3.5.86 \(\int \sec (c+d x) (a+b \sin (c+d x))^{3/2} \, dx\) [486]

Optimal. Leaf size=94 \[ -\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{d}+\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{d}-\frac {2 b \sqrt {a+b \sin (c+d x)}}{d} \]

[Out]

-(a-b)^(3/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a-b)^(1/2))/d+(a+b)^(3/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1
/2))/d-2*b*(a+b*sin(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.11, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2747, 718, 841, 1180, 212} \begin {gather*} -\frac {2 b \sqrt {a+b \sin (c+d x)}}{d}-\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{d}+\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 718

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[e*((d + e*x)^(m - 1)/(c*(m - 1))), x] +
Dist[1/c, Int[(d + e*x)^(m - 2)*(Simp[c*d^2 - a*e^2 + 2*c*d*e*x, x]/(a + c*x^2)), x], x] /; FreeQ[{a, c, d, e}
, x] && NeQ[c*d^2 + a*e^2, 0] && GtQ[m, 1]

Rule 841

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1180

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rule 2747

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*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]
time = 0.10, size = 89, normalized size = 0.95 \begin {gather*} \frac {-(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )+(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )-2 b \sqrt {a+b \sin (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(200\) vs. \(2(80)=160\).
time = 1.27, size = 201, normalized size = 2.14

method	result	size
default	\(\frac {-2 b \sqrt {a +b \sin \left (d x +c \right )}+\frac {\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{2}}{\sqrt {a +b}}+\frac {2 b \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{\sqrt {a +b}}+\frac {b^{2} \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{\sqrt {a +b}}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{2}}{\sqrt {-a +b}}-\frac {2 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{\sqrt {-a +b}}+\frac {b^{2} \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{\sqrt {-a +b}}}{d}\)	\(201\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-2*b*(a+b*sin(d*x+c))^(1/2)+1/(a+b)^(1/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a^2+2*b/(a+b)^(1/2)*arc
tanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))*a+b^2/(a+b)^(1/2)*arctanh((a+b*sin(d*x+c))^(1/2)/(a+b)^(1/2))+1/(-a+b
)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2))*a^2-2*b/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^
(1/2))*a+b^2/(-a+b)^(1/2)*arctan((a+b*sin(d*x+c))^(1/2)/(-a+b)^(1/2)))/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*a-4*b>0)', see `assume?` for
more detail

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (a + b \sin {\left (c + d x \right )}\right )^{\frac {3}{2}} \sec {\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+b\,\sin \left (c+d\,x\right )\right )}^{3/2}}{\cos \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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