3.571 \(\int \frac{(a+b \sin (c+d x))^4}{(e \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=237 \[ \frac{2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac{6 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{5 d e^3 \sqrt{e \cos (c+d x)}}+\frac{6 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^5}-\frac{6 \left (-4 a^2 b^2+a^4-4 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}} \]

[Out]

(2*a*b*(3*a^2 - 10*b^2)*(e*Cos[c + d*x])^(3/2))/(5*d*e^5) - (6*(a^4 - 4*a^2*b^2 - 4*b^4)*Sqrt[e*Cos[c + d*x]]*
EllipticE[(c + d*x)/2, 2])/(5*d*e^4*Sqrt[Cos[c + d*x]]) + (6*b*(a^2 - 2*b^2)*(e*Cos[c + d*x])^(3/2)*(a + b*Sin
[c + d*x]))/(5*d*e^5) + (2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^3)/(5*d*e*(e*Cos[c + d*x])^(5/2)) - (6*(a
+ b*Sin[c + d*x])^2*(a*b - (a^2 - 2*b^2)*Sin[c + d*x]))/(5*d*e^3*Sqrt[e*Cos[c + d*x]])

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Rubi [A]  time = 0.465307, antiderivative size = 237, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {2691, 2861, 2862, 2669, 2640, 2639} \[ \frac{2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac{6 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right ) (a+b \sin (c+d x))^2}{5 d e^3 \sqrt{e \cos (c+d x)}}+\frac{6 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^5}-\frac{6 \left (-4 a^2 b^2+a^4-4 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^4 \sqrt{\cos (c+d x)}}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x])^4/(e*Cos[c + d*x])^(7/2),x]

[Out]

(2*a*b*(3*a^2 - 10*b^2)*(e*Cos[c + d*x])^(3/2))/(5*d*e^5) - (6*(a^4 - 4*a^2*b^2 - 4*b^4)*Sqrt[e*Cos[c + d*x]]*
EllipticE[(c + d*x)/2, 2])/(5*d*e^4*Sqrt[Cos[c + d*x]]) + (6*b*(a^2 - 2*b^2)*(e*Cos[c + d*x])^(3/2)*(a + b*Sin
[c + d*x]))/(5*d*e^5) + (2*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x])^3)/(5*d*e*(e*Cos[c + d*x])^(5/2)) - (6*(a
+ b*Sin[c + d*x])^2*(a*b - (a^2 - 2*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 2862

