3.1.76 \(\int (a+b \sin ^2(x))^3 \, dx\) [76]

Optimal. Leaf size=87 \[ \frac {1}{16} (2 a+b) \left (8 a^2+8 a b+5 b^2\right ) x-\frac {1}{48} b \left (64 a^2+54 a b+15 b^2\right ) \cos (x) \sin (x)-\frac {5}{24} b^2 (2 a+b) \cos (x) \sin ^3(x)-\frac {1}{6} b \cos (x) \sin (x) \left (a+b \sin ^2(x)\right )^2 \]

[Out]

1/16*(2*a+b)*(8*a^2+8*a*b+5*b^2)*x-1/48*b*(64*a^2+54*a*b+15*b^2)*cos(x)*sin(x)-5/24*b^2*(2*a+b)*cos(x)*sin(x)^
3-1/6*b*cos(x)*sin(x)*(a+b*sin(x)^2)^2

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 87, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3259, 3248} \begin {gather*} \frac {1}{16} x (2 a+b) \left (8 a^2+8 a b+5 b^2\right )-\frac {1}{48} b \left (64 a^2+54 a b+15 b^2\right ) \sin (x) \cos (x)-\frac {5}{24} b^2 (2 a+b) \sin ^3(x) \cos (x)-\frac {1}{6} b \sin (x) \cos (x) \left (a+b \sin ^2(x)\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[x]^2)^3,x]

[Out]

((2*a + b)*(8*a^2 + 8*a*b + 5*b^2)*x)/16 - (b*(64*a^2 + 54*a*b + 15*b^2)*Cos[x]*Sin[x])/48 - (5*b^2*(2*a + b)*
Cos[x]*Sin[x]^3)/24 - (b*Cos[x]*Sin[x]*(a + b*Sin[x]^2)^2)/6

Rule 3248

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(4*A
*(2*a + b) + B*(4*a + 3*b))*(x/8), x] + (-Simp[b*B*Cos[e + f*x]*(Sin[e + f*x]^3/(4*f)), x] - Simp[(4*A*b + B*(
4*a + 3*b))*Cos[e + f*x]*(Sin[e + f*x]/(8*f)), x]) /; FreeQ[{a, b, e, f, A, B}, x]

Rule 3259

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Simp[(-b)*Cos[e + f*x]*Sin[e + f*x]*((a + b*Si
n[e + f*x]^2)^(p - 1)/(2*f*p)), x] + Dist[1/(2*p), Int[(a + b*Sin[e + f*x]^2)^(p - 2)*Simp[a*(b + 2*a*p) + b*(
2*a + b)*(2*p - 1)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a + b, 0] && GtQ[p, 1]

Rubi steps

\begin {align*} \int \left (a+b \sin ^2(x)\right )^3 \, dx &=-\frac {1}{6} b \cos (x) \sin (x) \left (a+b \sin ^2(x)\right )^2+\frac {1}{6} \int \left (a+b \sin ^2(x)\right ) \left (a (6 a+b)+5 b (2 a+b) \sin ^2(x)\right ) \, dx\\ &=\frac {1}{16} (2 a+b) \left (8 a^2+8 a b+5 b^2\right ) x-\frac {1}{48} b \left (64 a^2+54 a b+15 b^2\right ) \cos (x) \sin (x)-\frac {5}{24} b^2 (2 a+b) \cos (x) \sin ^3(x)-\frac {1}{6} b \cos (x) \sin (x) \left (a+b \sin ^2(x)\right )^2\\ \end {align*}

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.07, size = 80, normalized size = 0.92 \begin {gather*} \frac {1}{192} \left (12 (2 a+b) \left (8 a^2+8 a b+5 b^2\right ) x+9 i b (4 i a+(1+2 i) b) (4 a+(2+i) b) \sin (2 x)+9 b^2 (2 a+b) \sin (4 x)-b^3 \sin (6 x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[x]^2)^3,x]

[Out]

