3.54 \(\int \sin ^4(a+b x) \sqrt{d \tan (a+b x)} \, dx\)

Optimal. Leaf size=257 \[ -\frac{\cos ^4(a+b x) (d \tan (a+b x))^{7/2}}{4 b d^3}-\frac{21 \sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}+\frac{21 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}+1\right )}{32 \sqrt{2} b}+\frac{21 \sqrt{d} \log \left (\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{64 \sqrt{2} b}-\frac{21 \sqrt{d} \log \left (\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{64 \sqrt{2} b}-\frac{7 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d} \]

[Out]

(-21*Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) + (21*Sqrt[d]*ArcTan[1 + (Sqrt
[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) + (21*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*
Sqrt[d*Tan[a + b*x]]])/(64*Sqrt[2]*b) - (21*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a
+ b*x]]])/(64*Sqrt[2]*b) - (7*Cos[a + b*x]^2*(d*Tan[a + b*x])^(3/2))/(16*b*d) - (Cos[a + b*x]^4*(d*Tan[a + b*x
])^(7/2))/(4*b*d^3)

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Rubi [A]  time = 0.195679, antiderivative size = 257, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2591, 288, 329, 297, 1162, 617, 204, 1165, 628} \[ -\frac{\cos ^4(a+b x) (d \tan (a+b x))^{7/2}}{4 b d^3}-\frac{21 \sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}+\frac{21 \sqrt{d} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}+1\right )}{32 \sqrt{2} b}+\frac{21 \sqrt{d} \log \left (\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{64 \sqrt{2} b}-\frac{21 \sqrt{d} \log \left (\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{64 \sqrt{2} b}-\frac{7 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^4*Sqrt[d*Tan[a + b*x]],x]

[Out]

(-21*Sqrt[d]*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) + (21*Sqrt[d]*ArcTan[1 + (Sqrt
[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) + (21*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*
Sqrt[d*Tan[a + b*x]]])/(64*Sqrt[2]*b) - (21*Sqrt[d]*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a
+ b*x]]])/(64*Sqrt[2]*b) - (7*Cos[a + b*x]^2*(d*Tan[a + b*x])^(3/2))/(16*b*d) - (Cos[a + b*x]^4*(d*Tan[a + b*x
])^(7/2))/(4*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 288

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 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 \sin ^4(a+b x) \sqrt{d \tan (a+b x)} \, dx &=\frac{d \operatorname{Subst}\left (\int \frac{x^{9/2}}{\left (d^2+x^2\right )^3} \, dx,x,d \tan (a+b x)\right )}{b}\\ &=-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{7/2}}{4 b d^3}+\frac{(7 d) \operatorname{Subst}\left (\int \frac{x^{5/2}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{8 b}\\ &=-\frac{7 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{7/2}}{4 b d^3}+\frac{(21 d) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{32 b}\\ &=-\frac{7 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{7/2}}{4 b d^3}+\frac{(21 d) \operatorname{Subst}\left (\int \frac{x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (a+b x)}\right )}{16 b}\\ &=-\frac{7 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{7/2}}{4 b d^3}-\frac{(21 d) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (a+b x)}\right )}{32 b}+\frac{(21 d) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (a+b x)}\right )}{32 b}\\ &=-\frac{7 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{7/2}}{4 b d^3}+\frac{\left (21 \sqrt{d}\right ) \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 )}{64 \sqrt{2} b}+\frac{\left (21 \sqrt{d}\right ) \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 )}{64 \sqrt{2} b}+\frac{(21 d) \operatorname{Subst}\left (\int \frac{1}{d-\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (a+b x)}\right )}{64 b}+\frac{(21 d) \operatorname{Subst}\left (\int \frac{1}{d+\sqrt{2} \sqrt{d} x+x^2} \, dx,x,\sqrt{d \tan (a+b x)}\right )}{64 b}\\ &=\frac{21 \sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}\right )}{64 \sqrt{2} b}-\frac{21 \sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}\right )}{64 \sqrt{2} b}-\frac{7 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{7/2}}{4 b d^3}+\frac{\left (21 \sqrt{d}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}-\frac{\left (21 \sqrt{d}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}\\ &=-\frac{21 \sqrt{d} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}+\frac{21 \sqrt{d} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}+\frac{21 \sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}\right )}{64 \sqrt{2} b}-\frac{21 \sqrt{d} \log \left (\sqrt{d}+\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}\right )}{64 \sqrt{2} b}-\frac{7 \cos ^2(a+b x) (d \tan (a+b x))^{3/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{7/2}}{4 b d^3}\\ \end{align*}

Mathematica [A]  time = 0.216264, size = 122, normalized size = 0.47 \[ -\frac{\sqrt{d \tan (a+b x)} \left (18 \sin (2 (a+b x))-2 \sin (4 (a+b x))+21 \sqrt{\sin (2 (a+b x))} \csc (a+b x) \sin ^{-1}(\cos (a+b x)-\sin (a+b x))+21 \sqrt{\sin (2 (a+b x))} \csc (a+b x) \log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{64 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^4*Sqrt[d*Tan[a + b*x]],x]

[Out]

-((21*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Csc[a + b*x]*Sqrt[Sin[2*(a + b*x)]] + 21*Csc[a + b*x]*Log[Cos[a + b*
x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]*Sqrt[Sin[2*(a + b*x)]] + 18*Sin[2*(a + b*x)] - 2*Sin[4*(a + b*x)])
*Sqrt[d*Tan[a + b*x]])/(64*b)

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Maple [C]  time = 0.279, size = 542, normalized size = 2.1 \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)^4*(d*tan(b*x+a))^(1/2),x)

[Out]

-1/64/b*2^(1/2)*(cos(b*x+a)-1)*(21*I*((cos(b*x+a)-1)/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))-21*I*((cos(b*x+a)-1)/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))-8*cos(b*x+a)^4*2^(1/2)+8*cos(b*x+a)^3*2^(1/2)-21*((cos(b*x+a)-1)/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))/s
in(b*x+a))^(1/2),1/2-1/2*I,1/2*2^(1/2))-21*((cos(b*x+a)-1)/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))+22*cos(b*x+a)^2*2^(1/2)-22*cos(b*x+a)*2^(1/2))*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*
x+a))^(1/2)/sin(b*x+a)^3

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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)^4*(d*tan(b*x+a))^(1/2),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)^4*(d*tan(b*x+a))^(1/2),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)**4*(d*tan(b*x+a))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \tan \left (b x + a\right )} \sin \left (b x + a\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(b*x+a)^4*(d*tan(b*x+a))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*tan(b*x + a))*sin(b*x + a)^4, x)