3.489 \(\int \cos ^4(c+d x) (a+b \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=329 \[ \frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (a^2+28 a b \sin (c+d x)+3 b^2\right )}{231 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (-3 a b \left (a^2+31 b^2\right ) \sin (c+d x)-21 a^2 b^2+4 a^4-15 b^4\right )}{1155 b^3 d}+\frac{8 \left (-25 a^4 b^2+6 a^2 b^4+4 a^6+15 b^6\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{1155 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{32 a \left (-6 a^2 b^2+a^4-27 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{1155 b^4 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 b \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{11 d} \]

[Out]

(-2*b*Cos[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]])/(11*d) - (32*a*(a^4 - 6*a^2*b^2 - 27*b^4)*EllipticE[(c - Pi/2 +
d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(1155*b^4*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (8*(4*a^6
- 25*a^4*b^2 + 6*a^2*b^4 + 15*b^6)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a +
b)])/(1155*b^4*d*Sqrt[a + b*Sin[c + d*x]]) + (2*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(a^2 + 3*b^2 + 28*a*b
*Sin[c + d*x]))/(231*b*d) - (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(4*a^4 - 21*a^2*b^2 - 15*b^4 - 3*a*b*(a^2
+ 31*b^2)*Sin[c + d*x]))/(1155*b^3*d)

________________________________________________________________________________________

Rubi [A]  time = 0.691636, antiderivative size = 329, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (a^2+28 a b \sin (c+d x)+3 b^2\right )}{231 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (-3 a b \left (a^2+31 b^2\right ) \sin (c+d x)-21 a^2 b^2+4 a^4-15 b^4\right )}{1155 b^3 d}+\frac{8 \left (-25 a^4 b^2+6 a^2 b^4+4 a^6+15 b^6\right ) \sqrt{\frac{a+b \sin (c+d x)}{a+b}} F\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{1155 b^4 d \sqrt{a+b \sin (c+d x)}}-\frac{32 a \left (-6 a^2 b^2+a^4-27 b^4\right ) \sqrt{a+b \sin (c+d x)} E\left (\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )|\frac{2 b}{a+b}\right )}{1155 b^4 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}-\frac{2 b \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{11 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(-2*b*Cos[c + d*x]^5*Sqrt[a + b*Sin[c + d*x]])/(11*d) - (32*a*(a^4 - 6*a^2*b^2 - 27*b^4)*EllipticE[(c - Pi/2 +
d*x)/2, (2*b)/(a + b)]*Sqrt[a + b*Sin[c + d*x]])/(1155*b^4*d*Sqrt[(a + b*Sin[c + d*x])/(a + b)]) + (8*(4*a^6
- 25*a^4*b^2 + 6*a^2*b^4 + 15*b^6)*EllipticF[(c - Pi/2 + d*x)/2, (2*b)/(a + b)]*Sqrt[(a + b*Sin[c + d*x])/(a +
b)])/(1155*b^4*d*Sqrt[a + b*Sin[c + d*x]]) + (2*Cos[c + d*x]^3*Sqrt[a + b*Sin[c + d*x]]*(a^2 + 3*b^2 + 28*a*b
*Sin[c + d*x]))/(231*b*d) - (4*Cos[c + d*x]*Sqrt[a + b*Sin[c + d*x]]*(4*a^4 - 21*a^2*b^2 - 15*b^4 - 3*a*b*(a^2
+ 31*b^2)*Sin[c + d*x]))/(1155*b^3*d)

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 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]

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 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 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 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 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]

Rubi steps

\begin{align*} \int \cos ^4(c+d x) (a+b \sin (c+d x))^{3/2} \, dx &=-\frac{2 b \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{11 d}+\frac{2}{11} \int \frac{\cos ^4(c+d x) \left (\frac{11 a^2}{2}+\frac{b^2}{2}+6 a b \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx\\ &=-\frac{2 b \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{11 d}+\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (a^2+3 b^2+28 a b \sin (c+d x)\right )}{231 b d}+\frac{8 \int \frac{\cos ^2(c+d x) \left (\frac{3}{4} b^2 \left (29 a^2+3 b^2\right )+\frac{3}{4} a b \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{\sqrt{a+b \sin (c+d x)}} \, dx}{231 b^2}\\ &=-\frac{2 b \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{11 d}+\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (a^2+3 b^2+28 a b \sin (c+d x)\right )}{231 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a^4-21 a^2 b^2-15 b^4-3 a b \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{1155 b^3 d}+\frac{32 \int \frac{-\frac{3}{8} b^2 \left (a^4-114 a^2 b^2-15 b^4\right )-\frac{3}{2} a b \left (a^4-6 a^2 b^2-27 b^4\right ) \sin (c+d x)}{\sqrt{a+b \sin (c+d x)}} \, dx}{3465 b^4}\\ &=-\frac{2 b \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{11 d}+\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (a^2+3 b^2+28 a b \sin (c+d x)\right )}{231 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a^4-21 a^2 b^2-15 b^4-3 a b \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{1155 b^3 d}-\frac{\left (16 a \left (a^4-6 a^2 b^2-27 b^4\right )\right ) \int \sqrt{a+b \sin (c+d x)} \, dx}{1155 b^4}+\frac{\left (4 \left (4 a^6-25 a^4 b^2+6 a^2 b^4+15 b^6\right )\right ) \int \frac{1}{\sqrt{a+b \sin (c+d x)}} \, dx}{1155 b^4}\\ &=-\frac{2 b \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{11 d}+\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (a^2+3 b^2+28 a b \sin (c+d x)\right )}{231 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a^4-21 a^2 b^2-15 b^4-3 a b \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{1155 b^3 d}-\frac{\left (16 a \left (a^4-6 a^2 b^2-27 b^4\right ) \sqrt{a+b \sin (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \sin (c+d x)}{a+b}} \, dx}{1155 b^4 \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{\left (4 \left (4 a^6-25 a^4 b^2+6 a^2 b^4+15 b^6\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}{1155 b^4 \sqrt{a+b \sin (c+d x)}}\\ &=-\frac{2 b \cos ^5(c+d x) \sqrt{a+b \sin (c+d x)}}{11 d}-\frac{32 a \left (a^4-6 a^2 b^2-27 b^4\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)}}{1155 b^4 d \sqrt{\frac{a+b \sin (c+d x)}{a+b}}}+\frac{8 \left (4 a^6-25 a^4 b^2+6 a^2 b^4+15 b^6\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}}}{1155 b^4 d \sqrt{a+b \sin (c+d x)}}+\frac{2 \cos ^3(c+d x) \sqrt{a+b \sin (c+d x)} \left (a^2+3 b^2+28 a b \sin (c+d x)\right )}{231 b d}-\frac{4 \cos (c+d x) \sqrt{a+b \sin (c+d x)} \left (4 a^4-21 a^2 b^2-15 b^4-3 a b \left (a^2+31 b^2\right ) \sin (c+d x)\right )}{1155 b^3 d}\\ \end{align*}