(12*(2*a + b)*(8*a^2 + 8*a*b + 5*b^2)*x + (9*I)*b*((4*I)*a + (1 + 2*I)*b)*(4*a + (2 + I)*b)*Sin[2*x] + 9*b^2*(
2*a + b)*Sin[4*x] - b^3*Sin[6*x])/192

________________________________________________________________________________________

Maple [A]
time = 0.21, size = 73, normalized size = 0.84

method	result	size
default	\(b^{3} \left (-\frac {\left (\sin ^{5}\left (x \right )+\frac {5 \left (\sin ^{3}\left (x \right )\right )}{4}+\frac {15 \sin \left (x \right )}{8}\right ) \cos \left (x \right )}{6}+\frac {5 x}{16}\right )+3 a \,b^{2} \left (-\frac {\left (\sin ^{3}\left (x \right )+\frac {3 \sin \left (x \right )}{2}\right ) \cos \left (x \right )}{4}+\frac {3 x}{8}\right )+3 a^{2} b \left (-\frac {\sin \left (x \right ) \cos \left (x \right )}{2}+\frac {x}{2}\right )+a^{3} x\)	\(73\)
risch	\(a^{3} x +\frac {3 x \,a^{2} b}{2}+\frac {9 a \,b^{2} x}{8}+\frac {5 b^{3} x}{16}-\frac {b^{3} \sin \left (6 x \right )}{192}+\frac {3 \sin \left (4 x \right ) a \,b^{2}}{32}+\frac {3 \sin \left (4 x \right ) b^{3}}{64}-\frac {3 \sin \left (2 x \right ) a^{2} b}{4}-\frac {3 \sin \left (2 x \right ) a \,b^{2}}{4}-\frac {15 \sin \left (2 x \right ) b^{3}}{64}\)	\(84\)
norman	\(\frac {\left (-9 a^{2} b -\frac {51}{4} a \,b^{2}-\frac {85}{24} b^{3}\right ) \left (\tan ^{3}\left (\frac {x}{2}\right )\right )+\left (-6 a^{2} b -\frac {21}{2} a \,b^{2}-\frac {33}{4} b^{3}\right ) \left (\tan ^{5}\left (\frac {x}{2}\right )\right )+\left (-3 a^{2} b -\frac {9}{4} a \,b^{2}-\frac {5}{8} b^{3}\right ) \tan \left (\frac {x}{2}\right )+\left (3 a^{2} b +\frac {9}{4} a \,b^{2}+\frac {5}{8} b^{3}\right ) \left (\tan ^{11}\left (\frac {x}{2}\right )\right )+\left (6 a^{2} b +\frac {21}{2} a \,b^{2}+\frac {33}{4} b^{3}\right ) \left (\tan ^{7}\left (\frac {x}{2}\right )\right )+\left (9 a^{2} b +\frac {51}{4} a \,b^{2}+\frac {85}{24} b^{3}\right ) \left (\tan ^{9}\left (\frac {x}{2}\right )\right )+\left (a^{3}+\frac {3}{2} a^{2} b +\frac {9}{8} a \,b^{2}+\frac {5}{16} b^{3}\right ) x +\left (a^{3}+\frac {3}{2} a^{2} b +\frac {9}{8} a \,b^{2}+\frac {5}{16} b^{3}\right ) x \left (\tan ^{12}\left (\frac {x}{2}\right )\right )+\left (6 a^{3}+9 a^{2} b +\frac {27}{4} a \,b^{2}+\frac {15}{8} b^{3}\right ) x \left (\tan ^{2}\left (\frac {x}{2}\right )\right )+\left (6 a^{3}+9 a^{2} b +\frac {27}{4} a \,b^{2}+\frac {15}{8} b^{3}\right ) x \left (\tan ^{10}\left (\frac {x}{2}\right )\right )+\left (15 a^{3}+\frac {45}{2} a^{2} b +\frac {135}{8} a \,b^{2}+\frac {75}{16} b^{3}\right ) x \left (\tan ^{4}\left (\frac {x}{2}\right )\right )+\left (15 a^{3}+\frac {45}{2} a^{2} b +\frac {135}{8} a \,b^{2}+\frac {75}{16} b^{3}\right ) x \left (\tan ^{8}\left (\frac {x}{2}\right )\right )+\left (20 a^{3}+30 a^{2} b +\frac {45}{2} a \,b^{2}+\frac {25}{4} b^{3}\right ) x \left (\tan ^{6}\left (\frac {x}{2}\right )\right )}{\left (1+\tan ^{2}\left (\frac {x}{2}\right )\right )^{6}}\)	\(368\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(x)^2)^3,x,method=_RETURNVERBOSE)

