3.499 \(\int (a+b \sin ^2(e+f x))^{3/2} \tan ^5(e+f x) \, dx\)

Optimal. Leaf size=220 \[ -\frac{\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 f (a+b)^2}-\frac{\left (8 a^2+40 a b+35 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 f (a+b)}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f \sqrt{a+b}}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 f (a+b)}-\frac{(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 f (a+b)^2} \]

[Out]

((8*a^2 + 40*a*b + 35*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*Sqrt[a + b]*f) - ((8*a^2 + 40*a
*b + 35*b^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)*f) - ((8*a^2 + 40*a*b + 35*b^2)*(a + b*Sin[e + f*x]^2)^(3/
2))/(24*(a + b)^2*f) - ((8*a + 9*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2))/(8*(a + b)^2*f) + (Sec[e + f*
x]^4*(a + b*Sin[e + f*x]^2)^(5/2))/(4*(a + b)*f)

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Rubi [A]  time = 0.257801, antiderivative size = 220, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3194, 89, 78, 50, 63, 208} \[ -\frac{\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 f (a+b)^2}-\frac{\left (8 a^2+40 a b+35 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 f (a+b)}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f \sqrt{a+b}}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 f (a+b)}-\frac{(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]

[Out]

((8*a^2 + 40*a*b + 35*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*Sqrt[a + b]*f) - ((8*a^2 + 40*a
*b + 35*b^2)*Sqrt[a + b*Sin[e + f*x]^2])/(8*(a + b)*f) - ((8*a^2 + 40*a*b + 35*b^2)*(a + b*Sin[e + f*x]^2)^(3/
2))/(24*(a + b)^2*f) - ((8*a + 9*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2))/(8*(a + b)^2*f) + (Sec[e + f*
x]^4*(a + b*Sin[e + f*x]^2)^(5/2))/(4*(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 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

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 ^5(e+f x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2 (a+b x)^{3/2}}{(1-x)^3} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}-\frac{\operatorname{Subst}\left (\int \frac{(a+b x)^{3/2} \left (\frac{1}{2} (4 a+5 b)+2 (a+b) x\right )}{(1-x)^2} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=-\frac{(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{(a+b x)^{3/2}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b)^2 f}\\ &=-\frac{\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac{(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x}}{1-x} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b) f}\\ &=-\frac{\left (8 a^2+40 a b+35 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 (a+b) f}-\frac{\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac{(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 f}\\ &=-\frac{\left (8 a^2+40 a b+35 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 (a+b) f}-\frac{\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac{(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}+\frac{\left (8 a^2+40 a b+35 b^2\right ) \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 )}{8 b f}\\ &=\frac{\left (8 a^2+40 a b+35 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 \sqrt{a+b} f}-\frac{\left (8 a^2+40 a b+35 b^2\right ) \sqrt{a+b \sin ^2(e+f x)}}{8 (a+b) f}-\frac{\left (8 a^2+40 a b+35 b^2\right ) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 (a+b)^2 f}-\frac{(8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{8 (a+b)^2 f}+\frac{\sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{4 (a+b) f}\\ \end{align*}

Mathematica [A]  time = 2.18382, size = 160, normalized size = 0.73 \[ -\frac{\left (8 a^2+40 a b+35 b^2\right ) \left (\sqrt{a+b \sin ^2(e+f x)} \left (4 a+b \sin ^2(e+f x)+3 b\right )-3 (a+b)^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )\right )-6 (a+b) \sec ^4(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}+3 (8 a+9 b) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{24 f (a+b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x]^2)^(3/2)*Tan[e + f*x]^5,x]

[Out]

-(3*(8*a + 9*b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)^(5/2) - 6*(a + b)*Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2)^
(5/2) + (8*a^2 + 40*a*b + 35*b^2)*(-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)))/(24*(a + b)^2*f)

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Maple [B]  time = 3.613, size = 711, normalized size = 3.2 \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)^5,x)

[Out]

1/48*(16*b*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*cos(f*x+e)^6+(-64*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*a-1
60*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*b+24*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^4+168*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*b+369
*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*b^2+330*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^3+105*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^4+24*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^4+168*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*b+369*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*b^2+330*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^3+105*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^4)*cos(f*x+e)^4-6*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*(8*a+13*b)*cos(f*x+e)^
2+12*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*a+12*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(5/2)*b)/(a+b)^(5/2)/cos(f*x
+e)^4/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)^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 7.07125, size = 953, normalized size = 4.33 \begin{align*} \left [\frac{3 \,{\left (8 \, a^{2} + 40 \, a b + 35 \, b^{2}\right )} \sqrt{a + b} \cos \left (f x + e\right )^{4} \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 (8 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 16 \,{\left (2 \, a^{2} + 7 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 3 \,{\left (8 \, a^{2} + 21 \, a b + 13 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{48 \,{\left (a + b\right )} f \cos \left (f x + e\right )^{4}}, -\frac{3 \,{\left (8 \, a^{2} + 40 \, a b + 35 \, b^{2}\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 )^{4} -{\left (8 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{6} - 16 \,{\left (2 \, a^{2} + 7 \, a b + 5 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 3 \,{\left (8 \, a^{2} + 21 \, a b + 13 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 6 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{24 \,{\left (a + b\right )} f \cos \left (f x + e\right )^{4}}\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)^5,x, algorithm="fricas")

[Out]

[1/48*(3*(8*a^2 + 40*a*b + 35*b^2)*sqrt(a + b)*cos(f*x + e)^4*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*(8*(a*b + b^2)*cos(f*x + e)^6 - 16*(2*a^2 + 7*a*b + 5*b
^2)*cos(f*x + e)^4 - 3*(8*a^2 + 21*a*b + 13*b^2)*cos(f*x + e)^2 + 6*a^2 + 12*a*b + 6*b^2)*sqrt(-b*cos(f*x + e)
^2 + a + b))/((a + b)*f*cos(f*x + e)^4), -1/24*(3*(8*a^2 + 40*a*b + 35*b^2)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*
x + e)^2 + a + b)*sqrt(-a - b)/(a + b))*cos(f*x + e)^4 - (8*(a*b + b^2)*cos(f*x + e)^6 - 16*(2*a^2 + 7*a*b + 5
*b^2)*cos(f*x + e)^4 - 3*(8*a^2 + 21*a*b + 13*b^2)*cos(f*x + e)^2 + 6*a^2 + 12*a*b + 6*b^2)*sqrt(-b*cos(f*x +
e)^2 + a + b))/((a + b)*f*cos(f*x + e)^4)]

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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)**5,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 )^{5}\,{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)^5,x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*tan(f*x + e)^5, x)