3.1705 \(\int \frac{(a+b x)^{7/4}}{(c+d x)^{3/4}} \, dx\)

Optimal. Leaf size=167 \[ -\frac{7 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}{8 d^2}-\frac{21 (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 \sqrt [4]{b} d^{11/4}}+\frac{21 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 \sqrt [4]{b} d^{11/4}}+\frac{(a+b x)^{7/4} \sqrt [4]{c+d x}}{2 d} \]

[Out]

(-7*(b*c - a*d)*(a + b*x)^(3/4)*(c + d*x)^(1/4))/(8*d^2) + ((a + b*x)^(7/4)*(c + d*x)^(1/4))/(2*d) - (21*(b*c
- a*d)^2*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(1/4)*d^(11/4)) + (21*(b*c - a*d)^
2*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(1/4)*d^(11/4))

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Rubi [A]  time = 0.105018, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {50, 63, 331, 298, 205, 208} \[ -\frac{7 (a+b x)^{3/4} \sqrt [4]{c+d x} (b c-a d)}{8 d^2}-\frac{21 (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 \sqrt [4]{b} d^{11/4}}+\frac{21 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 \sqrt [4]{b} d^{11/4}}+\frac{(a+b x)^{7/4} \sqrt [4]{c+d x}}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(7/4)/(c + d*x)^(3/4),x]

[Out]

(-7*(b*c - a*d)*(a + b*x)^(3/4)*(c + d*x)^(1/4))/(8*d^2) + ((a + b*x)^(7/4)*(c + d*x)^(1/4))/(2*d) - (21*(b*c
- a*d)^2*ArcTan[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(1/4)*d^(11/4)) + (21*(b*c - a*d)^
2*ArcTanh[(d^(1/4)*(a + b*x)^(1/4))/(b^(1/4)*(c + d*x)^(1/4))])/(16*b^(1/4)*d^(11/4))

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 331

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^(p + (m + 1)/n), Subst[Int[x^m/(1 - b*x^n)^(
p + (m + 1)/n + 1), x], x, x/(a + b*x^n)^(1/n)], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[-1, p, 0] && NeQ[
p, -2^(-1)] && IntegersQ[m, p + (m + 1)/n]

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 205

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{(a+b x)^{7/4}}{(c+d x)^{3/4}} \, dx &=\frac{(a+b x)^{7/4} \sqrt [4]{c+d x}}{2 d}-\frac{(7 (b c-a d)) \int \frac{(a+b x)^{3/4}}{(c+d x)^{3/4}} \, dx}{8 d}\\ &=-\frac{7 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 d^2}+\frac{(a+b x)^{7/4} \sqrt [4]{c+d x}}{2 d}+\frac{\left (21 (b c-a d)^2\right ) \int \frac{1}{\sqrt [4]{a+b x} (c+d x)^{3/4}} \, dx}{32 d^2}\\ &=-\frac{7 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 d^2}+\frac{(a+b x)^{7/4} \sqrt [4]{c+d x}}{2 d}+\frac{\left (21 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (c-\frac{a d}{b}+\frac{d x^4}{b}\right )^{3/4}} \, dx,x,\sqrt [4]{a+b x}\right )}{8 b d^2}\\ &=-\frac{7 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 d^2}+\frac{(a+b x)^{7/4} \sqrt [4]{c+d x}}{2 d}+\frac{\left (21 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1-\frac{d x^4}{b}} \, dx,x,\frac{\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{8 b d^2}\\ &=-\frac{7 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 d^2}+\frac{(a+b x)^{7/4} \sqrt [4]{c+d x}}{2 d}+\frac{\left (21 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b}-\sqrt{d} x^2} \, dx,x,\frac{\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{16 d^{5/2}}-\frac{\left (21 (b c-a d)^2\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{b}+\sqrt{d} x^2} \, dx,x,\frac{\sqrt [4]{a+b x}}{\sqrt [4]{c+d x}}\right )}{16 d^{5/2}}\\ &=-\frac{7 (b c-a d) (a+b x)^{3/4} \sqrt [4]{c+d x}}{8 d^2}+\frac{(a+b x)^{7/4} \sqrt [4]{c+d x}}{2 d}-\frac{21 (b c-a d)^2 \tan ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 \sqrt [4]{b} d^{11/4}}+\frac{21 (b c-a d)^2 \tanh ^{-1}\left (\frac{\sqrt [4]{d} \sqrt [4]{a+b x}}{\sqrt [4]{b} \sqrt [4]{c+d x}}\right )}{16 \sqrt [4]{b} d^{11/4}}\\ \end{align*}

