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

Optimal. Leaf size=160 \[ \frac{2 \left (3 a^2-2 b^2\right ) \sin (c+d x)}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{2 \left (3 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 a b}{5 d e^3 \sqrt{e \cos (c+d x)}}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))}{5 d e (e \cos (c+d x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.172351, antiderivative size = 160, 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, 2669, 2636, 2640, 2639} \[ \frac{2 \left (3 a^2-2 b^2\right ) \sin (c+d x)}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{2 \left (3 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 a b}{5 d e^3 \sqrt{e \cos (c+d x)}}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))}{5 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 2636

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

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

Mathematica [A]  time = 0.535524, size = 105, normalized size = 0.66 \[ \frac{\left (7 a^2+2 b^2\right ) \sin (c+d x)-4 \left (3 a^2-2 b^2\right ) \cos ^{\frac{5}{2}}(c+d x) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+3 a^2 \sin (3 (c+d x))+8 a b-2 b^2 \sin (3 (c+d x))}{10 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a*b - 4*(3*a^2 - 2*b^2)*Cos[c + d*x]^(5/2)*EllipticE[(c + d*x)/2, 2] + (7*a^2 + 2*b^2)*Sin[c + d*x] + 3*a^2
*Sin[3*(c + d*x)] - 2*b^2*Sin[3*(c + d*x)])/(10*d*e*(e*Cos[c + d*x])^(5/2))

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Maple [B]  time = 3.381, size = 564, normalized size = 3.5 \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))^2/(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*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^
2*sin(1/2*d*x+1/2*c)^4-8*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1
/2*c),2^(1/2))*b^2*sin(1/2*d*x+1/2*c)^4-24*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+16*b^2*cos(1/2*d*x+1/2*
c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x
+1/2*c),2^(1/2))*a^2*sin(1/2*d*x+1/2*c)^2+8*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli
pticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2*sin(1/2*d*x+1/2*c)^2+24*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-16*b
^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+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)*Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2-2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2-8*a^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+2*b^2*cos(1/2*d*x+1/2*c)*sin(
1/2*d*x+1/2*c)^2-2*sin(1/2*d*x+1/2*c)*a*b)/d

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-\frac{{\left (b^{2} \cos \left (d x + c\right )^{2} - 2 \, a b \sin \left (d x + c\right ) - a^{2} - b^{2}\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))^2/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral(-(b^2*cos(d*x + c)^2 - 2*a*b*sin(d*x + c) - a^2 - b^2)*sqrt(e*cos(d*x + c))/(e^4*cos(d*x + c)^4), 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((a+b*sin(d*x+c))**2/(e*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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

[Out]

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