3.490 \(\int \cos ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=247 \[ \frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{105 b d}+\frac {4 a \left (3 a^2+29 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 \left (3 a^4+2 a^2 b^2-5 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {a+b \sin (c+d x)}}-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d} \]

[Out]

-2/7*b*cos(d*x+c)^3*(a+b*sin(d*x+c))^(1/2)/d+2/105*cos(d*x+c)*(3*a^2+5*b^2+24*a*b*sin(d*x+c))*(a+b*sin(d*x+c))
^(1/2)/b/d-4/105*a*(3*a^2+29*b^2)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(
1/2*c+1/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*(a+b*sin(d*x+c))^(1/2)/b^2/d/((a+b*sin(d*x+c))/(a+b))^(1/2)+4/1
05*(3*a^4+2*a^2*b^2-5*b^4)*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticF(cos(1/2*c+1
/4*Pi+1/2*d*x),2^(1/2)*(b/(a+b))^(1/2))*((a+b*sin(d*x+c))/(a+b))^(1/2)/b^2/d/(a+b*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.46, antiderivative size = 247, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {2692, 2865, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+24 a b \sin (c+d x)+5 b^2\right )}{105 b d}-\frac {4 \left (2 a^2 b^2+3 a^4-5 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {4 a \left (3 a^2+29 b^2\right ) \sqrt {a+b \sin (c+d x)} E\left (\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*b*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]])/(7*d) + (4*a*(3*a^2 + 29*b^2)*EllipticE[(c - Pi/2 + d*x)/2, (2*
b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(105*b^2*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) - (4*(3*a^4 + 2*a^2*b^2 -
5*b^4)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a + b)])/(105*b^2*d*Sqrt[a + b
*Sin[c + d*x]]) + (2*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(3*a^2 + 5*b^2 + 24*a*b*Sin[c + d*x]))/(105*b*d)

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2692

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

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2865

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Simp[(g*(g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1)*(b*c*(m + p + 1) -
a*d*p + b*d*(m + p)*Sin[e + f*x]))/(b^2*f*(m + p)*(m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(m + p)*(m + p +
1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*Simp[b*(a*d*m + b*c*(m + p + 1)) + (a*b*c*(m + p + 1
) - d*(a^2*p - b^2*(m + p)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && NeQ[a^2 - b^2,
0] && GtQ[p, 1] && NeQ[m + p, 0] && NeQ[m + p + 1, 0] && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}+\frac {2}{7} \int \frac {\cos ^2(c+d x) \left (\frac {7 a^2}{2}+\frac {b^2}{2}+4 a b \sin (c+d x)\right )}{\sqrt {a+b \sin (c+d x)}} \, dx\\ &=-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+5 b^2+24 a b \sin (c+d x)\right )}{105 b d}+\frac {8 \int \frac {\frac {1}{4} b^2 \left (27 a^2+5 b^2\right )+\frac {1}{4} a b \left (3 a^2+29 b^2\right ) \sin (c+d x)}{\sqrt {a+b \sin (c+d x)}} \, dx}{105 b^2}\\ &=-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+5 b^2+24 a b \sin (c+d x)\right )}{105 b d}+\frac {1}{105} \left (2 a \left (29+\frac {3 a^2}{b^2}\right )\right ) \int \sqrt {a+b \sin (c+d x)} \, dx-\frac {\left (2 \left (3 a^4+2 a^2 b^2-5 b^4\right )\right ) \int \frac {1}{\sqrt {a+b \sin (c+d x)}} \, dx}{105 b^2}\\ &=-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+5 b^2+24 a b \sin (c+d x)\right )}{105 b d}+\frac {\left (2 a \left (29+\frac {3 a^2}{b^2}\right ) \sqrt {a+b \sin (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}} \, dx}{105 \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {\left (2 \left (3 a^4+2 a^2 b^2-5 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \sin (c+d x)}{a+b}}} \, dx}{105 b^2 \sqrt {a+b \sin (c+d x)}}\\ &=-\frac {2 b \cos ^3(c+d x) \sqrt {a+b \sin (c+d x)}}{7 d}+\frac {4 a \left (29+\frac {3 a^2}{b^2}\right ) E\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {a+b \sin (c+d x)}}{105 d \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}-\frac {4 \left (3 a^4+2 a^2 b^2-5 b^4\right ) F\left (\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )|\frac {2 b}{a+b}\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}}}{105 b^2 d \sqrt {a+b \sin (c+d x)}}+\frac {2 \cos (c+d x) \sqrt {a+b \sin (c+d x)} \left (3 a^2+5 b^2+24 a b \sin (c+d x)\right )}{105 b d}\\ \end {align*}

