3.104 \(\int \frac {\sin ^2(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx\)

Optimal. Leaf size=227 \[ -\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{5/2}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b d^{5/2}}-\frac {3 \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{5/2}}+\frac {3 \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{5/2}}+\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d^3} \]

[Out]

-3/8*arctan(1-2^(1/2)*(d*tan(b*x+a))^(1/2)/d^(1/2))/b/d^(5/2)*2^(1/2)+3/8*arctan(1+2^(1/2)*(d*tan(b*x+a))^(1/2
)/d^(1/2))/b/d^(5/2)*2^(1/2)-3/16*ln(d^(1/2)-2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b/d^(5/2)*2^(1/2
)+3/16*ln(d^(1/2)+2^(1/2)*(d*tan(b*x+a))^(1/2)+d^(1/2)*tan(b*x+a))/b/d^(5/2)*2^(1/2)+1/2*cos(b*x+a)^2*(d*tan(b
*x+a))^(1/2)/b/d^3

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Rubi [A]  time = 0.16, 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 = {2591, 290, 329, 211, 1165, 628, 1162, 617, 204} \[ -\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{5/2}}+\frac {3 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}+1\right )}{4 \sqrt {2} b d^{5/2}}-\frac {3 \log \left (\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{5/2}}+\frac {3 \log \left (\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}+\sqrt {d}\right )}{8 \sqrt {2} b d^{5/2}}+\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d^3} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^2/(d*Tan[a + b*x])^(5/2),x]

[Out]

(-3*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(4*Sqrt[2]*b*d^(5/2)) + (3*ArcTan[1 + (Sqrt[2]*Sqrt[d*
Tan[a + b*x]])/Sqrt[d]])/(4*Sqrt[2]*b*d^(5/2)) - (3*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*Sqrt[d*Tan[a
+ b*x]]])/(8*Sqrt[2]*b*d^(5/2)) + (3*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a + b*x]]])/(8*Sq
rt[2]*b*d^(5/2)) + (Cos[a + b*x]^2*Sqrt[d*Tan[a + b*x]])/(2*b*d^3)

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 211

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 290

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

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 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 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]

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 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 2591

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])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rubi steps

\begin {align*} \int \frac {\sin ^2(a+b x)}{(d \tan (a+b x))^{5/2}} \, dx &=\frac {d \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \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^3}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (d^2+x^2\right )} \, dx,x,d \tan (a+b x)\right )}{4 b d}\\ &=\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d^3}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{2 b d}\\ &=\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d^3}+\frac {3 \operatorname {Subst}\left (\int \frac {d-x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b d^2}+\frac {3 \operatorname {Subst}\left (\int \frac {d+x^2}{d^2+x^4} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{4 b d^2}\\ &=\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d^3}-\frac {3 \operatorname {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 d^{5/2}}-\frac {3 \operatorname {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 d^{5/2}}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{d-\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 b d^2}+\frac {3 \operatorname {Subst}\left (\int \frac {1}{d+\sqrt {2} \sqrt {d} x+x^2} \, dx,x,\sqrt {d \tan (a+b x)}\right )}{8 b d^2}\\ &=-\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{5/2}}+\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{5/2}}+\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d^3}+\frac {3 \operatorname {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 d^{5/2}}-\frac {3 \operatorname {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 d^{5/2}}\\ &=-\frac {3 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{5/2}}+\frac {3 \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {d \tan (a+b x)}}{\sqrt {d}}\right )}{4 \sqrt {2} b d^{5/2}}-\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)-\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{5/2}}+\frac {3 \log \left (\sqrt {d}+\sqrt {d} \tan (a+b x)+\sqrt {2} \sqrt {d \tan (a+b x)}\right )}{8 \sqrt {2} b d^{5/2}}+\frac {\cos ^2(a+b x) \sqrt {d \tan (a+b x)}}{2 b d^3}\\ \end {align*}

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Mathematica [A]  time = 0.55, size = 113, normalized size = 0.50 \[ \frac {\csc (a+b x) \sqrt {d \tan (a+b x)} \left (\sin (a+b x)+\sin (3 (a+b x))-3 \sqrt {\sin (2 (a+b x))} \sin ^{-1}(\cos (a+b x)-\sin (a+b x))+3 \sqrt {\sin (2 (a+b x))} \log \left (\sin (a+b x)+\sqrt {\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{8 b d^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^2/(d*Tan[a + b*x])^(5/2),x]