Mathematica [C]  time = 0.0324898, size = 73, normalized size = 0.44 \[ \frac{4 (a+b x)^{11/4} \left (\frac{b (c+d x)}{b c-a d}\right )^{3/4} \, _2F_1\left (\frac{3}{4},\frac{11}{4};\frac{15}{4};\frac{d (a+b x)}{a d-b c}\right )}{11 b (c+d x)^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(7/4)/(c + d*x)^(3/4),x]

[Out]

(4*(a + b*x)^(11/4)*((b*(c + d*x))/(b*c - a*d))^(3/4)*Hypergeometric2F1[3/4, 11/4, 15/4, (d*(a + b*x))/(-(b*c)
+ a*d)])/(11*b*(c + d*x)^(3/4))

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Maple [F]  time = 0.018, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{7}{4}}} \left ( dx+c \right ) ^{-{\frac{3}{4}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(7/4)/(d*x+c)^(3/4),x)

[Out]

int((b*x+a)^(7/4)/(d*x+c)^(3/4),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{7}{4}}}{{\left (d x + c\right )}^{\frac{3}{4}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/4)/(d*x+c)^(3/4),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(7/4)/(d*x + c)^(3/4), x)

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Fricas [B]  time = 3.75017, size = 3069, normalized size = 18.38 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/4)/(d*x+c)^(3/4),x, algorithm="fricas")

[Out]

-1/32*(84*d^2*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^
5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b*d^11))^(1/4)*arctan(-((b^3*c^2*d^8 - 2*a*b^2*
c*d^9 + a^2*b*d^10)*(b*x + a)^(3/4)*(d*x + c)^(1/4)*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^
5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b*d^11))^
(3/4) - (b^2*d^8*x + a*b*d^8)*sqrt(((b^4*c^4 - 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 - 4*a^3*b*c*d^3 + a^4*d^4)*sq
rt(b*x + a)*sqrt(d*x + c) + (b*d^6*x + a*d^6)*sqrt((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*
c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b*d^11)))/(
b*x + a))*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^
3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b*d^11))^(3/4))/(a*b^8*c^8 - 8*a^2*b^7*c^7*d + 28*a
^3*b^6*c^6*d^2 - 56*a^4*b^5*c^5*d^3 + 70*a^5*b^4*c^4*d^4 - 56*a^6*b^3*c^3*d^5 + 28*a^7*b^2*c^2*d^6 - 8*a^8*b*c
*d^7 + a^9*d^8 + (b^9*c^8 - 8*a*b^8*c^7*d + 28*a^2*b^7*c^6*d^2 - 56*a^3*b^6*c^5*d^3 + 70*a^4*b^5*c^4*d^4 - 56*
a^5*b^4*c^3*d^5 + 28*a^6*b^3*c^2*d^6 - 8*a^7*b^2*c*d^7 + a^8*b*d^8)*x)) - 21*d^2*((b^8*c^8 - 8*a*b^7*c^7*d + 2
8*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*
b*c*d^7 + a^8*d^8)/(b*d^11))^(1/4)*log(21*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*x + a)^(3/4)*(d*x + c)^(1/4) + (
b*d^3*x + a*d^3)*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56
*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b*d^11))^(1/4))/(b*x + a)) + 21*d^2*((b^8*c^
8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6
*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b*d^11))^(1/4)*log(21*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*x + a)^(3/4
)*(d*x + c)^(1/4) - (b*d^3*x + a*d^3)*((b^8*c^8 - 8*a*b^7*c^7*d + 28*a^2*b^6*c^6*d^2 - 56*a^3*b^5*c^5*d^3 + 70
*a^4*b^4*c^4*d^4 - 56*a^5*b^3*c^3*d^5 + 28*a^6*b^2*c^2*d^6 - 8*a^7*b*c*d^7 + a^8*d^8)/(b*d^11))^(1/4))/(b*x +
a)) - 4*(4*b*d*x - 7*b*c + 11*a*d)*(b*x + a)^(3/4)*(d*x + c)^(1/4))/d^2

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{7}{4}}}{\left (c + d x\right )^{\frac{3}{4}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(7/4)/(d*x+c)**(3/4),x)

[Out]

Integral((a + b*x)**(7/4)/(c + d*x)**(3/4), x)

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Giac [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((b*x+a)^(7/4)/(d*x+c)^(3/4),x, algorithm="giac")

[Out]

Timed out