3.6.91 \(\int \frac {a+b \sin ^2(e+f x)}{(g \cos (e+f x))^{5/2} \sqrt {d \sin (e+f x)}} \, dx\) [591]

Optimal. Leaf size=107 \[ \frac {2 (a+b) \sqrt {d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}+\frac {(2 a-b) F\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {\sin (2 e+2 f x)}}{3 f g^2 \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \]

[Out]

2/3*(a+b)*(d*sin(f*x+e))^(1/2)/d/f/g/(g*cos(f*x+e))^(3/2)-1/3*(2*a-b)*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi
+f*x)*EllipticF(cos(e+1/4*Pi+f*x),2^(1/2))*sin(2*f*x+2*e)^(1/2)/f/g^2/(g*cos(f*x+e))^(1/2)/(d*sin(f*x+e))^(1/2
)

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Rubi [A]
time = 0.12, antiderivative size = 114, normalized size of antiderivative = 1.07, number of steps used = 7, number of rules used = 7, integrand size = 37, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.189, Rules used = {3281, 468, 335, 243, 342, 281, 238} \begin {gather*} \frac {2 (a+b) \sqrt {d \sin (e+f x)}}{3 d f g (g \cos (e+f x))^{3/2}}-\frac {2 (2 a-b) \left (1-\csc ^2(e+f x)\right )^{3/4} (d \sin (e+f x))^{3/2} F\left (\left .\frac {1}{2} \csc ^{-1}(\sin (e+f x))\right |2\right )}{3 d^2 f g (g \cos (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x]^2)/((g*Cos[e + f*x])^(5/2)*Sqrt[d*Sin[e + f*x]]),x]

[Out]

(2*(a + b)*Sqrt[d*Sin[e + f*x]])/(3*d*f*g*(g*Cos[e + f*x])^(3/2)) - (2*(2*a - b)*(1 - Csc[e + f*x]^2)^(3/4)*El
lipticF[ArcCsc[Sin[e + f*x]]/2, 2]*(d*Sin[e + f*x])^(3/2))/(3*d^2*f*g*(g*Cos[e + f*x])^(3/2))

Rule 238

Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[-b/a, 2]))*EllipticF[(1/2)*ArcSin[Rt[-b/a,
2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 243

Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Dist[x^3*((1 + a/(b*x^4))^(3/4)/(a + b*x^4)^(3/4)), Int[1/(x^3*
(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ[{a, b}, x]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rule 468

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(-(b*c - a*d
))*(e*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*b*e*n*(p + 1))), x] - Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a
*b*n*(p + 1)), Int[(e*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0]
&& LtQ[p, -1] && (( !IntegerQ[p + 1/2] && NeQ[p, -5/4]) ||  !RationalQ[m] || (IGtQ[n, 0] && ILtQ[p + 1/2, 0]
&& LeQ[-1, m, (-n)*(p + 1)]))

Rule 3281

Int[(cos[(e_.) + (f_.)*(x_)]*(c_.))^(m_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Dist[ff*c^(2*IntPart[(m - 1)/2] + 1)*(
(c*Cos[e + f*x])^(2*FracPart[(m - 1)/2])/(f*(Cos[e + f*x]^2)^FracPart[(m - 1)/2])), Subst[Int[(d*ff*x)^n*(1 -
ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x
] &&  !IntegerQ[m]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.14, size = 102, normalized size = 0.95 \begin {gather*} \frac {2 \cos ^2(e+f x)^{3/4} \left (5 a \, _2F_1\left (\frac {1}{4},\frac {7}{4};\frac {5}{4};\sin ^2(e+f x)\right ) \sin (e+f x)+b \, _2F_1\left (\frac {5}{4},\frac {7}{4};\frac {9}{4};\sin ^2(e+f x)\right ) \sin ^3(e+f x)\right )}{5 f g (g \cos (e+f x))^{3/2} \sqrt {d \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[e + f*x]^2)/((g*Cos[e + f*x])^(5/2)*Sqrt[d*Sin[e + f*x]]),x]