Mathematica [A]  time = 1.05822, size = 278, normalized size = 0.84 \[ \frac{64 \sqrt{\frac{a+b \sin (c+d x)}{a+b}} \left (b^2 \left (-114 a^2 b^2+a^4-15 b^4\right ) F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )+4 \left (-6 a^3 b^2+a^5-27 a b^4\right ) \left ((a+b) E\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{4} (-2 c-2 d x+\pi )|\frac{2 b}{a+b}\right )\right )\right )-b (a+b \sin (c+d x)) \left (-16 a b \left (3 a^2+128 b^2\right ) \sin (2 (c+d x))+5 b^2 \left (93 b^2-4 a^2\right ) \cos (3 (c+d x))+2 \left (-366 a^2 b^2+64 a^4+195 b^4\right ) \cos (c+d x)-280 a b^3 \sin (4 (c+d x))+105 b^4 \cos (5 (c+d x))\right )}{9240 b^4 d \sqrt{a+b \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^4*(a + b*Sin[c + d*x])^(3/2),x]

[Out]

(64*(b^2*(a^4 - 114*a^2*b^2 - 15*b^4)*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] + 4*(a^5 - 6*a^3*b^2 - 2
7*a*b^4)*((a + b)*EllipticE[(-2*c + Pi - 2*d*x)/4, (2*b)/(a + b)] - a*EllipticF[(-2*c + Pi - 2*d*x)/4, (2*b)/(
a + b)]))*Sqrt[(a + b*Sin[c + d*x])/(a + b)] - b*(a + b*Sin[c + d*x])*(2*(64*a^4 - 366*a^2*b^2 + 195*b^4)*Cos[
c + d*x] + 5*b^2*(-4*a^2 + 93*b^2)*Cos[3*(c + d*x)] + 105*b^4*Cos[5*(c + d*x)] - 16*a*b*(3*a^2 + 128*b^2)*Sin[
2*(c + d*x)] - 280*a*b^3*Sin[4*(c + d*x)]))/(9240*b^4*d*Sqrt[a + b*Sin[c + d*x]])

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Maple [B]  time = 0.607, size = 1355, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^4*(a+b*sin(d*x+c))^(3/2),x)

[Out]

-2/1155*(-245*a*b^6*sin(d*x+c)^6+24*((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^5+60*a*b^6-47*a^3*b^4
+8*a^5*b^2-105*b^7*sin(d*x+c)^7+300*b^7*sin(d*x+c)^5-255*b^7*sin(d*x+c)^3+60*b^7*sin(d*x+c)-145*a^2*b^5*sin(d*
x+c)^5+2*a^4*b^3*sin(d*x+c)-581*a*b^6*sin(d*x+c)^2-16*((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^7+60*((
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*s
in(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*b^7+16*((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^6*b-12*((
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*s
in(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a^5*b^2-100*((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^
3-360*((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^4+336*((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^4-432*((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)*E
llipticE(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^6-8*a^5*b^2*sin(d*x+c)^2+a^3*b^4*sin(d*x+c)^4
+46*a^3*b^4*sin(d*x+c)^2-2*a^4*b^3*sin(d*x+c)^3+518*a^2*b^5*sin(d*x+c)^3+766*a*b^6*sin(d*x+c)^4-373*a^2*b^5*si
n(d*x+c)+372*((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)*El
lipticF(((a+b*sin(d*x+c))/(a-b))^(1/2),((a-b)/(a+b))^(1/2))*a*b^6+112*((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*b^2)/b^5/cos(d*x+c)/(a+b*sin(d*x+c))^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(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)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) + a \cos \left (d x + c\right )^{4}\right )} \sqrt{b \sin \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(a+b*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

integral((b*cos(d*x + c)^4*sin(d*x + c) + a*cos(d*x + c)^4)*sqrt(b*sin(d*x + c) + a), x)

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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(cos(d*x+c)**4*(a+b*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cos \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^4*(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)^4, x)