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

Optimal. Leaf size=148 \[ \frac{(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 f (a+b)}+\frac{(2 a+5 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}-\frac{\sqrt{a+b} (2 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{2 f}+\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 f (a+b)} \]

[Out]

-(Sqrt[a + b]*(2*a + 5*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(2*f) + ((2*a + 5*b)*Sqrt[a + b*Sin
[e + f*x]^2])/(2*f) + ((2*a + 5*b)*(a + b*Sin[e + f*x]^2)^(3/2))/(6*(a + b)*f) + (Sec[e + f*x]^2*(a + b*Sin[e
+ f*x]^2)^(5/2))/(2*(a + b)*f)

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Rubi [A]  time = 0.136133, antiderivative size = 148, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3194, 78, 50, 63, 208} \[ \frac{(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 f (a+b)}+\frac{(2 a+5 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}-\frac{\sqrt{a+b} (2 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{2 f}+\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 f (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sqrt[a + b]*(2*a + 5*b)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(2*f) + ((2*a + 5*b)*Sqrt[a + b*Sin
[e + f*x]^2])/(2*f) + ((2*a + 5*b)*(a + b*Sin[e + f*x]^2)^(3/2))/(6*(a + b)*f) + (Sec[e + f*x]^2*(a + b*Sin[e
+ f*x]^2)^(5/2))/(2*(a + b)*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 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

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 ^3(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x (a+b x)^{3/2}}{(1-x)^2} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac{(2 a+5 b) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=\frac{(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac{(2 a+5 b) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{4 f}\\ &=\frac{(2 a+5 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}+\frac{(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac{((a+b) (2 a+5 b)) \operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{4 f}\\ &=\frac{(2 a+5 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}+\frac{(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}-\frac{((a+b) (2 a+5 b)) \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 )}{2 b f}\\ &=-\frac{\sqrt{a+b} (2 a+5 b) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{2 f}+\frac{(2 a+5 b) \sqrt{a+b \sin ^2(e+f x)}}{2 f}+\frac{(2 a+5 b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{6 (a+b) f}+\frac{\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{2 (a+b) f}\\ \end{align*}

Mathematica [A]  time = 0.534811, size = 116, normalized size = 0.78 \[ \frac{(2 a+5 b) \left (\sqrt{a+b \sin ^2(e+f x)} \left (4 a+b \sin ^2(e+f x)+3 b\right )-3 (a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )\right )+3 \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 f (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2) + (2*a + 5*b)*(-3*(a + b)^(3/2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]
^2]/Sqrt[a + b]] + Sqrt[a + b*Sin[e + f*x]^2]*(4*a + 3*b + b*Sin[e + f*x]^2)))/(6*(a + b)*f)

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Maple [B]  time = 3.349, size = 567, normalized size = 3.8 \begin{align*} \text{result too large to display} \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)^3,x)

[Out]

1/12*(-4*b*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*cos(f*x+e)^4-(-16*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a-2
8*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*b+6*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*si
n(f*x+e)+a))*a^3+27*a^2*b*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))+36*a*b
^2*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))+15*b^3*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))+6*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^3+27*a^2*b*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))+36*a*b^2*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))+15*b^3*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)))*cos(f*x+e)^2+6*(a+b)^(3/2)*(a+b-b*cos(f*x+e)
^2)^(1/2)*a+6*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*b)/(a+b)^(3/2)/cos(f*x+e)^2/f

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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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.52874, size = 672, normalized size = 4.54 \begin{align*} \left [\frac{3 \,{\left (2 \, a + 5 \, b\right )} \sqrt{a + b} \cos \left (f x + e\right )^{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 (2 \, b \cos \left (f x + e\right )^{4} - 2 \,{\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{12 \, f \cos \left (f x + e\right )^{2}}, \frac{3 \,{\left (2 \, a + 5 \, 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 ) \cos \left (f x + e\right )^{2} -{\left (2 \, b \cos \left (f x + e\right )^{4} - 2 \,{\left (4 \, a + 7 \, b\right )} \cos \left (f x + e\right )^{2} - 3 \, a - 3 \, b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{6 \, f \cos \left (f x + e\right )^{2}}\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)^3,x, algorithm="fricas")

[Out]

[1/12*(3*(2*a + 5*b)*sqrt(a + b)*cos(f*x + e)^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*(2*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + e)^2 - 3*a - 3*b)*sqrt(
-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2), 1/6*(3*(2*a + 5*b)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2
+ a + b)*sqrt(-a - b)/(a + b))*cos(f*x + e)^2 - (2*b*cos(f*x + e)^4 - 2*(4*a + 7*b)*cos(f*x + e)^2 - 3*a - 3*b
)*sqrt(-b*cos(f*x + e)^2 + a + b))/(f*cos(f*x + e)^2)]

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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)**3,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 )^{3}\,{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)^3,x, algorithm="giac")

[Out]

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