3.1.85 \(\int \frac {\sin ^2(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\) [85]

Optimal. Leaf size=227 \[ -\frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {\text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b \sqrt {d}}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b \sqrt {d}}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d} \]

[Out]

-1/8*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b*2^(1/2)/d^(1/2)+1/8*arctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2
)/d^(1/2))/b*2^(1/2)/d^(1/2)-1/16*ln(d^(1/2)-2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b*2^(1/2)/d^(1/2
)+1/16*ln(d^(1/2)+2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b*2^(1/2)/d^(1/2)-1/2*cos(b*x+a)^2*(d*tan(b
*x+a))^(1/2)/b/d

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Rubi [A]
time = 0.11, antiderivative size = 227, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {2671, 294, 335, 217, 1179, 642, 1176, 631, 210} \begin {gather*} -\frac {\text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {\text {ArcTan}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b \sqrt {d}}-\frac {\log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b \sqrt {d}}+\frac {\log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/Sqrt[d*Tan[a + b*x]],x]

[Out]

-1/4*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]]/(Sqrt[2]*b*Sqrt[d]) + ArcTan[1 + (Sqrt[2]*Sqrt[d*Tan[a
+ b*x]])/Sqrt[d]]/(4*Sqrt[2]*b*Sqrt[d]) - Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*Sqrt[d*Tan[a + b*x]]]/
(8*Sqrt[2]*b*Sqrt[d]) + Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a + b*x]]]/(8*Sqrt[2]*b*Sqrt[d
]) - (Cos[a + b*x]^2*Sqrt[d*Tan[a + b*x]])/(2*b*d)

Rule 210

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

Rule 217

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

Rule 294

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[c^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^
n)^(p + 1)/(b*n*(p + 1))), x] - Dist[c^n*((m - n + 1)/(b*n*(p + 1))), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 335

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 631

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 642

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]

Rule 1176

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 1179

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 2671

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[b*(ff/f), Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, b*(Tan[e + f*x]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\sin ^2(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx &=\frac {d \text {Subst}\left (\int \frac {x^{3/2}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d}+\frac {d \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \tan (a+b x)\right )}{4 b}\\ &=-\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d}+\frac {d \text {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{2 b}\\ &=-\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d}+\frac {\text {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b}+\frac {\text {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b}\\ &=-\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d}+\frac {\text {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 b}+\frac {\text {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 b}-\frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}+2 x}{-d-\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {\text {Subst}\left (\int \frac {\sqrt {2} \sqrt {d}-2 x}{-d+\sqrt {2} \sqrt {d} x-x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b \sqrt {d}}\\ &=-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b \sqrt {d}}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d}+\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b \sqrt {d}}-\frac {\text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b \sqrt {d}}\\ &=-\frac {\tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b \sqrt {d}}+\frac {\tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b \sqrt {d}}-\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b \sqrt {d}}+\frac {\log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b \sqrt {d}}-\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d}\\ \end {align*}

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Mathematica [A]
time = 0.67, size = 109, normalized size = 0.48 \begin {gather*} -\frac {\sec (a+b x) \left (\sin (a+b x)+\text {ArcSin}(\cos (a+b x)-\sin (a+b x)) \sqrt {\sin (2 (a+b x))}-\log \left (\cos (a+b x)+\sin (a+b x)+\sqrt {\sin (2 (a+b x))}\right ) \sqrt {\sin (2 (a+b x))}+\sin (3 (a+b x))\right )}{8 b \sqrt {d \tan (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2/Sqrt[d*Tan[a + b*x]],x]

[Out]

-1/8*(Sec[a + b*x]*(Sin[a + b*x] + ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Sqrt[Sin[2*(a + b*x)]] - Log[Cos[a + b*
x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]*Sqrt[Sin[2*(a + b*x)]] + Sin[3*(a + b*x)]))/(b*Sqrt[d*Tan[a + b*x]
])

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.33, size = 662, normalized size = 2.92

method	result	size
default	\(\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (i \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-i \EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-\EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+2 \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sin \left (b x +a \right ) \EllipticF \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )-\EllipticPi \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right ) \sin \left (b x +a \right ) \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-2 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+2 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \sqrt {2}}{8 b \sin \left (b x +a \right )^{3} \sqrt {\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}}\, \cos \left (b x +a \right )}\)	\(662\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^2/(d*tan(b*x+a))^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/8/b*(-1+cos(b*x+a))*(I*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos
(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(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))*sin(b*x+a)-I*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*
x+a)-1+sin(b*x+a))/sin(b*x+a))^(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))*sin(b*x+a)-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))*sin(b*x+a)*((-
1+cos(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*
x+a))^(1/2)+2*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+s
in(b*x+a))/sin(b*x+a))^(1/2)*sin(b*x+a)*EllipticF(((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-El
lipticPi(((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))*sin(b*x+a)*((-1+cos(b*x+a))/sin(b
*x+a))^(1/2)*((1-cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)-2*cos(b
*x+a)^3*2^(1/2)+2*cos(b*x+a)^2*2^(1/2))*(cos(b*x+a)+1)^2/sin(b*x+a)^3/(d*sin(b*x+a)/cos(b*x+a))^(1/2)/cos(b*x+
a)*2^(1/2)

