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*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)

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Rubi [A]  time = 0.162345, antiderivative size = 227, normalized size of antiderivative = 1., 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 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]

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 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 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]

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])

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*}

Mathematica [A]  time = 0.509314, size = 113, normalized size = 0.5 \[ \frac{\csc (a+b x) \sqrt{d \tan (a+b x)} \left (\sin (a+b x)+\sin (3 (a+b x))-3 \sqrt{\sin (2 (a+b x))} \sin ^{-1}(\cos (a+b x)-\sin (a+b x))+3 \sqrt{\sin (2 (a+b x))} \log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{8 b d^3} \]

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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Maple [C]  time = 0.129, size = 672, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

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*2^(1/2)*(cos(b*x+a)-1)*(-3*I*sin(b*x+a)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1)/si
n(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a)
)^(1/2),1/2-1/2*I,1/2*2^(1/2))+3*I*sin(b*x+a)*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+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((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a)
)^(1/2),1/2+1/2*I,1/2*2^(1/2))-3*sin(b*x+a)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2-1/2*I
,1/2*2^(1/2))*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+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)+6*sin(b*x+a)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))
*((cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/si
n(b*x+a))^(1/2)-3*sin(b*x+a)*EllipticPi((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2+1/2*I,1/2*2^(1/2))*(
(cos(b*x+a)-1)/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+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/cos(b*x+a)^3/sin(b*x+a)/(d*sin(b
*x+a)/cos(b*x+a))^(5/2)

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [A]  time = 1.21178, size = 301, normalized size = 1.33 \begin{align*} \frac{1}{16} \, d{\left (\frac{6 \, \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 )}{b d^{4}} + \frac{6 \, \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 )}{b d^{4}} + \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 )}{b d^{4}} - \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 )}{b d^{4}} + \frac{8 \, \sqrt{d \tan \left (b x + a\right )}}{{\left (d^{2} \tan \left (b x + a\right )^{2} + d^{2}\right )} b d^{2}}\right )} \end{align*}

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]

1/16*d*(6*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^4) + 6*sqrt(2)*sqrt(abs(d))*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(abs(d)) - 2*sqrt(d*tan(b*x + a)))/sqrt(ab
s(d)))/(b*d^4) + 3*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^4) - 3*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^4) + 8*sqrt(d*tan(b*x + a))/((d^2*tan(b*x + a)^2 + d^2)*b*d^2))