3.1278 \(\int \frac{(a+b \sin (e+f x))^2}{(g \cos (e+f x))^{7/2} \sqrt{d \sin (e+f x)}} \, dx\)

Optimal. Leaf size=193 \[ \frac{8 a^2 \sqrt{d \sin (e+f x)}}{5 d f g^3 \sqrt{g \cos (e+f x)}}+\frac{8 a b (d \sin (e+f x))^{3/2}}{5 d^2 f g^3 \sqrt{g \cos (e+f x)}}-\frac{8 a b E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}}{5 d f g^4 \sqrt{\sin (2 e+2 f x)}}+\frac{2 \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^2}{5 d f g (g \cos (e+f x))^{5/2}} \]

[Out]

(8*a^2*Sqrt[d*Sin[e + f*x]])/(5*d*f*g^3*Sqrt[g*Cos[e + f*x]]) + (8*a*b*(d*Sin[e + f*x])^(3/2))/(5*d^2*f*g^3*Sq
rt[g*Cos[e + f*x]]) + (2*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^2)/(5*d*f*g*(g*Cos[e + f*x])^(5/2)) - (8*a*
b*Sqrt[g*Cos[e + f*x]]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Sin[e + f*x]])/(5*d*f*g^4*Sqrt[Sin[2*e + 2*f*x]])

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Rubi [A]  time = 0.473507, antiderivative size = 193, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 37, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.162, Rules used = {2888, 2838, 2563, 2571, 2572, 2639} \[ \frac{8 a^2 \sqrt{d \sin (e+f x)}}{5 d f g^3 \sqrt{g \cos (e+f x)}}+\frac{8 a b (d \sin (e+f x))^{3/2}}{5 d^2 f g^3 \sqrt{g \cos (e+f x)}}-\frac{8 a b E\left (\left .e+f x-\frac{\pi }{4}\right |2\right ) \sqrt{d \sin (e+f x)} \sqrt{g \cos (e+f x)}}{5 d f g^4 \sqrt{\sin (2 e+2 f x)}}+\frac{2 \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^2}{5 d f g (g \cos (e+f x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*a^2*Sqrt[d*Sin[e + f*x]])/(5*d*f*g^3*Sqrt[g*Cos[e + f*x]]) + (8*a*b*(d*Sin[e + f*x])^(3/2))/(5*d^2*f*g^3*Sq
rt[g*Cos[e + f*x]]) + (2*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^2)/(5*d*f*g*(g*Cos[e + f*x])^(5/2)) - (8*a*
b*Sqrt[g*Cos[e + f*x]]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Sin[e + f*x]])/(5*d*f*g^4*Sqrt[Sin[2*e + 2*f*x]])

Rule 2888

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_))/Sqrt[(d_.)*sin[(e_.) +
(f_.)*(x_)]], x_Symbol] :> Simp[(2*(g*Cos[e + f*x])^(p + 1)*Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x])^m)/(d*f*
g*(2*m + 1)), x] + Dist[(2*a*m)/(g^2*(2*m + 1)), Int[((g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1))/S
qrt[d*Sin[e + f*x]], x], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && EqQ[m + p + 3/2,
0]

Rule 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 2563

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

Rule 2571

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[((b*Sin[e +
f*x])^(n + 1)*(a*Cos[e + f*x])^(m + 1))/(a*b*f*(m + 1)), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Sin[e + f
*x])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n]

Rule 2572

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(Sqrt[a*Sin[e +
f*x]]*Sqrt[b*Cos[e + f*x]])/Sqrt[Sin[2*e + 2*f*x]], Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f},
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 (e+f x))^2}{(g \cos (e+f x))^{7/2} \sqrt{d \sin (e+f x)}} \, dx &=\frac{2 \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^2}{5 d f g (g \cos (e+f x))^{5/2}}+\frac{(4 a) \int \frac{a+b \sin (e+f x)}{(g \cos (e+f x))^{3/2} \sqrt{d \sin (e+f x)}} \, dx}{5 g^2}\\ &=\frac{2 \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^2}{5 d f g (g \cos (e+f x))^{5/2}}+\frac{\left (4 a^2\right ) \int \frac{1}{(g \cos (e+f x))^{3/2} \sqrt{d \sin (e+f x)}} \, dx}{5 g^2}+\frac{(4 a b) \int \frac{\sqrt{d \sin (e+f x)}}{(g \cos (e+f x))^{3/2}} \, dx}{5 d g^2}\\ &=\frac{8 a^2 \sqrt{d \sin (e+f x)}}{5 d f g^3 \sqrt{g \cos (e+f x)}}+\frac{8 a b (d \sin (e+f x))^{3/2}}{5 d^2 f g^3 \sqrt{g \cos (e+f x)}}+\frac{2 \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^2}{5 d f g (g \cos (e+f x))^{5/2}}-\frac{(8 a b) \int \sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)} \, dx}{5 d g^4}\\ &=\frac{8 a^2 \sqrt{d \sin (e+f x)}}{5 d f g^3 \sqrt{g \cos (e+f x)}}+\frac{8 a b (d \sin (e+f x))^{3/2}}{5 d^2 f g^3 \sqrt{g \cos (e+f x)}}+\frac{2 \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^2}{5 d f g (g \cos (e+f x))^{5/2}}-\frac{\left (8 a b \sqrt{g \cos (e+f x)} \sqrt{d \sin (e+f x)}\right ) \int \sqrt{\sin (2 e+2 f x)} \, dx}{5 d g^4 \sqrt{\sin (2 e+2 f x)}}\\ &=\frac{8 a^2 \sqrt{d \sin (e+f x)}}{5 d f g^3 \sqrt{g \cos (e+f x)}}+\frac{8 a b (d \sin (e+f x))^{3/2}}{5 d^2 f g^3 \sqrt{g \cos (e+f x)}}+\frac{2 \sqrt{d \sin (e+f x)} (a+b \sin (e+f x))^2}{5 d f g (g \cos (e+f x))^{5/2}}-\frac{8 a b \sqrt{g \cos (e+f x)} E\left (\left .e-\frac{\pi }{4}+f x\right |2\right ) \sqrt{d \sin (e+f x)}}{5 d f g^4 \sqrt{\sin (2 e+2 f x)}}\\ \end{align*}

