3.110 \(\int (a+a \sec (c+d x)) \sqrt {e \sin (c+d x)} \, dx\)

Optimal. Leaf size=104 \[ -\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}} \]

[Out]

-a*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*e^(1/2)/d+a*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))*e^(1/2)/d-2*a*(sin(1
/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticE(cos(1/2*c+1/4*Pi+1/2*d*x),2^(1/2))*(e*sin(d*
x+c))^(1/2)/d/sin(d*x+c)^(1/2)

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Rubi [A]  time = 0.15, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.391, Rules used = {3872, 2838, 2564, 329, 298, 203, 206, 2640, 2639} \[ -\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a E\left (\left .\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])*Sqrt[e*Sin[c + d*x]],x]

[Out]

-((a*Sqrt[e]*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d) + (a*Sqrt[e]*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d +
(2*a*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*Sqrt[Sin[c + d*x]])

Rule 203

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

Rule 206

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

Rule 298

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b),
2]]}, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &
&  !GtQ[a/b, 0]

Rule 329

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 2564

Int[cos[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(a*f), Subst[Int[
x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Sin[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2] &&
!(IntegerQ[(m - 1)/2] && LtQ[0, m, n])

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]

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 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 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[e + f*x])^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x)) \sqrt {e \sin (c+d x)} \, dx &=-\int (-a-a \cos (c+d x)) \sec (c+d x) \sqrt {e \sin (c+d x)} \, dx\\ &=a \int \sqrt {e \sin (c+d x)} \, dx+a \int \sec (c+d x) \sqrt {e \sin (c+d x)} \, dx\\ &=\frac {a \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{e^2}} \, dx,x,e \sin (c+d x)\right )}{d e}+\frac {\left (a \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{\sqrt {\sin (c+d x)}}\\ &=\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}+\frac {(2 a) \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{e^2}} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e}\\ &=\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}+\frac {(a e) \operatorname {Subst}\left (\int \frac {1}{e-x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}-\frac {(a e) \operatorname {Subst}\left (\int \frac {1}{e+x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}\\ &=-\frac {a \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 69, normalized size = 0.66 \[ \frac {a \sqrt {e \sin (c+d x)} \left (-2 E\left (\left .\frac {1}{4} (-2 c-2 d x+\pi )\right |2\right )-\tan ^{-1}\left (\sqrt {\sin (c+d x)}\right )+\tanh ^{-1}\left (\sqrt {\sin (c+d x)}\right )\right )}{d \sqrt {\sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])*Sqrt[e*Sin[c + d*x]],x]

[Out]

(a*(-ArcTan[Sqrt[Sin[c + d*x]]] + ArcTanh[Sqrt[Sin[c + d*x]]] - 2*EllipticE[(-2*c + Pi - 2*d*x)/4, 2])*Sqrt[e*
Sin[c + d*x]])/(d*Sqrt[Sin[c + d*x]])

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fricas [F]  time = 1.06, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(e*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

integral((a*sec(d*x + c) + a)*sqrt(e*sin(d*x + c)), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(e*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)*sqrt(e*sin(d*x + c)), x)

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maple [A]  time = 3.64, size = 198, normalized size = 1.90 \[ -\frac {a \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e}}{d}+\frac {a \arctanh \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right ) \sqrt {e}}{d}-\frac {2 a e \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{d \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}+\frac {a e \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right )}{d \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))*(e*sin(d*x+c))^(1/2),x)

[Out]

-a*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*e^(1/2)/d+a*arctanh((e*sin(d*x+c))^(1/2)/e^(1/2))*e^(1/2)/d-2/d*a*e*(-
sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*EllipticE((-sin(d*
x+c)+1)^(1/2),1/2*2^(1/2))+1/d*a*e*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)/cos(d*x+c)/(e
*sin(d*x+c))^(1/2)*EllipticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (a \sec \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(e*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)*sqrt(e*sin(d*x + c)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \sqrt {e\,\sin \left (c+d\,x\right )}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(c + d*x))^(1/2)*(a + a/cos(c + d*x)),x)

[Out]

int((e*sin(c + d*x))^(1/2)*(a + a/cos(c + d*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \sqrt {e \sin {\left (c + d x \right )}}\, dx + \int \sqrt {e \sin {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(e*sin(d*x+c))**(1/2),x)

[Out]

a*(Integral(sqrt(e*sin(c + d*x)), x) + Integral(sqrt(e*sin(c + d*x))*sec(c + d*x), x))

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