Optimal. Leaf size=712 \[ \frac{3 \sqrt [4]{b} (b c-a d)^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right ),\frac{1}{2}\right )}{\sqrt{2} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}+\frac{6 \sqrt{b} \sqrt{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \sqrt{(a d+b (c+2 d x))^2}}{d^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d) (a d+b c+2 b d x) \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )}-\frac{3 \sqrt{2} \sqrt [4]{b} (b c-a d)^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}} \]
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Rubi [A] time = 0.608653, antiderivative size = 712, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {47, 62, 623, 305, 220, 1196} \[ \frac{6 \sqrt{b} \sqrt{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \sqrt{(a d+b (c+2 d x))^2}}{d^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c-a d) (a d+b c+2 b d x) \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )}+\frac{3 \sqrt [4]{b} (b c-a d)^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{\sqrt{2} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{3 \sqrt{2} \sqrt [4]{b} (b c-a d)^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(a d+b c+2 b d x)^2} \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}+1\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (a d+b c+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}-\frac{4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}} \]
Antiderivative was successfully verified.
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Rule 47
Rule 62
Rule 623
Rule 305
Rule 220
Rule 1196
Rubi steps
\begin{align*} \int \frac{(a+b x)^{3/4}}{(c+d x)^{5/4}} \, dx &=-\frac{4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}+\frac{(3 b) \int \frac{1}{\sqrt [4]{a+b x} \sqrt [4]{c+d x}} \, dx}{d}\\ &=-\frac{4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}+\frac{\left (3 b \sqrt [4]{(a+b x) (c+d x)}\right ) \int \frac{1}{\sqrt [4]{a c+(b c+a d) x+b d x^2}} \, dx}{d \sqrt [4]{a+b x} \sqrt [4]{c+d x}}\\ &=-\frac{4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}+\frac{\left (12 b \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2}\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{d \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=-\frac{4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}+\frac{\left (6 \sqrt{b} (b c-a d) \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{d^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}-\frac{\left (6 \sqrt{b} (b c-a d) \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{2 \sqrt{b} \sqrt{d} x^2}{b c-a d}}{\sqrt{-4 a b c d+(b c+a d)^2+4 b d x^4}} \, dx,x,\sqrt [4]{(a+b x) (c+d x)}\right )}{d^{3/2} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x)}\\ &=-\frac{4 (a+b x)^{3/4}}{d \sqrt [4]{c+d x}}+\frac{6 \sqrt{b} \sqrt{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2} \sqrt{(a d+b (c+2 d x))^2}}{d^{3/2} (b c-a d) \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right )}-\frac{3 \sqrt{2} \sqrt [4]{b} (b c-a d)^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2} \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}+\frac{3 \sqrt [4]{b} (b c-a d)^{3/2} \sqrt [4]{(a+b x) (c+d x)} \sqrt{(b c+a d+2 b d x)^2} \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right ) \sqrt{\frac{(a d+b (c+2 d x))^2}{(b c-a d)^2 \left (1+\frac{2 \sqrt{b} \sqrt{d} \sqrt{(a+b x) (c+d x)}}{b c-a d}\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt [4]{b} \sqrt [4]{d} \sqrt [4]{(a+b x) (c+d x)}}{\sqrt{b c-a d}}\right )|\frac{1}{2}\right )}{\sqrt{2} d^{7/4} \sqrt [4]{a+b x} \sqrt [4]{c+d x} (b c+a d+2 b d x) \sqrt{(a d+b (c+2 d x))^2}}\\ \end{align*}
Mathematica [C] time = 0.0421649, size = 73, normalized size = 0.1 \[ \frac{4 (a+b x)^{7/4} \left (\frac{b (c+d x)}{b c-a d}\right )^{5/4} \, _2F_1\left (\frac{5}{4},\frac{7}{4};\frac{11}{4};\frac{d (a+b x)}{a d-b c}\right )}{7 b (c+d x)^{5/4}} \]
Antiderivative was successfully verified.
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Maple [F] time = 0.03, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{3}{4}}} \left ( dx+c \right ) ^{-{\frac{5}{4}}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{3}{4}}}{{\left (d x + c\right )}^{\frac{5}{4}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x + a\right )}^{\frac{3}{4}}{\left (d x + c\right )}^{\frac{3}{4}}}{d^{2} x^{2} + 2 \, c d x + c^{2}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{3}{4}}}{\left (c + d x\right )^{\frac{5}{4}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{3}{4}}}{{\left (d x + c\right )}^{\frac{5}{4}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
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