[Out]

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

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fricas [B]  time = 66.16, size = 1545, normalized size = 6.81 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/32*(6*sqrt(2)*b*d^3*(1/(b^4*d^10))^(1/4)*arctan(1/2*(sqrt(4*b^2*d^5*sqrt(1/(b^4*d^10))*cos(b*x + a)*sin(b*x
+ a) - 2*(sqrt(2)*b^3*d^7*(1/(b^4*d^10))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*d^2*(1/(b^4*d^10))^(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^7*(1/(b^4*d^10))^(3/4)*sin(b*x + a) +
sqrt(2)*b*d^2*(1/(b^4*d^10))^(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^7*(1/(b^4*d^10))^(3/4)*sin(b*x + a) - sqrt(2)*b*d^2*(1/(b^4*d^10))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x
+ a)/cos(b*x + a)))/sin(b*x + a)) + 6*sqrt(2)*b*d^3*(1/(b^4*d^10))^(1/4)*arctan(1/2*(sqrt(4*b^2*d^5*sqrt(1/(b^
4*d^10))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d^7*(1/(b^4*d^10))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*d^2*(1
/(b^4*d^10))^(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^7*(1/(b^4
*d^10))^(3/4)*sin(b*x + a) + sqrt(2)*b*d^2*(1/(b^4*d^10))^(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^7*(1/(b^4*d^10))^(3/4)*sin(b*x + a) - sqrt(2)*b*d^2*(1/(b^4*d^10))^(1/4)*
cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)))/sin(b*x + a)) + 6*sqrt(2)*b*d^3*(1/(b^4*d^10))^(1/4)*arctan(1
/2*((sqrt(2)*b^3*d^7*(1/(b^4*d^10))^(3/4)*sin(b*x + a) + sqrt(2)*b*d^2*(1/(b^4*d^10))^(1/4)*cos(b*x + a))*sqrt
(4*b^2*d^5*sqrt(1/(b^4*d^10))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d^7*(1/(b^4*d^10))^(3/4)*cos(b*x + a)
^2 + sqrt(2)*b*d^2*(1/(b^4*d^10))^(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^7*(1/(b^4*d^10))^(3/4)*sin(b*x + a) + sqrt(2)*b*d^2*(1/(b^4*d^1
0))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) - 4*(b^2*d^5*cos(b*x + a)^3 - b^2*d^5*cos(b*x + a))*
sqrt(1/(b^4*d^10)) + 2*sin(b*x + a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) + 6*sqrt(2)*b*d^3*(1/(b^4*d^10))^(
1/4)*arctan(1/2*((sqrt(2)*b^3*d^7*(1/(b^4*d^10))^(3/4)*sin(b*x + a) + sqrt(2)*b*d^2*(1/(b^4*d^10))^(1/4)*cos(b
*x + a))*sqrt(4*b^2*d^5*sqrt(1/(b^4*d^10))*cos(b*x + a)*sin(b*x + a) - 2*(sqrt(2)*b^3*d^7*(1/(b^4*d^10))^(3/4)
*cos(b*x + a)^2 + sqrt(2)*b*d^2*(1/(b^4*d^10))^(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^7*(1/(b^4*d^10))^(3/4)*sin(b*x + a) + sqrt(2)*b*d^
2*(1/(b^4*d^10))^(1/4)*cos(b*x + a))*sqrt(d*sin(b*x + a)/cos(b*x + a)) + 4*(b^2*d^5*cos(b*x + a)^3 - b^2*d^5*c
os(b*x + a))*sqrt(1/(b^4*d^10)) - 2*sin(b*x + a))/((2*cos(b*x + a)^2 - 1)*sin(b*x + a))) - 3*sqrt(2)*b*d^3*(1/
(b^4*d^10))^(1/4)*log(4*b^2*d^5*sqrt(1/(b^4*d^10))*cos(b*x + a)*sin(b*x + a) + 2*(sqrt(2)*b^3*d^7*(1/(b^4*d^10
))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*d^2*(1/(b^4*d^10))^(1/4)*cos(b*x + a)*sin(b*x + a))*sqrt(d*sin(b*x + a)/co
s(b*x + a)) + 1) + 3*sqrt(2)*b*d^3*(1/(b^4*d^10))^(1/4)*log(4*b^2*d^5*sqrt(1/(b^4*d^10))*cos(b*x + a)*sin(b*x
+ a) - 2*(sqrt(2)*b^3*d^7*(1/(b^4*d^10))^(3/4)*cos(b*x + a)^2 + sqrt(2)*b*d^2*(1/(b^4*d^10))^(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^3)