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Mathematica [A]  time = 1.04, size = 222, normalized size = 0.90 \[ \frac {8 \left (3 a^4+2 a^2 b^2-5 b^4\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} F\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )+b \cos (c+d x) \left (12 a^3+b \left (108 a^2+5 b^2\right ) \sin (c+d x)-78 a b^2 \cos (2 (c+d x))+38 a b^2-15 b^3 \sin (3 (c+d x))\right )-8 a \left (3 a^3+3 a^2 b+29 a b^2+29 b^3\right ) \sqrt {\frac {a+b \sin (c+d x)}{a+b}} E\left (\frac {1}{4} (-2 c-2 d x+\pi )|\frac {2 b}{a+b}\right )}{210 b^2 d \sqrt {a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-8*a*(3*a^3 + 3*a^2*b + 29*a*b^2 + 29*b^3)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c
+ d*x])/(a + b)] + 8*(3*a^4 + 2*a^2*b^2 - 5*b^4)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)]*Sqrt[(a + b*S
in[c + d*x])/(a + b)] + b*Cos[c + d*x]*(12*a^3 + 38*a*b^2 - 78*a*b^2*Cos[2*(c + d*x)] + b*(108*a^2 + 5*b^2)*Si
n[c + d*x] - 15*b^3*Sin[3*(c + d*x)]))/(210*b^2*d*Sqrt[a + b*Sin[c + d*x]])

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fricas [F]  time = 0.78, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a \cos \left (d x + c\right )^{2}\right )} \sqrt {b \sin \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*cos(d*x + c)^2*sin(d*x + c) + a*cos(d*x + c)^2)*sqrt(b*sin(d*x + c) + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^2, x)

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maple [B]  time = 0.72, size = 943, normalized size = 3.82 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+b*sin(d*x+c))^(3/2),x)

[Out]

2/105*(-15*b^5*sin(d*x+c)^5+6*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*
b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^4*b+48*((a+b*sin(d*x+c))/(a-b))
^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2
),((a-b)/(a+b))^(1/2))*a^3*b^2+4*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c
))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^2*b^3-48*((a+b*sin(d*x+c))/(
a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))
^(1/2),((a-b)/(a+b))^(1/2))*a*b^4-10*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d
*x+c))*b/(a-b))^(1/2)*EllipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^5-6*((a+b*sin(d*x+c))/(a
-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^
(1/2),((a-b)/(a+b))^(1/2))*a^5-52*((a+b*sin(d*x+c))/(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+
c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^3*b^2+58*((a+b*sin(d*x+c))/
(a-b))^(1/2)*(-(sin(d*x+c)-1)*b/(a+b))^(1/2)*(-(1+sin(d*x+c))*b/(a-b))^(1/2)*EllipticE(((a+b*sin(d*x+c))/(a-b)
)^(1/2),((a-b)/(a+b))^(1/2))*a*b^4-39*a*b^4*sin(d*x+c)^4-27*a^2*b^3*sin(d*x+c)^3+25*b^5*sin(d*x+c)^3-3*a^3*b^2
*sin(d*x+c)^2+49*a*b^4*sin(d*x+c)^2+27*a^2*b^3*sin(d*x+c)-10*b^5*sin(d*x+c)+3*a^3*b^2-10*a*b^4)/b^3/cos(d*x+c)
/(a+b*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(d*x + c) + a)^(3/2)*cos(d*x + c)^2, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(a + b*sin(c + d*x))^(3/2),x)

[Out]

int(cos(c + d*x)^2*(a + b*sin(c + d*x))^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}} \cos ^{2}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+b*sin(d*x+c))**(3/2),x)

[Out]

Integral((a + b*sin(c + d*x))**(3/2)*cos(c + d*x)**2, x)

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