Mathematica [C]  time = 0.652753, size = 105, normalized size = 0.54 \[ \frac{2 \tan (e+f x) \left (3 \left (b^2-4 a^2\right ) \sin ^2(e+f x)+15 a^2+10 a b \sin (e+f x) \cos ^2(e+f x)^{5/4} \, _2F_1\left (\frac{3}{4},\frac{9}{4};\frac{7}{4};\sin ^2(e+f x)\right )\right )}{15 f g^2 \sqrt{d \sin (e+f x)} (g \cos (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(15*a^2 + 10*a*b*(Cos[e + f*x]^2)^(5/4)*Hypergeometric2F1[3/4, 9/4, 7/4, Sin[e + f*x]^2]*Sin[e + f*x] + 3*(
-4*a^2 + b^2)*Sin[e + f*x]^2)*Tan[e + f*x])/(15*f*g^2*(g*Cos[e + f*x])^(3/2)*Sqrt[d*Sin[e + f*x]])

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Maple [B]  time = 0.398, size = 616, normalized size = 3.2 \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(f*x+e))^2/(g*cos(f*x+e))^(7/2)/(d*sin(f*x+e))^(1/2),x)

[Out]

-1/5/f*2^(1/2)*(-8*cos(f*x+e)^3*EllipticE((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-(-1+co
s(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+
e))^(1/2)*a*b+4*cos(f*x+e)^3*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*
x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^
(1/2))*a*b-8*cos(f*x+e)^2*EllipticE((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))*(-(-1+cos(f*x+
e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1
/2)*a*b+4*cos(f*x+e)^2*(-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2)*((-1+cos(f*x+e)+sin(f*x+e))/sin(f*x+e))^
(1/2)*((-1+cos(f*x+e))/sin(f*x+e))^(1/2)*EllipticF((-(-1+cos(f*x+e)-sin(f*x+e))/sin(f*x+e))^(1/2),1/2*2^(1/2))
*a*b+4*2^(1/2)*cos(f*x+e)^3*a*b-4*2^(1/2)*cos(f*x+e)^2*sin(f*x+e)*a^2+2^(1/2)*cos(f*x+e)^2*sin(f*x+e)*b^2-2*2^
(1/2)*cos(f*x+e)^2*a*b-2^(1/2)*sin(f*x+e)*a^2-2^(1/2)*sin(f*x+e)*b^2-2*2^(1/2)*a*b)*cos(f*x+e)/(g*cos(f*x+e))^
(7/2)/(d*sin(f*x+e))^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(7/2)/(d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e) + a)^2/((g*cos(f*x + e))^(7/2)*sqrt(d*sin(f*x + e))), 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 (f x + e\right )^{2} - 2 \, a b \sin \left (f x + e\right ) - a^{2} - b^{2}\right )} \sqrt{g \cos \left (f x + e\right )} \sqrt{d \sin \left (f x + e\right )}}{d g^{4} \cos \left (f x + e\right )^{4} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(7/2)/(d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

integral(-(b^2*cos(f*x + e)^2 - 2*a*b*sin(f*x + e) - a^2 - b^2)*sqrt(g*cos(f*x + e))*sqrt(d*sin(f*x + e))/(d*g
^4*cos(f*x + e)^4*sin(f*x + e)), 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(f*x+e))**2/(g*cos(f*x+e))**(7/2)/(d*sin(f*x+e))**(1/2),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 (f x + e\right ) + a\right )}^{2}}{\left (g \cos \left (f x + e\right )\right )^{\frac{7}{2}} \sqrt{d \sin \left (f x + e\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e))^2/(g*cos(f*x+e))^(7/2)/(d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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