3.296 \(\int \frac{\sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}} \, dx\)

Optimal. Leaf size=280 \[ \frac{\sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{d} \sqrt{c \sin (a+b x)}}\right )}{\sqrt{2} b \sqrt{c}}-\frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{d} \sqrt{c \sin (a+b x)}}+1\right )}{\sqrt{2} b \sqrt{c}}-\frac{\sqrt{d} \log \left (-\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}+\sqrt{d} \cot (a+b x)+\sqrt{d}\right )}{2 \sqrt{2} b \sqrt{c}}+\frac{\sqrt{d} \log \left (\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}+\sqrt{d} \cot (a+b x)+\sqrt{d}\right )}{2 \sqrt{2} b \sqrt{c}} \]

[Out]

(Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/(Sqrt[d]*Sqrt[c*Sin[a + b*x]])])/(Sqrt[2]*b*Sqrt[c]
) - (Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/(Sqrt[d]*Sqrt[c*Sin[a + b*x]])])/(Sqrt[2]*b*Sqr
t[c]) - (Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Cot[a + b*x] - (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/Sqrt[c*Sin[a + b*
x]]])/(2*Sqrt[2]*b*Sqrt[c]) + (Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Cot[a + b*x] + (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*
x]])/Sqrt[c*Sin[a + b*x]]])/(2*Sqrt[2]*b*Sqrt[c])

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Rubi [A]  time = 0.176786, antiderivative size = 280, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2575, 297, 1162, 617, 204, 1165, 628} \[ \frac{\sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{d} \sqrt{c \sin (a+b x)}}\right )}{\sqrt{2} b \sqrt{c}}-\frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{d} \sqrt{c \sin (a+b x)}}+1\right )}{\sqrt{2} b \sqrt{c}}-\frac{\sqrt{d} \log \left (-\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}+\sqrt{d} \cot (a+b x)+\sqrt{d}\right )}{2 \sqrt{2} b \sqrt{c}}+\frac{\sqrt{d} \log \left (\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}+\sqrt{d} \cot (a+b x)+\sqrt{d}\right )}{2 \sqrt{2} b \sqrt{c}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[d*Cos[a + b*x]]/Sqrt[c*Sin[a + b*x]],x]

[Out]

(Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/(Sqrt[d]*Sqrt[c*Sin[a + b*x]])])/(Sqrt[2]*b*Sqrt[c]
) - (Sqrt[d]*ArcTan[1 + (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/(Sqrt[d]*Sqrt[c*Sin[a + b*x]])])/(Sqrt[2]*b*Sqr
t[c]) - (Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Cot[a + b*x] - (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*x]])/Sqrt[c*Sin[a + b*
x]]])/(2*Sqrt[2]*b*Sqrt[c]) + (Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Cot[a + b*x] + (Sqrt[2]*Sqrt[c]*Sqrt[d*Cos[a + b*
x]])/Sqrt[c*Sin[a + b*x]]])/(2*Sqrt[2]*b*Sqrt[c])

Rule 2575

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> With[{k = Denomina
tor[m]}, -Dist[(k*a*b)/f, Subst[Int[x^(k*(m + 1) - 1)/(a^2 + b^2*x^(2*k)), x], x, (a*Cos[e + f*x])^(1/k)/(b*Si
n[e + f*x])^(1/k)], x]] /; FreeQ[{a, b, e, f}, x] && EqQ[m + n, 0] && GtQ[m, 0] && LtQ[m, 1]

Rule 297

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]},
Dist[1/(2*s), Int[(r + s*x^2)/(a + b*x^4), x], x] - Dist[1/(2*s), Int[(r - s*x^2)/(a + b*x^4), x], x]] /; Free
Q[{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ,
b]]))