[Out]

(2*(Cos[e + f*x]^2)^(3/4)*(5*a*Hypergeometric2F1[1/4, 7/4, 5/4, Sin[e + f*x]^2]*Sin[e + f*x] + b*Hypergeometri
c2F1[5/4, 7/4, 9/4, Sin[e + f*x]^2]*Sin[e + f*x]^3))/(5*f*g*(g*Cos[e + f*x])^(3/2)*Sqrt[d*Sin[e + f*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(323\) vs. \(2(118)=236\).
time = 183.30, size = 324, normalized size = 3.03

method	result	size
default	\(-\frac {\left (2 \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) a -\sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \sqrt {\frac {-1+\cos \left (f x +e \right )}{\sin \left (f x +e \right )}}\, \EllipticF \left (\sqrt {\frac {1-\cos \left (f x +e \right )+\sin \left (f x +e \right )}{\sin \left (f x +e \right )}}, \frac {\sqrt {2}}{2}\right ) b -\cos \left (f x +e \right ) \sqrt {2}\, a -\cos \left (f x +e \right ) \sqrt {2}\, b +\sqrt {2}\, a +\sqrt {2}\, b \right ) \sin \left (f x +e \right ) \cos \left (f x +e \right ) \sqrt {2}}{3 f \left (-1+\cos \left (f x +e \right )\right ) \sqrt {d \sin \left (f x +e \right )}\, \left (g \cos \left (f x +e \right )\right )^{\frac {5}{2}}}\)	\(324\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sin(f*x+e)^2)/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/f*(2*sin(f*x+e)*cos(f*x+e)*((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-sin(f*x+e)*cos(f*x+e)*((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))*b-cos(f*x+e)*2^(1/2)*a-cos(f*x+e)*2^(1/2)*b+2^(1/2)*a+2^(1/2)*b)*sin(f*x+e)*cos(f*x+e)/(-1+cos(f*x+e))/(d*
sin(f*x+e))^(1/2)/(g*cos(f*x+e))^(5/2)*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e)^2)/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)/((g*cos(f*x + e))^(5/2)*sqrt(d*sin(f*x + e))), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.11, size = 125, normalized size = 1.17 \begin {gather*} -\frac {\sqrt {i \, d g} {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} F(\arcsin \left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )\,|\,-1) + \sqrt {-i \, d g} {\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} F(\arcsin \left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )\,|\,-1) - 2 \, \sqrt {g \cos \left (f x + e\right )} \sqrt {d \sin \left (f x + e\right )} {\left (a + b\right )}}{3 \, d f g^{3} \cos \left (f x + e\right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e)^2)/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

-1/3*(sqrt(I*d*g)*(2*a - b)*cos(f*x + e)^2*elliptic_f(arcsin(cos(f*x + e) + I*sin(f*x + e)), -1) + sqrt(-I*d*g
)*(2*a - b)*cos(f*x + e)^2*elliptic_f(arcsin(cos(f*x + e) - I*sin(f*x + e)), -1) - 2*sqrt(g*cos(f*x + e))*sqrt
(d*sin(f*x + e))*(a + b))/(d*f*g^3*cos(f*x + e)^2)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e)**2)/(g*cos(f*x+e))**(5/2)/(d*sin(f*x+e))**(1/2),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3007 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(f*x+e)^2)/(g*cos(f*x+e))^(5/2)/(d*sin(f*x+e))^(1/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)/((g*cos(f*x + e))^(5/2)*sqrt(d*sin(f*x + e))), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {b\,{\sin \left (e+f\,x\right )}^2+a}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}\,\sqrt {d\,\sin \left (e+f\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*sin(e + f*x)^2)/((g*cos(e + f*x))^(5/2)*(d*sin(e + f*x))^(1/2)),x)

[Out]

int((a + b*sin(e + f*x)^2)/((g*cos(e + f*x))^(5/2)*(d*sin(e + f*x))^(1/2)), x)

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