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giac [A]  time = 2.01, size = 220, normalized size = 0.97 \[ \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{8 \, b d^{3}} + \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {{\left | d \right |}} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {{\left | d \right |}}}\right )}{8 \, b d^{3}} + \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{16 \, b d^{3}} - \frac {3 \, \sqrt {2} \sqrt {{\left | d \right |}} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {{\left | d \right |}} + {\left | d \right |}\right )}{16 \, b d^{3}} + \frac {\sqrt {d \tan \left (b x + a\right )}}{2 \, {\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )} b d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.43, size = 662, normalized size = 2.92 \[ \frac {\left (-1+\cos \left (b x +a \right )\right ) \left (-3 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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+3 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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-3 \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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}-3 \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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}+6 \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 ) \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 {1-\cos \left (b x +a \right )+\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 \cos \left (b x +a \right )^{3} \sin \left (b x +a \right ) \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(b*x+a)^2/(d*tan(b*x+a))^(5/2),x)

[Out]

1/8/b*(-1+cos(b*x+a))*(-3*I*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)*((1-cos(b*x+a)+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))+3*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))*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)*((1-cos(b*x+a)+sin(b*
x+a))/sin(b*x+a))^(1/2)-3*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)*((1-cos(b*x+a)+sin(b*x+
a))/sin(b*x+a))^(1/2)-3*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)*((1-cos(b*x+a)+sin(b*x+a)
)/sin(b*x+a))^(1/2)+6*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))*((-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)*((1-cos(b*x+a)+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/cos(b*x+a)^3/sin(b*x+a)/(d*sin(b*x+a)/c
os(b*x+a))^(5/2)*2^(1/2)

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maxima [A]  time = 0.65, size = 189, normalized size = 0.83 \[ \frac {6 \, \sqrt {2} \sqrt {d} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} + 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 6 \, \sqrt {2} \sqrt {d} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {d} - 2 \, \sqrt {d \tan \left (b x + a\right )}\right )}}{2 \, \sqrt {d}}\right ) + 3 \, \sqrt {2} \sqrt {d} \log \left (d \tan \left (b x + a\right ) + \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) - 3 \, \sqrt {2} \sqrt {d} \log \left (d \tan \left (b x + a\right ) - \sqrt {2} \sqrt {d \tan \left (b x + a\right )} \sqrt {d} + d\right ) + \frac {8 \, \sqrt {d \tan \left (b x + a\right )} d^{2}}{d^{2} \tan \left (b x + a\right )^{2} + d^{2}}}{16 \, b d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(6*sqrt(2)*sqrt(d)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(d) + 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + 6*sqrt(2)*sqr
t(d)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(d) - 2*sqrt(d*tan(b*x + a)))/sqrt(d)) + 3*sqrt(2)*sqrt(d)*log(d*tan(b*x
+ a) + sqrt(2)*sqrt(d*tan(b*x + a))*sqrt(d) + d) - 3*sqrt(2)*sqrt(d)*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^2/(d^2*tan(b*x + a)^2 + d^2))/(b*d^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\sin \left (a+b\,x\right )}^2}{{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(a + b*x)^2/(d*tan(a + b*x))^(5/2),x)

[Out]

int(sin(a + b*x)^2/(d*tan(a + b*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin ^{2}{\left (a + b x \right )}}{\left (d \tan {\left (a + b x \right )}\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)**2/(d*tan(b*x+a))**(5/2),x)

[Out]

Integral(sin(a + b*x)**2/(d*tan(a + b*x))**(5/2), x)

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