3.562 \(\int \frac{(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=187 \[ \frac{2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac{2 \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{6 a \left (a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}} \]

[Out]

(2*b*(3*a^2 - 4*b^2)*(e*Cos[c + d*x])^(3/2))/(5*d*e^5) - (6*a*(a^2 - 2*b^2)*Sqrt[e*Cos[c + d*x]]*EllipticE[(c
+ d*x)/2, 2])/(5*d*e^4*Sqrt[Cos[c + d*x]]) + (2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(5*d*e*(e*Cos[c +
d*x])^(5/2)) - (2*(a + b*Sin[c + d*x])*(a*b - (3*a^2 - 4*b^2)*Sin[c + d*x]))/(5*d*e^3*Sqrt[e*Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 0.264079, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2691, 2861, 2669, 2640, 2639} \[ \frac{2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac{2 \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{6 a \left (a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]

[Out]

(2*b*(3*a^2 - 4*b^2)*(e*Cos[c + d*x])^(3/2))/(5*d*e^5) - (6*a*(a^2 - 2*b^2)*Sqrt[e*Cos[c + d*x]]*EllipticE[(c
+ d*x)/2, 2])/(5*d*e^4*Sqrt[Cos[c + d*x]]) + (2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^2)/(5*d*e*(e*Cos[c +
d*x])^(5/2)) - (2*(a + b*Sin[c + d*x])*(a*b - (3*a^2 - 4*b^2)*Sin[c + d*x]))/(5*d*e^3*Sqrt[e*Cos[c + d*x]])

Rule 2691

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

Rule 2861

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

Rule 2669

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rubi steps

\begin{align*} \int \frac{(a+b \sin (c+d x))^3}{(e \cos (c+d x))^{7/2}} \, dx &=\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac{2 \int \frac{(a+b \sin (c+d x)) \left (-\frac{3 a^2}{2}+2 b^2+\frac{1}{2} a b \sin (c+d x)\right )}{(e \cos (c+d x))^{3/2}} \, dx}{5 e^2}\\ &=\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac{2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt{e \cos (c+d x)}}+\frac{4 \int \sqrt{e \cos (c+d x)} \left (-\frac{3}{4} a \left (a^2-2 b^2\right )-\frac{3}{4} b \left (3 a^2-4 b^2\right ) \sin (c+d x)\right ) \, dx}{5 e^4}\\ &=\frac{2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac{2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{\left (3 a \left (a^2-2 b^2\right )\right ) \int \sqrt{e \cos (c+d x)} \, dx}{5 e^4}\\ &=\frac{2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac{2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{\left (3 a \left (a^2-2 b^2\right ) \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 e^4 \sqrt{\cos (c+d x)}}\\ &=\frac{2 b \left (3 a^2-4 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac{6 a \left (a^2-2 b^2\right ) \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^2}{5 d e (e \cos (c+d x))^{5/2}}-\frac{2 (a+b \sin (c+d x)) \left (a b-\left (3 a^2-4 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt{e \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.704941, size = 126, normalized size = 0.67 \[ \frac{2 \left (b \left (3 a^2+b^2\right ) \sec ^2(c+d x)-3 a \left (a^2-2 b^2\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+a \left (a^2+3 b^2\right ) \tan (c+d x) \sec (c+d x)+3 a^3 \sin (c+d x)-6 a b^2 \sin (c+d x)-5 b^3\right )}{5 d e^3 \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^3/(e*Cos[c + d*x])^(7/2),x]

[Out]

(2*(-5*b^3 - 3*a*(a^2 - 2*b^2)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + b*(3*a^2 + b^2)*Sec[c + d*x]^2 +
3*a^3*Sin[c + d*x] - 6*a*b^2*Sin[c + d*x] + a*(a^2 + 3*b^2)*Sec[c + d*x]*Tan[c + d*x]))/(5*d*e^3*Sqrt[e*Cos[c
+ d*x]])

________________________________________________________________________________________

Maple [B]  time = 3.974, size = 618, normalized size = 3.3 \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(d*x+c))^3/(e*cos(d*x+c))^(7/2),x)

[Out]

-2/5/(4*sin(1/2*d*x+1/2*c)^4-4*sin(1/2*d*x+1/2*c)^2+1)/sin(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/
e^3*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a^
3*sin(1/2*d*x+1/2*c)^4-24*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+
1/2*c)^2)^(1/2)*a*b^2*sin(1/2*d*x+1/2*c)^4-24*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+48*a*b^2*cos(1/2*d*x
+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*a^3*sin(1/2*d*x+1/2*c)^2+24*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^
2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a*b^2*sin(1/2*d*x+1/2*c)^2+24*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^4-48*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+20*b^3*sin(1/2*d*x+1/2*c)^5+3*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-6*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2
*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^2-8*a^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^2+6*a*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-20*sin(1/2*d*x+1/2*c)^3*b^3-3*a^2*b*sin(1/2*d*x+1/2
*c)+4*b^3*sin(1/2*d*x+1/2*c))/d

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(7/2), x)

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(-(3*a*b^2*cos(d*x + c)^2 - a^3 - 3*a*b^2 + (b^3*cos(d*x + c)^2 - 3*a^2*b - b^3)*sin(d*x + c))*sqrt(e*
cos(d*x + c))/(e^4*cos(d*x + c)^4), x)

________________________________________________________________________________________

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(d*x+c))**3/(e*cos(d*x+c))**(7/2),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \sin \left (d x + c\right ) + a\right )}^{3}}{\left (e \cos \left (d x + c\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c))^3/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c) + a)^3/(e*cos(d*x + c))^(7/2), x)