3.235 \(\int \frac{\sqrt{c \sec (a+b x)}}{\sqrt{d \csc (a+b x)}} \, dx\)

Optimal. Leaf size=270 \[ -\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (a+b x)}\right ) \sqrt{c \sec (a+b x)}}{\sqrt{2} b \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (a+b x)}+1\right ) \sqrt{c \sec (a+b x)}}{\sqrt{2} b \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}+\frac{\sqrt{c \sec (a+b x)} \log \left (\tan (a+b x)-\sqrt{2} \sqrt{\tan (a+b x)}+1\right )}{2 \sqrt{2} b \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{\sqrt{c \sec (a+b x)} \log \left (\tan (a+b x)+\sqrt{2} \sqrt{\tan (a+b x)}+1\right )}{2 \sqrt{2} b \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}} \]

[Out]

-((ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sqrt[c*Sec[a + b*x]])/(Sqrt[2]*b*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b
*x]])) + (ArcTan[1 + Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sqrt[c*Sec[a + b*x]])/(Sqrt[2]*b*Sqrt[d*Csc[a + b*x]]*Sqrt[Ta
n[a + b*x]]) + (Log[1 - Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]*Sqrt[c*Sec[a + b*x]])/(2*Sqrt[2]*b*Sqrt[d*C
sc[a + b*x]]*Sqrt[Tan[a + b*x]]) - (Log[1 + Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]*Sqrt[c*Sec[a + b*x]])/(
2*Sqrt[2]*b*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]])

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Rubi [A]  time = 0.140481, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 12, number of rules used = 9, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.36, Rules used = {2629, 3476, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (a+b x)}\right ) \sqrt{c \sec (a+b x)}}{\sqrt{2} b \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}+\frac{\tan ^{-1}\left (\sqrt{2} \sqrt{\tan (a+b x)}+1\right ) \sqrt{c \sec (a+b x)}}{\sqrt{2} b \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}+\frac{\sqrt{c \sec (a+b x)} \log \left (\tan (a+b x)-\sqrt{2} \sqrt{\tan (a+b x)}+1\right )}{2 \sqrt{2} b \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}}-\frac{\sqrt{c \sec (a+b x)} \log \left (\tan (a+b x)+\sqrt{2} \sqrt{\tan (a+b x)}+1\right )}{2 \sqrt{2} b \sqrt{\tan (a+b x)} \sqrt{d \csc (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[c*Sec[a + b*x]]/Sqrt[d*Csc[a + b*x]],x]

[Out]

-((ArcTan[1 - Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sqrt[c*Sec[a + b*x]])/(Sqrt[2]*b*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b
*x]])) + (ArcTan[1 + Sqrt[2]*Sqrt[Tan[a + b*x]]]*Sqrt[c*Sec[a + b*x]])/(Sqrt[2]*b*Sqrt[d*Csc[a + b*x]]*Sqrt[Ta
n[a + b*x]]) + (Log[1 - Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]*Sqrt[c*Sec[a + b*x]])/(2*Sqrt[2]*b*Sqrt[d*C
sc[a + b*x]]*Sqrt[Tan[a + b*x]]) - (Log[1 + Sqrt[2]*Sqrt[Tan[a + b*x]] + Tan[a + b*x]]*Sqrt[c*Sec[a + b*x]])/(
2*Sqrt[2]*b*Sqrt[d*Csc[a + b*x]]*Sqrt[Tan[a + b*x]])

Rule 2629

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[((a*Csc[e + f
*x])^m*(b*Sec[e + f*x])^n)/Tan[e + f*x]^n, Int[Tan[e + f*x]^n, x], x] /; FreeQ[{a, b, e, f, m, n}, x] &&  !Int
egerQ[n] && EqQ[m + n, 0]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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

Mathematica [A]  time = 1.32641, size = 171, normalized size = 0.63 \[ \frac{\cot (a+b x) \sqrt{c \sec (a+b x)} \left (\log \left (\sqrt{\cot ^2(a+b x)}-\sqrt{2} \sqrt [4]{\cot ^2(a+b x)}+1\right )-\log \left (\sqrt{\cot ^2(a+b x)}+\sqrt{2} \sqrt [4]{\cot ^2(a+b x)}+1\right )+2 \tan ^{-1}\left (1-\sqrt{2} \sqrt [4]{\cot ^2(a+b x)}\right )-2 \tan ^{-1}\left (\sqrt{2} \sqrt [4]{\cot ^2(a+b x)}+1\right )\right )}{2 \sqrt{2} b \sqrt [4]{\cot ^2(a+b x)} \sqrt{d \csc (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[c*Sec[a + b*x]]/Sqrt[d*Csc[a + b*x]],x]

[Out]

(Cot[a + b*x]*(2*ArcTan[1 - Sqrt[2]*(Cot[a + b*x]^2)^(1/4)] - 2*ArcTan[1 + Sqrt[2]*(Cot[a + b*x]^2)^(1/4)] + L
og[1 - Sqrt[2]*(Cot[a + b*x]^2)^(1/4) + Sqrt[Cot[a + b*x]^2]] - Log[1 + Sqrt[2]*(Cot[a + b*x]^2)^(1/4) + Sqrt[
Cot[a + b*x]^2]])*Sqrt[c*Sec[a + b*x]])/(2*Sqrt[2]*b*(Cot[a + b*x]^2)^(1/4)*Sqrt[d*Csc[a + b*x]])

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Maple [C]  time = 0.148, size = 276, normalized size = 1. \begin{align*}{\frac{\sqrt{2}\sin \left ( bx+a \right ) }{2\,b \left ( -1+\cos \left ( bx+a \right ) \right ) }\sqrt{{\frac{c}{\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 ) \right ){\frac{1}{\sqrt{{\frac{d}{\sin \left ( bx+a \right ) }}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*sec(b*x+a))^(1/2)/(d*csc(b*x+a))^(1/2),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*sec(b*x + a))/sqrt(d*csc(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((c*sec(b*x+a))^(1/2)/(d*csc(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{c \sec{\left (a + b x \right )}}}{\sqrt{d \csc{\left (a + b x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*sec(b*x+a))**(1/2)/(d*csc(b*x+a))**(1/2),x)

[Out]

Integral(sqrt(c*sec(a + b*x))/sqrt(d*csc(a + b*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(c*sec(b*x + a))/sqrt(d*csc(b*x + a)), x)