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

Optimal. Leaf size=218 \[ \frac{2 a b \left (15 a^2+62 b^2\right ) (e \cos (c+d x))^{3/2}}{15 d e^3}+\frac{2 b \left (5 a^2+6 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^3}-\frac{2 \left (60 a^2 b^2+5 a^4+12 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^2 \sqrt{\cos (c+d x)}}+\frac{2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{d e^3}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{d e \sqrt{e \cos (c+d x)}} \]

[Out]

(2*a*b*(15*a^2 + 62*b^2)*(e*Cos[c + d*x])^(3/2))/(15*d*e^3) - (2*(5*a^4 + 60*a^2*b^2 + 12*b^4)*Sqrt[e*Cos[c +
d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]) + (2*b*(5*a^2 + 6*b^2)*(e*Cos[c + d*x])^(3/2)*(a
+ b*Sin[c + d*x]))/(5*d*e^3) + (2*a*b*(e*Cos[c + d*x])^(3/2)*(a + b*Sin[c + d*x])^2)/(d*e^3) + (2*(b + a*Sin[
c + d*x])*(a + b*Sin[c + d*x])^3)/(d*e*Sqrt[e*Cos[c + d*x]])

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Rubi [A]  time = 0.438102, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2691, 2862, 2669, 2640, 2639} \[ \frac{2 a b \left (15 a^2+62 b^2\right ) (e \cos (c+d x))^{3/2}}{15 d e^3}+\frac{2 b \left (5 a^2+6 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^3}-\frac{2 \left (60 a^2 b^2+5 a^4+12 b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right ) \sqrt{e \cos (c+d x)}}{5 d e^2 \sqrt{\cos (c+d x)}}+\frac{2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{d e^3}+\frac{2 (a \sin (c+d x)+b) (a+b \sin (c+d x))^3}{d e \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*(15*a^2 + 62*b^2)*(e*Cos[c + d*x])^(3/2))/(15*d*e^3) - (2*(5*a^4 + 60*a^2*b^2 + 12*b^4)*Sqrt[e*Cos[c +
d*x]]*EllipticE[(c + d*x)/2, 2])/(5*d*e^2*Sqrt[Cos[c + d*x]]) + (2*b*(5*a^2 + 6*b^2)*(e*Cos[c + d*x])^(3/2)*(a
+ b*Sin[c + d*x]))/(5*d*e^3) + (2*a*b*(e*Cos[c + d*x])^(3/2)*(a + b*Sin[c + d*x])^2)/(d*e^3) + (2*(b + a*Sin[
c + d*x])*(a + b*Sin[c + d*x])^3)/(d*e*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 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))^{3/2}} \, dx &=\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{d e \sqrt{e \cos (c+d x)}}-\frac{2 \int \sqrt{e \cos (c+d x)} (a+b \sin (c+d x))^2 \left (\frac{a^2}{2}+3 b^2+\frac{7}{2} a b \sin (c+d x)\right ) \, dx}{e^2}\\ &=\frac{2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{d e \sqrt{e \cos (c+d x)}}-\frac{4 \int \sqrt{e \cos (c+d x)} (a+b \sin (c+d x)) \left (\frac{7}{4} a \left (a^2+10 b^2\right )+\frac{7}{4} b \left (5 a^2+6 b^2\right ) \sin (c+d x)\right ) \, dx}{7 e^2}\\ &=\frac{2 b \left (5 a^2+6 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^3}+\frac{2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{d e \sqrt{e \cos (c+d x)}}-\frac{8 \int \sqrt{e \cos (c+d x)} \left (\frac{7}{8} \left (5 a^4+60 a^2 b^2+12 b^4\right )+\frac{7}{8} a b \left (15 a^2+62 b^2\right ) \sin (c+d x)\right ) \, dx}{35 e^2}\\ &=\frac{2 a b \left (15 a^2+62 b^2\right ) (e \cos (c+d x))^{3/2}}{15 d e^3}+\frac{2 b \left (5 a^2+6 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^3}+\frac{2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{d e \sqrt{e \cos (c+d x)}}-\frac{\left (5 a^4+60 a^2 b^2+12 b^4\right ) \int \sqrt{e \cos (c+d x)} \, dx}{5 e^2}\\ &=\frac{2 a b \left (15 a^2+62 b^2\right ) (e \cos (c+d x))^{3/2}}{15 d e^3}+\frac{2 b \left (5 a^2+6 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^3}+\frac{2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{d e \sqrt{e \cos (c+d x)}}-\frac{\left (\left (5 a^4+60 a^2 b^2+12 b^4\right ) \sqrt{e \cos (c+d x)}\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 e^2 \sqrt{\cos (c+d x)}}\\ &=\frac{2 a b \left (15 a^2+62 b^2\right ) (e \cos (c+d x))^{3/2}}{15 d e^3}-\frac{2 \left (5 a^4+60 a^2 b^2+12 b^4\right ) \sqrt{e \cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 d e^2 \sqrt{\cos (c+d x)}}+\frac{2 b \left (5 a^2+6 b^2\right ) (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))}{5 d e^3}+\frac{2 a b (e \cos (c+d x))^{3/2} (a+b \sin (c+d x))^2}{d e^3}+\frac{2 (b+a \sin (c+d x)) (a+b \sin (c+d x))^3}{d e \sqrt{e \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.57541, size = 135, normalized size = 0.62 \[ \frac{\frac{1}{2} \left (\left (360 a^2 b^2+60 a^4+63 b^4\right ) \sin (c+d x)+240 a^3 b+40 a b^3 \cos (2 (c+d x))+280 a b^3+3 b^4 \sin (3 (c+d x))\right )-6 \left (60 a^2 b^2+5 a^4+12 b^4\right ) \sqrt{\cos (c+d x)} E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{15 d e \sqrt{e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(5*a^4 + 60*a^2*b^2 + 12*b^4)*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2] + (240*a^3*b + 280*a*b^3 + 40*a
*b^3*Cos[2*(c + d*x)] + (60*a^4 + 360*a^2*b^2 + 63*b^4)*Sin[c + d*x] + 3*b^4*Sin[3*(c + d*x)])/2)/(15*d*e*Sqrt
[e*Cos[c + d*x]])

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Maple [A]  time = 2.007, size = 378, normalized size = 1.7 \begin{align*} -{\frac{2}{15\,de} \left ( -24\,{b}^{4}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}-80\,a{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}+24\,{b}^{4}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+15\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{4}+180\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){a}^{2}{b}^{2}+36\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ){b}^{4}-30\,{a}^{4}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-180\,{a}^{2}{b}^{2}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+80\,a{b}^{3} \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-36\,{b}^{4}\cos \left ( 1/2\,dx+c/2 \right ) \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-60\,{a}^{3}b\sin \left ( 1/2\,dx+c/2 \right ) -80\,a{b}^{3}\sin \left ( 1/2\,dx+c/2 \right ) \right ){\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}e+e}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}} \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))^(3/2),x)

[Out]

-2/15/e/(-2*sin(1/2*d*x+1/2*c)^2*e+e)^(1/2)/sin(1/2*d*x+1/2*c)*(-24*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^
6-80*a*b^3*sin(1/2*d*x+1/2*c)^5+24*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^4+15*(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+180*(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+36*(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^4-30*a^4*cos(1/2*d*x+1/2*c)*sin(1/2*d
*x+1/2*c)^2-180*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)^3-36*b^4*cos(1/2*d
*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-60*a^3*b*sin(1/2*d*x+1/2*c)-80*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{3}{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))^(3/2),x, algorithm="maxima")

[Out]

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

[Out]

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