sqrt{{\frac{-1+\cos \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}}{\it EllipticF} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{\sqrt{2}}{2}} \right ) + \left ( \cos \left ( bx+a \right ) \right ) ^{2}\sqrt{2}-\cos \left ( bx+a \right ) \sqrt{2} \right ) \left ( c\sin \left ( bx+a \right ) \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{d\cos \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (c \sin \left (b x + a\right )\right )^{\frac{3}{2}}}{\sqrt{d \cos \left (b x + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((c*sin(b*x + a))^(3/2)/sqrt(d*cos(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 \sin \left (b x + a\right )}{d \cos \left (b x + a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(d*cos(b*x + a))*sqrt(c*sin(b*x + a))*c*sin(b*x + a)/(d*cos(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((c*sin(b*x+a))**(3/2)/(d*cos(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{\left (c \sin \left (b x + a\right )\right )^{\frac{3}{2}}}{\sqrt{d \cos \left (b x + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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