3.121 \(\int \sec (c+d x) (a+a \sin (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=62 \[ \frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a \sqrt{a \sin (c+d x)+a}}{d} \]

[Out]

(2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (2*a*Sqrt[a + a*Sin[c + d*x]])/d

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Rubi [A]  time = 0.0682787, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2667, 50, 63, 206} \[ \frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a \sqrt{a \sin (c+d x)+a}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (2*a*Sqrt[a + a*Sin[c + d*x]])/d

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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])

Rubi steps

\begin{align*} \int \sec (c+d x) (a+a \sin (c+d x))^{3/2} \, dx &=\frac{a \operatorname{Subst}\left (\int \frac{\sqrt{a+x}}{a-x} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{2 a \sqrt{a+a \sin (c+d x)}}{d}+\frac{\left (2 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac{2 a \sqrt{a+a \sin (c+d x)}}{d}+\frac{\left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+a \sin (c+d x)}\right )}{d}\\ &=\frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+a \sin (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a \sqrt{a+a \sin (c+d x)}}{d}\\ \end{align*}

Mathematica [A]  time = 0.100852, size = 60, normalized size = 0.97 \[ \frac{2 a \left (\sqrt{2} \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a \sin (c+d x)+a}}{\sqrt{2} \sqrt{a}}\right )-\sqrt{a \sin (c+d x)+a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^(3/2),x]

[Out]

(2*a*(Sqrt[2]*Sqrt[a]*ArcTanh[Sqrt[a + a*Sin[c + d*x]]/(Sqrt[2]*Sqrt[a])] - Sqrt[a + a*Sin[c + d*x]]))/d

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Maple [A]  time = 0.075, size = 49, normalized size = 0.8 \begin{align*} -2\,{\frac{a}{d} \left ( \sqrt{a+a\sin \left ( dx+c \right ) }-\sqrt{a}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a+a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)*(a+a*sin(d*x+c))^(3/2),x)

[Out]

-2*a*((a+a*sin(d*x+c))^(1/2)-a^(1/2)*2^(1/2)*arctanh(1/2*(a+a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2)))/d

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.69702, size = 196, normalized size = 3.16 \begin{align*} \frac{\sqrt{2} a^{\frac{3}{2}} \log \left (-\frac{a \sin \left (d x + c\right ) + 2 \, \sqrt{2} \sqrt{a \sin \left (d x + c\right ) + a} \sqrt{a} + 3 \, a}{\sin \left (d x + c\right ) - 1}\right ) - 2 \, \sqrt{a \sin \left (d x + c\right ) + a} a}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

(sqrt(2)*a^(3/2)*log(-(a*sin(d*x + c) + 2*sqrt(2)*sqrt(a*sin(d*x + c) + a)*sqrt(a) + 3*a)/(sin(d*x + c) - 1))
- 2*sqrt(a*sin(d*x + c) + a)*a)/d

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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(sec(d*x+c)*(a+a*sin(d*x+c))**(3/2),x)

[Out]

Timed out

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Giac [B]  time = 8.41583, size = 910, normalized size = 14.68 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

2*(2*sqrt(2)*(a^2*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c)^2 + a^2*sgn(tan(1/2*
d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1))*arctan(1/2*sqrt(2)*(sqrt((a*tan(1/2*c)^2 + a)*tan(1/2*d*x)^2
+ a*tan(1/2*c)^2 + a)*a*tan(1/2*c)^3 + sqrt((a*tan(1/2*c)^2 + a)*tan(1/2*d*x)^2 + a*tan(1/2*c)^2 + a)*a*tan(1
/2*c)^2 + 2*sqrt((a*tan(1/2*c)^2 + a)*tan(1/2*d*x)^2 + a*tan(1/2*c)^2 + a)*a*tan(1/2*c) + 2*sqrt((a*tan(1/2*c)
^2 + a)*tan(1/2*d*x)^2 + a*tan(1/2*c)^2 + a)*a + sqrt(a^3*tan(1/2*c)^8 + 2*a^3*tan(1/2*c)^7 + 6*a^3*tan(1/2*c)
^6 + 10*a^3*tan(1/2*c)^5 + 13*a^3*tan(1/2*c)^4 + 16*a^3*tan(1/2*c)^3 + 12*a^3*tan(1/2*c)^2 + 8*a^3*tan(1/2*c)
+ 4*a^3)*tan(1/2*d*x) + sqrt(a^3*tan(1/2*c)^8 - 2*a^3*tan(1/2*c)^7 + 6*a^3*tan(1/2*c)^6 - 10*a^3*tan(1/2*c)^5
+ 13*a^3*tan(1/2*c)^4 - 16*a^3*tan(1/2*c)^3 + 12*a^3*tan(1/2*c)^2 - 8*a^3*tan(1/2*c) + 4*a^3))/sqrt(-a^3*tan(1
/2*c)^8 - 6*a^3*tan(1/2*c)^6 - 13*a^3*tan(1/2*c)^4 - 12*a^3*tan(1/2*c)^2 - 4*a^3))/sqrt(-a*tan(1/2*c)^4 - 2*a*
tan(1/2*c)^2 - a) + (a^2*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1)*tan(1/2*c) + a^2*sgn(tan
(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1) - (a^2*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(
1/2*c) - 1)*tan(1/2*c) - a^2*sgn(tan(1/2*d*x)*tan(1/2*c) - tan(1/2*d*x) - tan(1/2*c) - 1))*tan(1/2*d*x))/sqrt(
a*tan(1/2*d*x)^2*tan(1/2*c)^2 + a*tan(1/2*d*x)^2 + a*tan(1/2*c)^2 + a))/d