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

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

[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 = 0.17, 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 = {2668, 704, 827, 1166, 206} \[ -\frac {2 b \sqrt {a+b \sin (c+d x)}}{d}-\frac {(a-b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )}{d}+\frac {(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{d} \]

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 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 704

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 827

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 1166

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 2668

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 \operatorname {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 \operatorname {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) \operatorname {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 \operatorname {Subst}\left (\int \frac {1}{a-b-x^2} \, dx,x,\sqrt {a+b \sin (c+d x)}\right )}{d}+\frac {(a+b)^2 \operatorname {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 \[ \frac {-2 b \sqrt {a+b \sin (c+d x)}+(a-b)^{3/2} \left (-\tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a-b}}\right )\right )+(a+b)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+b \sin (c+d x)}}{\sqrt {a+b}}\right )}{d} \]

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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fricas [F]  time = 1.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (d x + c\right ) \sin \left (d x + c\right ) + a \sec \left (d x + c\right )\right )} \sqrt {b \sin \left (d x + c\right ) + a}, x\right ) \]

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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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maple [B]  time = 0.48, size = 218, normalized size = 2.32 \[ -\frac {2 b \sqrt {a +b \sin \left (d x +c \right )}}{d}+\frac {\arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a^{2}}{d \sqrt {-a +b}}-\frac {2 b \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right ) a}{d \sqrt {-a +b}}+\frac {b^{2} \arctan \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {-a +b}}\right )}{d \sqrt {-a +b}}+\frac {\arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a^{2}}{d \sqrt {a +b}}+\frac {2 b \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right ) a}{d \sqrt {a +b}}+\frac {b^{2} \arctanh \left (\frac {\sqrt {a +b \sin \left (d x +c \right )}}{\sqrt {a +b}}\right )}{d \sqrt {a +b}} \]

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)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

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 details)Is 4*a-4*b positive or negative?

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

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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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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]

Timed out

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