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[(d*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(f*g*(m + p + 1)), x]
+ Dist[1/(m + p + 1), Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1)*Simp[a*c*(m + p + 1) + b*d*m + (a*d*
m + b*c*(m + p + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && Gt
Q[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))^4}{(e \cos (c+d x))^{7/2}} \, dx &=\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}-\frac{2 \int \frac{(a+b \sin (c+d x))^2 \left (-\frac{3 a^2}{2}+3 b^2+\frac{3}{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))^3}{5 d e (e \cos (c+d x))^{5/2}}-\frac{6 (a+b \sin (c+d x))^2 \left (a b-\left (a^2-2 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)} (a+b \sin (c+d x)) \left (-\frac{3}{4} a \left (a^2-6 b^2\right )-\frac{15}{4} b \left (a^2-2 b^2\right ) \sin (c+d x)\right ) \, dx}{5 e^4}\\ &=\frac{6 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}-\frac{6 (a+b \sin (c+d x))^2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt{e \cos (c+d x)}}+\frac{8 \int \sqrt{e \cos (c+d x)} \left (-\frac{15}{8} \left (a^4-4 a^2 b^2-4 b^4\right )-\frac{15}{8} a b \left (3 a^2-10 b^2\right ) \sin (c+d x)\right ) \, dx}{25 e^4}\\ &=\frac{2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}+\frac{6 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}-\frac{6 (a+b \sin (c+d x))^2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{\left (3 \left (a^4-4 a^2 b^2-4 b^4\right )\right ) \int \sqrt{e \cos (c+d x)} \, dx}{5 e^4}\\ &=\frac{2 a b \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}+\frac{6 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}-\frac{6 (a+b \sin (c+d x))^2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt{e \cos (c+d x)}}-\frac{\left (3 \left (a^4-4 a^2 b^2-4 b^4\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 \left (3 a^2-10 b^2\right ) (e \cos (c+d x))^{3/2}}{5 d e^5}-\frac{6 \left (a^4-4 a^2 b^2-4 b^4\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{6 b \left (a^2-2 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^5}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{5 d e (e \cos (c+d x))^{5/2}}-\frac{6 (a+b \sin (c+d x))^2 \left (a b-\left (a^2-2 b^2\right ) \sin (c+d x)\right )}{5 d e^3 \sqrt{e \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.568214, size = 152, normalized size = 0.64 \[ \frac{2 \left (-12 a^2 b^2 \sin (c+d x)+4 a b \left (a^2+b^2\right ) \sec ^2(c+d x)-3 \left (-4 a^2 b^2+a^4-4 b^4\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\left (6 a^2 b^2+a^4+b^4\right ) \tan (c+d x) \sec (c+d x)+3 a^4 \sin (c+d x)-20 a b^3-7 b^4 \sin (c+d x)\right )}{5 d e^3 \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x])^4/(e*Cos[c + d*x])^(7/2),x]

[Out]

(2*(-20*a*b^3 - 3*(a^4 - 4*a^2*b^2 - 4*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + 4*a*b*(a^2 + b^2)*S
ec[c + d*x]^2 + 3*a^4*Sin[c + d*x] - 12*a^2*b^2*Sin[c + d*x] - 7*b^4*Sin[c + d*x] + (a^4 + 6*a^2*b^2 + b^4)*Se
c[c + d*x]*Tan[c + d*x]))/(5*d*e^3*Sqrt[e*Cos[c + d*x]])

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Maple [B]  time = 4.376, size = 874, normalized size = 3.7 \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))^4/(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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^
4*sin(1/2*d*x+1/2*c)^4-48*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/
2*c)^2-1)^(1/2)*a^2*b^2*sin(1/2*d*x+1/2*c)^4-48*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(
1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*b^4*sin(1/2*d*x+1/2*c)^4-24*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+
96*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+56*b^4*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))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a^4*sin(1/2*d*x+1/2
*c)^2+48*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*a
^2*b^2*sin(1/2*d*x+1/2*c)^2+48*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d
*x+1/2*c)^2-1)^(1/2)*b^4*sin(1/2*d*x+1/2*c)^2+24*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4-96*a^2*b^2*cos(1/
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+80*a*b^3*sin(1/2*d*x+1/2*c)^5-56*b^4*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^4-12*
(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^2*b^2-12
*(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))*b^4-8*a^4
*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+12*a^2*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-80*a*b^3*sin(1/2*d
*x+1/2*c)^3+12*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-4*a^3*b*sin(1/2*d*x+1/2*c)+16*a*b^3*sin(1/2*d*x+1/2
*c))/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 )}^{4}}{\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))^4/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c) + a)^4/(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^{4} \cos \left (d x + c\right )^{4} + a^{4} + 6 \, a^{2} b^{2} + b^{4} - 2 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \cos \left (d x + c\right )^{2} - 4 \,{\left (a b^{3} \cos \left (d x + c\right )^{2} - a^{3} b - a 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))^4/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

integral((b^4*cos(d*x + c)^4 + a^4 + 6*a^2*b^2 + b^4 - 2*(3*a^2*b^2 + b^4)*cos(d*x + c)^2 - 4*(a*b^3*cos(d*x +
c)^2 - a^3*b - a*b^3)*sin(d*x + c))*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))**4/(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 )}^{4}}{\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))^4/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

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