[Out]

b^3*(-1/6*(sin(x)^5+5/4*sin(x)^3+15/8*sin(x))*cos(x)+5/16*x)+3*a*b^2*(-1/4*(sin(x)^3+3/2*sin(x))*cos(x)+3/8*x)
+3*a^2*b*(-1/2*sin(x)*cos(x)+1/2*x)+a^3*x

________________________________________________________________________________________

Maxima [A]
time = 0.29, size = 71, normalized size = 0.82 \begin {gather*} \frac {1}{192} \, {\left (4 \, \sin \left (2 \, x\right )^{3} + 60 \, x + 9 \, \sin \left (4 \, x\right ) - 48 \, \sin \left (2 \, x\right )\right )} b^{3} + \frac {3}{32} \, a b^{2} {\left (12 \, x + \sin \left (4 \, x\right ) - 8 \, \sin \left (2 \, x\right )\right )} + \frac {3}{4} \, a^{2} b {\left (2 \, x - \sin \left (2 \, x\right )\right )} + a^{3} x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)^2)^3,x, algorithm="maxima")

[Out]

1/192*(4*sin(2*x)^3 + 60*x + 9*sin(4*x) - 48*sin(2*x))*b^3 + 3/32*a*b^2*(12*x + sin(4*x) - 8*sin(2*x)) + 3/4*a
^2*b*(2*x - sin(2*x)) + a^3*x

________________________________________________________________________________________

Fricas [A]
time = 0.40, size = 81, normalized size = 0.93 \begin {gather*} \frac {1}{16} \, {\left (16 \, a^{3} + 24 \, a^{2} b + 18 \, a b^{2} + 5 \, b^{3}\right )} x - \frac {1}{48} \, {\left (8 \, b^{3} \cos \left (x\right )^{5} - 2 \, {\left (18 \, a b^{2} + 13 \, b^{3}\right )} \cos \left (x\right )^{3} + 3 \, {\left (24 \, a^{2} b + 30 \, a b^{2} + 11 \, b^{3}\right )} \cos \left (x\right )\right )} \sin \left (x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a^3 + 24*a^2*b + 18*a*b^2 + 5*b^3)*x - 1/48*(8*b^3*cos(x)^5 - 2*(18*a*b^2 + 13*b^3)*cos(x)^3 + 3*(24*
a^2*b + 30*a*b^2 + 11*b^3)*cos(x))*sin(x)

________________________________________________________________________________________

Sympy [B] Leaf count of result is larger than twice the leaf count of optimal. 246 vs. \(2 (88) = 176\).
time = 0.32, size = 246, normalized size = 2.83 \begin {gather*} a^{3} x + \frac {3 a^{2} b x \sin ^{2}{\left (x \right )}}{2} + \frac {3 a^{2} b x \cos ^{2}{\left (x \right )}}{2} - \frac {3 a^{2} b \sin {\left (x \right )} \cos {\left (x \right )}}{2} + \frac {9 a b^{2} x \sin ^{4}{\left (x \right )}}{8} + \frac {9 a b^{2} x \sin ^{2}{\left (x \right )} \cos ^{2}{\left (x \right )}}{4} + \frac {9 a b^{2} x \cos ^{4}{\left (x \right )}}{8} - \frac {15 a b^{2} \sin ^{3}{\left (x \right )} \cos {\left (x \right )}}{8} - \frac {9 a b^{2} \sin {\left (x \right )} \cos ^{3}{\left (x \right )}}{8} + \frac {5 b^{3} x \sin ^{6}{\left (x \right )}}{16} + \frac {15 b^{3} x \sin ^{4}{\left (x \right )} \cos ^{2}{\left (x \right )}}{16} + \frac {15 b^{3} x \sin ^{2}{\left (x \right )} \cos ^{4}{\left (x \right )}}{16} + \frac {5 b^{3} x \cos ^{6}{\left (x \right )}}{16} - \frac {11 b^{3} \sin ^{5}{\left (x \right )} \cos {\left (x \right )}}{16} - \frac {5 b^{3} \sin ^{3}{\left (x \right )} \cos ^{3}{\left (x \right )}}{6} - \frac {5 b^{3} \sin {\left (x \right )} \cos ^{5}{\left (x \right )}}{16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(x)**2)**3,x)

