3.538 \(\int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx\)

Optimal. Leaf size=138 \[ -\frac {64 a^3 (7 c+5 d) \cos (e+f x)}{105 f \sqrt {a \sin (e+f x)+a}}-\frac {16 a^2 (7 c+5 d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{105 f}-\frac {2 a (7 c+5 d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f} \]

[Out]

-2/35*a*(7*c+5*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(3/2)/f-2/7*d*cos(f*x+e)*(a+a*sin(f*x+e))^(5/2)/f-64/105*a^3*(7*
c+5*d)*cos(f*x+e)/f/(a+a*sin(f*x+e))^(1/2)-16/105*a^2*(7*c+5*d)*cos(f*x+e)*(a+a*sin(f*x+e))^(1/2)/f

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Rubi [A]  time = 0.11, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2751, 2647, 2646} \[ -\frac {64 a^3 (7 c+5 d) \cos (e+f x)}{105 f \sqrt {a \sin (e+f x)+a}}-\frac {16 a^2 (7 c+5 d) \cos (e+f x) \sqrt {a \sin (e+f x)+a}}{105 f}-\frac {2 a (7 c+5 d) \cos (e+f x) (a \sin (e+f x)+a)^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a \sin (e+f x)+a)^{5/2}}{7 f} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x]),x]

[Out]

(-64*a^3*(7*c + 5*d)*Cos[e + f*x])/(105*f*Sqrt[a + a*Sin[e + f*x]]) - (16*a^2*(7*c + 5*d)*Cos[e + f*x]*Sqrt[a
+ a*Sin[e + f*x]])/(105*f) - (2*a*(7*c + 5*d)*Cos[e + f*x]*(a + a*Sin[e + f*x])^(3/2))/(35*f) - (2*d*Cos[e + f
*x]*(a + a*Sin[e + f*x])^(5/2))/(7*f)

Rule 2646

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

Rule 2647

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^(n -
1))/(d*n), x] + Dist[(a*(2*n - 1))/n, Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && Eq
Q[a^2 - b^2, 0] && IGtQ[n - 1/2, 0]

Rule 2751

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

Rubi steps

\begin {align*} \int (a+a \sin (e+f x))^{5/2} (c+d \sin (e+f x)) \, dx &=-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}+\frac {1}{7} (7 c+5 d) \int (a+a \sin (e+f x))^{5/2} \, dx\\ &=-\frac {2 a (7 c+5 d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}+\frac {1}{35} (8 a (7 c+5 d)) \int (a+a \sin (e+f x))^{3/2} \, dx\\ &=-\frac {16 a^2 (7 c+5 d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 a (7 c+5 d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}+\frac {1}{105} \left (32 a^2 (7 c+5 d)\right ) \int \sqrt {a+a \sin (e+f x)} \, dx\\ &=-\frac {64 a^3 (7 c+5 d) \cos (e+f x)}{105 f \sqrt {a+a \sin (e+f x)}}-\frac {16 a^2 (7 c+5 d) \cos (e+f x) \sqrt {a+a \sin (e+f x)}}{105 f}-\frac {2 a (7 c+5 d) \cos (e+f x) (a+a \sin (e+f x))^{3/2}}{35 f}-\frac {2 d \cos (e+f x) (a+a \sin (e+f x))^{5/2}}{7 f}\\ \end {align*}

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Mathematica [A]  time = 1.50, size = 119, normalized size = 0.86 \[ -\frac {a^2 \sqrt {a (\sin (e+f x)+1)} \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) ((392 c+505 d) \sin (e+f x)-6 (7 c+20 d) \cos (2 (e+f x))+1246 c-15 d \sin (3 (e+f x))+1040 d)}{210 f \left (\sin \left (\frac {1}{2} (e+f x)\right )+\cos \left (\frac {1}{2} (e+f x)\right )\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[e + f*x])^(5/2)*(c + d*Sin[e + f*x]),x]

[Out]

-1/210*(a^2*(Cos[(e + f*x)/2] - Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])]*(1246*c + 1040*d - 6*(7*c + 20*d)
*Cos[2*(e + f*x)] + (392*c + 505*d)*Sin[e + f*x] - 15*d*Sin[3*(e + f*x)]))/(f*(Cos[(e + f*x)/2] + Sin[(e + f*x
)/2]))

