3.2.16 \(\int (a+a \sec (c+d x))^2 \sqrt {e \sin (c+d x)} \, dx\) [116]

Optimal. Leaf size=138 \[ -\frac {2 a^2 \sqrt {e} \text {ArcTan}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a^2 E\left (\left .\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right )\right |2\right ) \sqrt {e \sin (c+d x)}}{d \sqrt {\sin (c+d x)}}+\frac {a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e} \]

[Out]

a^2*sec(d*x+c)*(e*sin(d*x+c))^(3/2)/d/e-2*a^2*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))*e^(1/2)/d+2*a^2*arctanh((e*
sin(d*x+c))^(1/2)/e^(1/2))*e^(1/2)/d-a^2*(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)*Ellipti
cE(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 = 0.22, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 10, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3957, 2952, 2721, 2719, 2644, 335, 304, 209, 212, 2651} \begin {gather*} -\frac {2 a^2 \sqrt {e} \text {ArcTan}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}+\frac {a^2 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)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^2*Sqrt[e*Sin[c + d*x]],x]

[Out]

(-2*a^2*Sqrt[e]*ArcTan[Sqrt[e*Sin[c + d*x]]/Sqrt[e]])/d + (2*a^2*Sqrt[e]*ArcTanh[Sqrt[e*Sin[c + d*x]]/Sqrt[e]]
)/d + (a^2*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d*x]])/(d*Sqrt[Sin[c + d*x]]) + (a^2*Sec[c + d*x]*(
e*Sin[c + d*x])^(3/2))/(d*e)

Rule 209

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

Rule 212

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

Rule 304

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 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 2644

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 2651

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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2952

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[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))^2 \sqrt {e \sin (c+d x)} \, dx &=\int (-a-a \cos (c+d x))^2 \sec ^2(c+d x) \sqrt {e \sin (c+d x)} \, dx\\ &=\int \left (a^2 \sqrt {e \sin (c+d x)}+2 a^2 \sec (c+d x) \sqrt {e \sin (c+d x)}+a^2 \sec ^2(c+d x) \sqrt {e \sin (c+d x)}\right ) \, dx\\ &=a^2 \int \sqrt {e \sin (c+d x)} \, dx+a^2 \int \sec ^2(c+d x) \sqrt {e \sin (c+d x)} \, dx+\left (2 a^2\right ) \int \sec (c+d x) \sqrt {e \sin (c+d x)} \, dx\\ &=\frac {a^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}-\frac {1}{2} a^2 \int \sqrt {e \sin (c+d x)} \, dx+\frac {\left (2 a^2\right ) \text {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^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{\sqrt {\sin (c+d x)}}\\ &=\frac {2 a^2 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^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}+\frac {\left (4 a^2\right ) \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{e^2}} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d e}-\frac {\left (a^2 \sqrt {e \sin (c+d x)}\right ) \int \sqrt {\sin (c+d x)} \, dx}{2 \sqrt {\sin (c+d x)}}\\ &=\frac {a^2 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^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}+\frac {\left (2 a^2 e\right ) \text {Subst}\left (\int \frac {1}{e-x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}-\frac {\left (2 a^2 e\right ) \text {Subst}\left (\int \frac {1}{e+x^2} \, dx,x,\sqrt {e \sin (c+d x)}\right )}{d}\\ &=-\frac {2 a^2 \sqrt {e} \tan ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {2 a^2 \sqrt {e} \tanh ^{-1}\left (\frac {\sqrt {e \sin (c+d x)}}{\sqrt {e}}\right )}{d}+\frac {a^2 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^2 \sec (c+d x) (e \sin (c+d x))^{3/2}}{d e}\\ \end {align*}

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Mathematica [A]
time = 11.60, size = 206, normalized size = 1.49 \begin {gather*} \frac {16 a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sqrt {e \sin (c+d x)} \left (-2 \text {ArcTan}\left (\sqrt {\sin (c+d x)}\right ) \sqrt {\cos ^2(c+d x)}+\sqrt {\cos ^2(c+d x)} E\left (\left .\text {ArcSin}\left (\sqrt {\sin (c+d x)}\right )\right |-1\right )-\sqrt {\cos ^2(c+d x)} F\left (\left .\text {ArcSin}\left (\sqrt {\sin (c+d x)}\right )\right |-1\right )-\sqrt {\cos ^2(c+d x)} \log \left (1-\sqrt {\sin (c+d x)}\right )+\sqrt {\cos ^2(c+d x)} \log \left (1+\sqrt {\sin (c+d x)}\right )+\sin ^{\frac {3}{2}}(c+d x)\right ) \sin ^4\left (\frac {1}{2} \text {ArcSin}(\sin (c+d x))\right )}{d \sin ^{\frac {9}{2}}(c+d x)} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(a + a*Sec[c + d*x])^2*Sqrt[e*Sin[c + d*x]],x]

[Out]

(16*a^2*Cos[(c + d*x)/2]^4*Sec[c + d*x]*Sqrt[e*Sin[c + d*x]]*(-2*ArcTan[Sqrt[Sin[c + d*x]]]*Sqrt[Cos[c + d*x]^
2] + Sqrt[Cos[c + d*x]^2]*EllipticE[ArcSin[Sqrt[Sin[c + d*x]]], -1] - Sqrt[Cos[c + d*x]^2]*EllipticF[ArcSin[Sq
rt[Sin[c + d*x]]], -1] - Sqrt[Cos[c + d*x]^2]*Log[1 - Sqrt[Sin[c + d*x]]] + Sqrt[Cos[c + d*x]^2]*Log[1 + Sqrt[
Sin[c + d*x]]] + Sin[c + d*x]^(3/2))*Sin[ArcSin[Sin[c + d*x]]/2]^4)/(d*Sin[c + d*x]^(9/2))