[Out]

a**3*x + 3*a**2*b*x*sin(x)**2/2 + 3*a**2*b*x*cos(x)**2/2 - 3*a**2*b*sin(x)*cos(x)/2 + 9*a*b**2*x*sin(x)**4/8 +
9*a*b**2*x*sin(x)**2*cos(x)**2/4 + 9*a*b**2*x*cos(x)**4/8 - 15*a*b**2*sin(x)**3*cos(x)/8 - 9*a*b**2*sin(x)*co
s(x)**3/8 + 5*b**3*x*sin(x)**6/16 + 15*b**3*x*sin(x)**4*cos(x)**2/16 + 15*b**3*x*sin(x)**2*cos(x)**4/16 + 5*b*
*3*x*cos(x)**6/16 - 11*b**3*sin(x)**5*cos(x)/16 - 5*b**3*sin(x)**3*cos(x)**3/6 - 5*b**3*sin(x)*cos(x)**5/16

________________________________________________________________________________________

Giac [A]
time = 0.42, size = 76, normalized size = 0.87 \begin {gather*} -\frac {1}{192} \, b^{3} \sin \left (6 \, x\right ) + \frac {1}{16} \, {\left (16 \, a^{3} + 24 \, a^{2} b + 18 \, a b^{2} + 5 \, b^{3}\right )} x + \frac {3}{64} \, {\left (2 \, a b^{2} + b^{3}\right )} \sin \left (4 \, x\right ) - \frac {3}{64} \, {\left (16 \, a^{2} b + 16 \, a b^{2} + 5 \, b^{3}\right )} \sin \left (2 \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/192*b^3*sin(6*x) + 1/16*(16*a^3 + 24*a^2*b + 18*a*b^2 + 5*b^3)*x + 3/64*(2*a*b^2 + b^3)*sin(4*x) - 3/64*(16
*a^2*b + 16*a*b^2 + 5*b^3)*sin(2*x)

________________________________________________________________________________________

Mupad [B]
time = 14.13, size = 118, normalized size = 1.36 \begin {gather*} a^3\,x+\frac {5\,b^3\,x}{16}-\frac {\left (72\,a^2\,b+90\,a\,b^2+33\,b^3\right )\,{\mathrm {tan}\left (x\right )}^5+\left (144\,a^2\,b+144\,a\,b^2+40\,b^3\right )\,{\mathrm {tan}\left (x\right )}^3+\left (72\,a^2\,b+54\,a\,b^2+15\,b^3\right )\,\mathrm {tan}\left (x\right )}{48\,{\mathrm {tan}\left (x\right )}^6+144\,{\mathrm {tan}\left (x\right )}^4+144\,{\mathrm {tan}\left (x\right )}^2+48}+\frac {9\,a\,b^2\,x}{8}+\frac {3\,a^2\,b\,x}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(x)^2)^3,x)

[Out]

a^3*x + (5*b^3*x)/16 - (tan(x)^5*(90*a*b^2 + 72*a^2*b + 33*b^3) + tan(x)^3*(144*a*b^2 + 144*a^2*b + 40*b^3) +
tan(x)*(54*a*b^2 + 72*a^2*b + 15*b^3))/(144*tan(x)^2 + 144*tan(x)^4 + 48*tan(x)^6 + 48) + (9*a*b^2*x)/8 + (3*a
^2*b*x)/2

________________________________________________________________________________________