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fricas [A]  time = 0.44, size = 206, normalized size = 1.49 \[ \frac {2 \, {\left (15 \, a^{2} d \cos \left (f x + e\right )^{4} + 3 \, {\left (7 \, a^{2} c + 20 \, a^{2} d\right )} \cos \left (f x + e\right )^{3} - 224 \, a^{2} c - 160 \, a^{2} d - {\left (77 \, a^{2} c + 85 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (161 \, a^{2} c + 145 \, a^{2} d\right )} \cos \left (f x + e\right ) + {\left (15 \, a^{2} d \cos \left (f x + e\right )^{3} + 224 \, a^{2} c + 160 \, a^{2} d - 3 \, {\left (7 \, a^{2} c + 15 \, a^{2} d\right )} \cos \left (f x + e\right )^{2} - 2 \, {\left (49 \, a^{2} c + 65 \, a^{2} d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a}}{105 \, {\left (f \cos \left (f x + e\right ) + f \sin \left (f x + e\right ) + f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e)),x, algorithm="fricas")

[Out]

2/105*(15*a^2*d*cos(f*x + e)^4 + 3*(7*a^2*c + 20*a^2*d)*cos(f*x + e)^3 - 224*a^2*c - 160*a^2*d - (77*a^2*c + 8
5*a^2*d)*cos(f*x + e)^2 - 2*(161*a^2*c + 145*a^2*d)*cos(f*x + e) + (15*a^2*d*cos(f*x + e)^3 + 224*a^2*c + 160*
a^2*d - 3*(7*a^2*c + 15*a^2*d)*cos(f*x + e)^2 - 2*(49*a^2*c + 65*a^2*d)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin
(f*x + e) + a)/(f*cos(f*x + e) + f*sin(f*x + e) + f)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)sqrt(2*a)*(2*f*(6*a^2*c*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+4*a
^2*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*sin(1/4*(2*f*x-pi)+1/2*exp(1))/(2*f)^2-8*f*(16*a^2*c*sign(cos(1/2*(f*
x+exp(1))-1/4*pi))+14*a^2*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(2*f*x+2*exp(1)+pi))/(8*f)^2-24*f*(16*
a^2*c*sign(cos(1/2*(f*x+exp(1))-1/4*pi))+14*a^2*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*cos(1/4*(6*f*x+6*exp(1)-
pi))/(24*f)^2+12*f*(-2*a^2*c*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-4*a^2*d*sign(cos(1/2*(f*x+exp(1))-1/4*pi)))*si
n(1/4*(6*f*x+6*exp(1)+pi))/(12*f)^2+20*f*(-2*a^2*c*sign(cos(1/2*(f*x+exp(1))-1/4*pi))-4*a^2*d*sign(cos(1/2*(f*
x+exp(1))-1/4*pi)))*sin(1/4*(10*f*x+10*exp(1)-pi))/(20*f)^2+80*a^2*d*f*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(
1/4*(10*f*x+10*exp(1)+pi))/(40*f)^2+112*a^2*d*f*sign(cos(1/2*(f*x+exp(1))-1/4*pi))*cos(1/4*(14*f*x+14*exp(1)-p
i))/(56*f)^2)

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maple [A]  time = 0.78, size = 99, normalized size = 0.72 \[ \frac {2 \left (1+\sin \left (f x +e \right )\right ) a^{3} \left (\sin \left (f x +e \right )-1\right ) \left (-15 \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right ) d +\left (98 c +130 d \right ) \sin \left (f x +e \right )+\left (-21 c -60 d \right ) \left (\cos ^{2}\left (f x +e \right )\right )+322 c +290 d \right )}{105 \cos \left (f x +e \right ) \sqrt {a +a \sin \left (f x +e \right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e)),x)

[Out]

2/105*(1+sin(f*x+e))*a^3*(sin(f*x+e)-1)*(-15*cos(f*x+e)^2*sin(f*x+e)*d+(98*c+130*d)*sin(f*x+e)+(-21*c-60*d)*co
s(f*x+e)^2+322*c+290*d)/cos(f*x+e)/(a+a*sin(f*x+e))^(1/2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sin \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (d \sin \left (f x + e\right ) + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))^(5/2)*(c+d*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^(5/2)*(d*sin(f*x + e) + c), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+a\,\sin \left (e+f\,x\right )\right )}^{5/2}\,\left (c+d\,\sin \left (e+f\,x\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(e + f*x))^(5/2)*(c + d*sin(e + f*x)),x)

[Out]

int((a + a*sin(e + f*x))^(5/2)*(c + d*sin(e + f*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {5}{2}} \left (c + d \sin {\left (e + f x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(f*x+e))**(5/2)*(c+d*sin(f*x+e)),x)

[Out]

Integral((a*(sin(e + f*x) + 1))**(5/2)*(c + d*sin(e + f*x)), x)

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