3.501 \(\int (a+b \sin ^2(e+f x))^{3/2} \tan (e+f x) \, dx\)

Optimal. Leaf size=84 \[ -\frac{(a+b) \sqrt{a+b \sin ^2(e+f x)}}{f}-\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}+\frac{(a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{f} \]

[Out]

((a + b)^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/f - ((a + b)*Sqrt[a + b*Sin[e + f*x]^2])/f - (
a + b*Sin[e + f*x]^2)^(3/2)/(3*f)

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Rubi [A]  time = 0.0775713, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3194, 50, 63, 208} \[ -\frac{(a+b) \sqrt{a+b \sin ^2(e+f x)}}{f}-\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}+\frac{(a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b)^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/f - ((a + b)*Sqrt[a + b*Sin[e + f*x]^2])/f - (
a + b*Sin[e + f*x]^2)^(3/2)/(3*f)

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

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

Rubi steps

\begin{align*} \int \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}+\frac{(a+b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac{(a+b) \sqrt{a+b \sin ^2(e+f x)}}{f}-\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}+\frac{(a+b)^2 \operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=-\frac{(a+b) \sqrt{a+b \sin ^2(e+f x)}}{f}-\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}+\frac{(a+b)^2 \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}-\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{b f}\\ &=\frac{(a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{f}-\frac{(a+b) \sqrt{a+b \sin ^2(e+f x)}}{f}-\frac{\left (a+b \sin ^2(e+f x)\right )^{3/2}}{3 f}\\ \end{align*}

Mathematica [A]  time = 0.158502, size = 79, normalized size = 0.94 \[ \frac{\sqrt{a-b \cos ^2(e+f x)+b} \left (b \cos ^2(e+f x)-4 (a+b)\right )+3 (a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b \cos ^2(e+f x)+b}}{\sqrt{a+b}}\right )}{3 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b)^(3/2)*ArcTanh[Sqrt[a + b - b*Cos[e + f*x]^2]/Sqrt[a + b]] + Sqrt[a + b - b*Cos[e + f*x]^2]*(-4*(a +
b) + b*Cos[e + f*x]^2))/(3*f)

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Maple [B]  time = 2.803, size = 423, normalized size = 5. \begin{align*}{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{3\,f}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}-{\frac{4\,a}{3\,f}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}-{\frac{4\,b}{3\,f}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{{a}^{2}}{2\,f}\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{ab}{f}\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{{b}^{2}}{2\,f}\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}-b\sin \left ( fx+e \right ) +a}{1+\sin \left ( fx+e \right ) }} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{{a}^{2}}{2\,f}\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{ab}{f}\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ){\frac{1}{\sqrt{a+b}}}}+{\frac{{b}^{2}}{2\,f}\ln \left ( 2\,{\frac{\sqrt{a+b}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}+b\sin \left ( fx+e \right ) +a}{-1+\sin \left ( fx+e \right ) }} \right ){\frac{1}{\sqrt{a+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e),x)

[Out]

1/3/f*b*(a+b-b*cos(f*x+e)^2)^(1/2)*cos(f*x+e)^2-4/3/f*a*(a+b-b*cos(f*x+e)^2)^(1/2)-4/3/f*b*(a+b-b*cos(f*x+e)^2
)^(1/2)+1/2/(a+b)^(1/2)/f*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2+1/(
a+b)^(1/2)/f*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b+1/2/(a+b)^(1/2)/
f*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^2+1/2/(a+b)^(1/2)/f*ln(2/(-1+
sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2+1/(a+b)^(1/2)/f*ln(2/(-1+sin(f*x+e))*
((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b+1/2/(a+b)^(1/2)/f*ln(2/(-1+sin(f*x+e))*((a+b)^(1/
2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^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((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.04154, size = 478, normalized size = 5.69 \begin{align*} \left [\frac{3 \,{\left (a + b\right )}^{\frac{3}{2}} \log \left (\frac{b \cos \left (f x + e\right )^{2} - 2 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \,{\left (b \cos \left (f x + e\right )^{2} - 4 \, a - 4 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, f}, -\frac{3 \,{\left (a + b\right )} \sqrt{-a - b} \arctan \left (\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a - b}}{a + b}\right ) -{\left (b \cos \left (f x + e\right )^{2} - 4 \, a - 4 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{3 \, f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e),x, algorithm="fricas")

[Out]

[1/6*(3*(a + b)^(3/2)*log((b*cos(f*x + e)^2 - 2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b) - 2*a - 2*b)/cos(f
*x + e)^2) + 2*(b*cos(f*x + e)^2 - 4*a - 4*b)*sqrt(-b*cos(f*x + e)^2 + a + b))/f, -1/3*(3*(a + b)*sqrt(-a - b)
*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-a - b)/(a + b)) - (b*cos(f*x + e)^2 - 4*a - 4*b)*sqrt(-b*cos(f*x
+ e)^2 + a + b))/f]

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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((a+b*sin(f*x+e)**2)**(3/2)*tan(f*x+e),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e)^2)^(3/2)*tan(f*x+e),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*tan(f*x + e), x)