3.64 \(\int \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx\)

Optimal. Leaf size=277 \[ \frac{45 d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}-\frac{45 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}+1\right )}{32 \sqrt{2} b}+\frac{45 d^{3/2} \log \left (\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{64 \sqrt{2} b}-\frac{45 d^{3/2} \log \left (\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{64 \sqrt{2} b}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac{45 d \sqrt{d \tan (a+b x)}}{16 b}-\frac{9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d} \]

[Out]

(45*d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) - (45*d^(3/2)*ArcTan[1 + (Sqrt[
2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) + (45*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*S
qrt[d*Tan[a + b*x]]])/(64*Sqrt[2]*b) - (45*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a +
b*x]]])/(64*Sqrt[2]*b) + (45*d*Sqrt[d*Tan[a + b*x]])/(16*b) - (9*Cos[a + b*x]^2*(d*Tan[a + b*x])^(5/2))/(16*b
*d) - (Cos[a + b*x]^4*(d*Tan[a + b*x])^(9/2))/(4*b*d^3)

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Rubi [A]  time = 0.19601, antiderivative size = 277, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 10, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.476, Rules used = {2591, 288, 321, 329, 211, 1165, 628, 1162, 617, 204} \[ \frac{45 d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}-\frac{45 d^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}+1\right )}{32 \sqrt{2} b}+\frac{45 d^{3/2} \log \left (\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{64 \sqrt{2} b}-\frac{45 d^{3/2} \log \left (\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}+\sqrt{d}\right )}{64 \sqrt{2} b}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac{45 d \sqrt{d \tan (a+b x)}}{16 b}-\frac{9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[a + b*x]^4*(d*Tan[a + b*x])^(3/2),x]

[Out]

(45*d^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) - (45*d^(3/2)*ArcTan[1 + (Sqrt[
2]*Sqrt[d*Tan[a + b*x]])/Sqrt[d]])/(32*Sqrt[2]*b) + (45*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] - Sqrt[2]*S
qrt[d*Tan[a + b*x]]])/(64*Sqrt[2]*b) - (45*d^(3/2)*Log[Sqrt[d] + Sqrt[d]*Tan[a + b*x] + Sqrt[2]*Sqrt[d*Tan[a +
b*x]]])/(64*Sqrt[2]*b) + (45*d*Sqrt[d*Tan[a + b*x]])/(16*b) - (9*Cos[a + b*x]^2*(d*Tan[a + b*x])^(5/2))/(16*b
*d) - (Cos[a + b*x]^4*(d*Tan[a + b*x])^(9/2))/(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 321