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rubi steps

\begin{align*} \int \frac{\sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}} \, dx &=-\frac{(2 c d) \operatorname{Subst}\left (\int \frac{x^2}{d^2+c^2 x^4} \, dx,x,\frac{\sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{b}\\ &=\frac{d \operatorname{Subst}\left (\int \frac{d-c x^2}{d^2+c^2 x^4} \, dx,x,\frac{\sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{b}-\frac{d \operatorname{Subst}\left (\int \frac{d+c x^2}{d^2+c^2 x^4} \, dx,x,\frac{\sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{b}\\ &=-\frac{\sqrt{d} \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{d}}{\sqrt{c}}+2 x}{-\frac{d}{c}-\frac{\sqrt{2} \sqrt{d} x}{\sqrt{c}}-x^2} \, dx,x,\frac{\sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{2 \sqrt{2} b \sqrt{c}}-\frac{\sqrt{d} \operatorname{Subst}\left (\int \frac{\frac{\sqrt{2} \sqrt{d}}{\sqrt{c}}-2 x}{-\frac{d}{c}+\frac{\sqrt{2} \sqrt{d} x}{\sqrt{c}}-x^2} \, dx,x,\frac{\sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{2 \sqrt{2} b \sqrt{c}}-\frac{d \operatorname{Subst}\left (\int \frac{1}{\frac{d}{c}-\frac{\sqrt{2} \sqrt{d} x}{\sqrt{c}}+x^2} \, dx,x,\frac{\sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{2 b c}-\frac{d \operatorname{Subst}\left (\int \frac{1}{\frac{d}{c}+\frac{\sqrt{2} \sqrt{d} x}{\sqrt{c}}+x^2} \, dx,x,\frac{\sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{2 b c}\\ &=-\frac{\sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \cot (a+b x)-\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{2 \sqrt{2} b \sqrt{c}}+\frac{\sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \cot (a+b x)+\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{2 \sqrt{2} b \sqrt{c}}-\frac{\sqrt{d} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{d} \sqrt{c \sin (a+b x)}}\right )}{\sqrt{2} b \sqrt{c}}+\frac{\sqrt{d} \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{d} \sqrt{c \sin (a+b x)}}\right )}{\sqrt{2} b \sqrt{c}}\\ &=\frac{\sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{d} \sqrt{c \sin (a+b x)}}\right )}{\sqrt{2} b \sqrt{c}}-\frac{\sqrt{d} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{d} \sqrt{c \sin (a+b x)}}\right )}{\sqrt{2} b \sqrt{c}}-\frac{\sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \cot (a+b x)-\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{2 \sqrt{2} b \sqrt{c}}+\frac{\sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \cot (a+b x)+\frac{\sqrt{2} \sqrt{c} \sqrt{d \cos (a+b x)}}{\sqrt{c \sin (a+b x)}}\right )}{2 \sqrt{2} b \sqrt{c}}\\ \end{align*}

Mathematica [C]  time = 0.0609574, size = 65, normalized size = 0.23 \[ \frac{2 \sqrt [4]{\cos ^2(a+b x)} \tan (a+b x) \sqrt{d \cos (a+b x)} \, _2F_1\left (\frac{1}{4},\frac{1}{4};\frac{5}{4};\sin ^2(a+b x)\right )}{b \sqrt{c \sin (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[d*Cos[a + b*x]]/Sqrt[c*Sin[a + b*x]],x]

[Out]

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

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Maple [C]  time = 0.096, size = 312, normalized size = 1.1 \begin{align*} -{\frac{\sqrt{2} \left ( \sin \left ( bx+a \right ) \right ) ^{2}}{2\,b\cos \left ( bx+a \right ) \left ( -1+\cos \left ( bx+a \right ) \right ) }\sqrt{d\cos \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 ) }}} \left ( i{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -i{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +{\it EllipticPi} \left ( \sqrt{{\frac{1-\cos \left ( bx+a \right ) +\sin \left ( bx+a \right ) }{\sin \left ( bx+a \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -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 ) \right ){\frac{1}{\sqrt{c\sin \left ( bx+a \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b*2^(1/2)*(d*cos(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))/s
in(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*(I*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)
,1/2-1/2*I,1/2*2^(1/2))-I*EllipticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))+Ellip
ticPi(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))+EllipticPi(((1-cos(b*x+a)+sin(b*x+a)
)/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*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
)))*sin(b*x+a)^2/(c*sin(b*x+a))^(1/2)/cos(b*x+a)/(-1+cos(b*x+a))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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