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Maple [A]
time = 0.24, size = 219, normalized size = 1.59

method	result	size
default	\(\frac {a^{2} \left (-2 \sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticE \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e +\sqrt {-\sin \left (d x +c \right )+1}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \EllipticF \left (\sqrt {-\sin \left (d x +c \right )+1}, \frac {\sqrt {2}}{2}\right ) e +4 \cos \left (d x +c \right ) \sqrt {e}\, \sqrt {e \sin \left (d x +c \right )}\, \arctanh \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-4 \cos \left (d x +c \right ) \sqrt {e}\, \sqrt {e \sin \left (d x +c \right )}\, \arctan \left (\frac {\sqrt {e \sin \left (d x +c \right )}}{\sqrt {e}}\right )-2 \left (\cos ^{2}\left (d x +c \right )\right ) e +2 e \right )}{2 \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}\, d}\)	\(219\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2/cos(d*x+c)/(e*sin(d*x+c))^(1/2)*a^2*(-2*(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Elli
pticE((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*e+(-sin(d*x+c)+1)^(1/2)*(2*sin(d*x+c)+2)^(1/2)*sin(d*x+c)^(1/2)*Ellip
ticF((-sin(d*x+c)+1)^(1/2),1/2*2^(1/2))*e+4*cos(d*x+c)*e^(1/2)*(e*sin(d*x+c))^(1/2)*arctanh((e*sin(d*x+c))^(1/
2)/e^(1/2))-4*cos(d*x+c)*e^(1/2)*(e*sin(d*x+c))^(1/2)*arctan((e*sin(d*x+c))^(1/2)/e^(1/2))-2*cos(d*x+c)^2*e+2*
e)/d

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.48, size = 255, normalized size = 1.85 \begin {gather*} \frac {i \, \sqrt {2} \sqrt {-i} a^{2} \cos \left (d x + c\right ) e^{\frac {1}{2}} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - i \, \sqrt {2} \sqrt {i} a^{2} \cos \left (d x + c\right ) e^{\frac {1}{2}} {\rm weierstrassZeta}\left (4, 0, {\rm weierstrassPInverse}\left (4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - 2 \, a^{2} \arctan \left (\frac {2 \, {\left (76 \, \cos \left (d x + c\right )^{2} + 425 \, {\left (\sin \left (d x + c\right ) - 1\right )} \sqrt {\sin \left (d x + c\right )} - 152 \, \sin \left (d x + c\right ) - 152\right )}}{361 \, \cos \left (d x + c\right )^{2} + 978 \, \sin \left (d x + c\right ) - 722}\right ) \cos \left (d x + c\right ) e^{\frac {1}{2}} + a^{2} \cos \left (d x + c\right ) e^{\frac {1}{2}} \log \left (\frac {\cos \left (d x + c\right )^{2} - 4 \, {\left (\sin \left (d x + c\right ) + 1\right )} \sqrt {\sin \left (d x + c\right )} - 6 \, \sin \left (d x + c\right ) - 2}{\cos \left (d x + c\right )^{2} + 2 \, \sin \left (d x + c\right ) - 2}\right ) + 2 \, a^{2} e^{\frac {1}{2}} \sin \left (d x + c\right )^{\frac {3}{2}}}{2 \, d \cos \left (d x + c\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(I*sqrt(2)*sqrt(-I)*a^2*cos(d*x + c)*e^(1/2)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0, cos(d*x + c)
+ I*sin(d*x + c))) - I*sqrt(2)*sqrt(I)*a^2*cos(d*x + c)*e^(1/2)*weierstrassZeta(4, 0, weierstrassPInverse(4, 0
, cos(d*x + c) - I*sin(d*x + c))) - 2*a^2*arctan(2*(76*cos(d*x + c)^2 + 425*(sin(d*x + c) - 1)*sqrt(sin(d*x +
c)) - 152*sin(d*x + c) - 152)/(361*cos(d*x + c)^2 + 978*sin(d*x + c) - 722))*cos(d*x + c)*e^(1/2) + a^2*cos(d*
x + c)*e^(1/2)*log((cos(d*x + c)^2 - 4*(sin(d*x + c) + 1)*sqrt(sin(d*x + c)) - 6*sin(d*x + c) - 2)/(cos(d*x +
c)^2 + 2*sin(d*x + c) - 2)) + 2*a^2*e^(1/2)*sin(d*x + c)^(3/2))/(d*cos(d*x + c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \sqrt {e \sin {\left (c + d x \right )}}\, dx + \int 2 \sqrt {e \sin {\left (c + d x \right )}} \sec {\left (c + d x \right )}\, dx + \int \sqrt {e \sin {\left (c + d x \right )}} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(sqrt(e*sin(c + d*x)), x) + Integral(2*sqrt(e*sin(c + d*x))*sec(c + d*x), x) + Integral(sqrt(e*s
in(c + d*x))*sec(c + d*x)**2, x))

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

[Out]

integrate((a*sec(d*x + c) + a)^2*sqrt(e*sin(d*x + c)), x)

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

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))^2,x)

[Out]

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

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