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*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 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 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 \sin ^4(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac{d \operatorname{Subst}\left (\int \frac{x^{11/2}}{\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))^{9/2}}{4 b d^3}+\frac{(9 d) \operatorname{Subst}\left (\int \frac{x^{7/2}}{\left (d^2+x^2\right )^2} \, dx,x,d \tan (a+b x)\right )}{8 b}\\ &=-\frac{9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac{(45 d) \operatorname{Subst}\left (\int \frac{x^{3/2}}{d^2+x^2} \, dx,x,d \tan (a+b x)\right )}{32 b}\\ &=\frac{45 d \sqrt{d \tan (a+b x)}}{16 b}-\frac{9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}-\frac{\left (45 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (d^2+x^2\right )} \, dx,x,d \tan (a+b x)\right )}{32 b}\\ &=\frac{45 d \sqrt{d \tan (a+b x)}}{16 b}-\frac{9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}-\frac{\left (45 d^3\right ) \operatorname{Subst}\left (\int \frac{1}{d^2+x^4} \, dx,x,\sqrt{d \tan (a+b x)}\right )}{16 b}\\ &=\frac{45 d \sqrt{d \tan (a+b x)}}{16 b}-\frac{9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}-\frac{\left (45 d^2\right ) \operatorname{Subst}\left (\int \frac{d-x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (a+b x)}\right )}{32 b}-\frac{\left (45 d^2\right ) \operatorname{Subst}\left (\int \frac{d+x^2}{d^2+x^4} \, dx,x,\sqrt{d \tan (a+b x)}\right )}{32 b}\\ &=\frac{45 d \sqrt{d \tan (a+b x)}}{16 b}-\frac{9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}+\frac{\left (45 d^{3/2}\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 (45 d^{3/2}\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 (45 d^2\right ) \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{\left (45 d^2\right ) \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{45 d^{3/2} \log \left (\sqrt{d}+\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}\right )}{64 \sqrt{2} b}-\frac{45 d^{3/2} \log \left (\sqrt{d}+\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}\right )}{64 \sqrt{2} b}+\frac{45 d \sqrt{d \tan (a+b x)}}{16 b}-\frac{9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}-\frac{\left (45 d^{3/2}\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 (45 d^{3/2}\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{45 d^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}-\frac{45 d^{3/2} \tan ^{-1}\left (1+\frac{\sqrt{2} \sqrt{d \tan (a+b x)}}{\sqrt{d}}\right )}{32 \sqrt{2} b}+\frac{45 d^{3/2} \log \left (\sqrt{d}+\sqrt{d} \tan (a+b x)-\sqrt{2} \sqrt{d \tan (a+b x)}\right )}{64 \sqrt{2} b}-\frac{45 d^{3/2} \log \left (\sqrt{d}+\sqrt{d} \tan (a+b x)+\sqrt{2} \sqrt{d \tan (a+b x)}\right )}{64 \sqrt{2} b}+\frac{45 d \sqrt{d \tan (a+b x)}}{16 b}-\frac{9 \cos ^2(a+b x) (d \tan (a+b x))^{5/2}}{16 b d}-\frac{\cos ^4(a+b x) (d \tan (a+b x))^{9/2}}{4 b d^3}\\ \end{align*}

Mathematica [A]  time = 0.805176, size = 123, normalized size = 0.44 \[ -\frac{d \csc (a+b x) \sqrt{d \tan (a+b x)} \left (-143 \sin (a+b x)-14 \sin (3 (a+b x))+\sin (5 (a+b x))-45 \sqrt{\sin (2 (a+b x))} \sin ^{-1}(\cos (a+b x)-\sin (a+b x))+45 \sqrt{\sin (2 (a+b x))} \log \left (\sin (a+b x)+\sqrt{\sin (2 (a+b x))}+\cos (a+b x)\right )\right )}{64 b} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[a + b*x]^4*(d*Tan[a + b*x])^(3/2),x]

[Out]

-(d*Csc[a + b*x]*(-143*Sin[a + b*x] - 45*ArcSin[Cos[a + b*x] - Sin[a + b*x]]*Sqrt[Sin[2*(a + b*x)]] + 45*Log[C
os[a + b*x] + Sin[a + b*x] + Sqrt[Sin[2*(a + b*x)]]]*Sqrt[Sin[2*(a + b*x)]] - 14*Sin[3*(a + b*x)] + Sin[5*(a +
b*x)])*Sqrt[d*Tan[a + b*x]])/(64*b)

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Maple [C]  time = 0.227, size = 712, normalized size = 2.6 \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))^(3/2),x)

[Out]

-1/64/b*2^(1/2)*(cos(b*x+a)-1)*(8*2^(1/2)*cos(b*x+a)^5+45*I*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)-45*I*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)-8*cos(b*x+a)^4*2^(1/2)-45*sin(b*x+a)*EllipticPi((-(co
s(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)-45*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)+90*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))/sin(b*x+a))^(1/2)-34*cos(b*x+a)^3*2^(1/2)+34*cos(b*x+a)^2*2^(1
/2)-64*cos(b*x+a)*2^(1/2)+64*2^(1/2))*cos(b*x+a)*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(3/2)/sin(b*x+a)^5

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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))^(3/2),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))^(3/2),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))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \tan \left (b x + a\right )\right )^{\frac{3}{2}} \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))^(3/2),x, algorithm="giac")

[Out]

integrate((d*tan(b*x + a))^(3/2)*sin(b*x + a)^4, x)