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Maxima [A]
time = 0.48, size = 188, normalized size = 0.83 \begin {gather*} \frac {2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 2 \, \sqrt {2} d^{\frac {5}{2}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \sqrt {2} d^{\frac {5}{2}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - \frac {8 \, \sqrt {d \tan \left (b x + a\right )} d^{4}}{d^{2} \tan \left (b x + a\right )^{2} + d^{2}}}{16 \, b d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*sqrt(2)*d^(5/2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + 2*sqrt(2)*d^(
5/2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + sqrt(2)*d^(5/2)*log(d*tan(b*x +
a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - sqrt(2)*d^(5/2)*log(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x
+ a))*sqrt(d) + d) - 8*sqrt(d*tan(b*x + a))*d^4/(d^2*tan(b*x + a)^2 + d^2))/(b*d^3)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 1442 vs. \(2 (171) = 342\).
time = 37.86, size = 1442, normalized size = 6.35 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(2*sqrt(2)*b*d*(1/(b^4*d^2))^(1/4)*arctan(1/2*(sqrt(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a)
- 2*(sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a
))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1)*((sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(2)*b*(1/(b^4
*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 2*sin(b*x + a)) - (sqrt(2)*b^3*d*(1/(b^4*d^2))^
(3/4)*sin(b*x + a) - sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/sin(b*x +
a)) + 2*sqrt(2)*b*d*(1/(b^4*d^2))^(1/4)*arctan(1/2*(sqrt(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a) +
2*(sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a)
)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1)*((sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(2)*b*(1/(b^4*
d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 2*sin(b*x + a)) - (sqrt(2)*b^3*d*(1/(b^4*d^2))^(
3/4)*sin(b*x + a) - sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/sin(b*x + a
)) + 2*sqrt(2)*b*d*(1/(b^4*d^2))^(1/4)*arctan(1/2*((sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(2)*b
*(1/(b^4*d^2))^(1/4)*cos(b*x + a))*sqrt(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d
*(1/(b^4*d^2))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x
+ a)/cos(b*x + a)) + 1)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + (sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) +
sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2*d*cos(b*x + a)^3 - b^2*
d*cos(b*x + a))*sqrt(1/(b^4*d^2)) + 2*sin(b*x + a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) + 2*sqrt(2)*b*d*(1/
(b^4*d^2))^(1/4)*arctan(1/2*((sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*c
os(b*x + a))*sqrt(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*c
os(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1
)*sqrt(d*sin(b*x + a)/cos(b*x + a)) + (sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*sin(b*x + a) + sqrt(2)*b*(1/(b^4*d^2)
)^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 4*(b^2*d*cos(b*x + a)^3 - b^2*d*cos(b*x + a))*sqrt(1
/(b^4*d^2)) - 2*sin(b*x + a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) - sqrt(2)*b*d*(1/(b^4*d^2))^(1/4)*log(4*b
^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*cos(b*x + a)^2 + sqrt(
2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 1) + sqrt(2)*b*d*(1/(b
^4*d^2))^(1/4)*log(4*b^2*d*sqrt(1/(b^4*d^2))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*d*(1/(b^4*d^2))^(3/4)*
cos(b*x + a)^2 + sqrt(2)*b*(1/(b^4*d^2))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) +
1) + 16*sqrt(d*sin(b*x + a)/cos(b*x + a))*cos(b*x + a)^2)/(b*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{2}{\left (a + b x \right )}}{\sqrt {d \tan {\left (a + b x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sin(a + b*x)**2/sqrt(d*tan(a + b*x)), x)

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Giac [A]
time = 0.70, size = 218, normalized size = 0.96 \begin {gather*} \frac {\sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{8 \, b d} + \frac {\sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{8 \, b d} + \frac {\sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{16 \, b d} - \frac {\sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{16 \, b d} - \frac {\sqrt {d \tan \left (b x + a\right )} d}{2 \, {\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )} b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*sqrt(2)*sqrt(abs(d))*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) + 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d)))/(b*d
) + 1/8*sqrt(2)*sqrt(abs(d))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(abs(d)))
/(b*d) + 1/16*sqrt(2)*sqrt(abs(d))*log(d*tan(b*x + a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/(b
*d) - 1/16*sqrt(2)*sqrt(abs(d))*log(d*tan(b*x + a) - sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(abs(d)) + abs(d))/(b*d)
- 1/2*sqrt(d*tan(b*x + a))*d/((d^2*tan(b*x + a)^2 + d^2)*b)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\sin \left (a+b\,x\right )}^2}{\sqrt {d\,\mathrm {tan}\left (a+b\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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